(i) 17 × 21 Using (a-b)(a+b) = a^2-b^2 with a=19, b=2. 17 × 21 = (19-2)(19+2) = 19^2 - 2^2 = 361 - 4 = 357 Final Answer: 357 (ii) 104 × 96 Using (a+b)(a-b) = a^2-b^2 with a=100, b=4. 104 × 96 = (100+4)(100-4) = 100^2 - 4^2 = 10000 - 16 = 9984 Final Answer: 9984 (iii) 24 × 16 Using (a+b)(a-b) =…

Q3 (i): 4 y^2+1+frac116 y^2

(i) 4 y^2+1+frac116 y^2 This is of the form a^2+2ab+b^2. Let a=2y, b=1/4y. (2y)^2+2(2y)left(1/4yright)+left(1/4yright)^2 = 4y^2+1+1/16y^2 This matches the given expression. So it factors to (2y+1/4y)^2. Final Answer: (2y+1/4y)^2 (ii) 9 m^2-frac125 n^2 Using a^2-b^2=(a-b)(a+b). (3m)^2 -…

Q3 (ii): 9 m^2-frac125 n^2

(i) 4 y^2+1+frac116 y^2 This is of the form a^2+2ab+b^2. Let a=2y, b=1/4y. (2y)^2+2(2y)left(1/4yright)+left(1/4yright)^2 = 4y^2+1+1/16y^2 This matches the given expression. So it factors to (2y+1/4y)^2. Final Answer: (2y+1/4y)^2 (ii) 9 m^2-frac125 n^2 Using a^2-b^2=(a-b)(a+b). (3m)^2 -…

Q3 (iii): 27 b^3-frac164 b^3

(i) 4 y^2+1+frac116 y^2 This is of the form a^2+2ab+b^2. Let a=2y, b=1/4y. (2y)^2+2(2y)left(1/4yright)+left(1/4yright)^2 = 4y^2+1+1/16y^2 This matches the given expression. So it factors to (2y+1/4y)^2. Final Answer: (2y+1/4y)^2 (ii) 9 m^2-frac125 n^2 Using a^2-b^2=(a-b)(a+b). (3m)^2 -…

Exploring Algebraic Identities Class 9 Mathematics NCERT Solutions

Q1End-of-Chapter Exercises

Use suitable identities to find the following products:

(i)

$(-3 x+4)^{2}$

(ii)

$(2 s+7)(2 s-7)$

(iii)

$\left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)$

(iv)

$(2 n+7)(2 n-7)$

(v)

$(s-2 t)\left(s^{2}+2 s t+4 t^{2}\right)$

(vi)

$\left(\frac{1}{2 r}-4 r\right)^{2}$

(vii)

$(-3 m+4 k-l)^{2}$

(viii)

$\left(x-\frac{1}{3} y\right)^{3}$

(ix)

$\left(\frac{7}{2} k-\frac{2}{3} m\right)^{3}$

Solution

(i) $(-3 x+4)^{2}$ Using

(a-b)^2 = a^2-2ab+b^2

with

a=4, b=3x

.

(4-3x)^2 = 4^2 - 2(4)(3x) + (3x)^2 = 16 - 24x + 9x^2

Final Answer:

9x^2 - 24x + 16

(ii) $(2 s+7)(2 s-7)$ Using

(a+b)(a-b) = a^2-b^2

.

(2s)^2 - 7^2 = 4s^2 - 49

Final Answer:

4s^2 - 49

(iii) $\left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)$ Using

(a+b)(a-b) = a^2-b^2

.

(p^2)^2 - \left(\frac{1}{2}\right)^2 = p^4 - \frac{1}{4}

Final Answer:

p^4 - \frac{1}{4}

(iv) $(2 n+7)(2 n-7)$ Using

(a+b)(a-b) = a^2-b^2

.

(2n)^2 - 7^2 = 4n^2 - 49

Final Answer:

4n^2 - 49

(v) $(s-2 t)\left(s^{2}+2 s t+4 t^{2}\right)$ This is of the form

(a-b)(a^2+ab+b^2) = a^3-b^3

with

a=s, b=2t

.

s^3 - (2t)^3 = s^3 - 8t^3

Final Answer:

s^3 - 8t^3

(vi) $\left(\frac{1}{2 r}-4 r\right)^{2}$ Using

(a-b)^2 = a^2-2ab+b^2

.

\left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2 = \frac{1}{4r^2} - 4 + 16r^2

Final Answer:

\frac{1}{4r^2} - 4 + 16r^2

(vii) $(-3 m+4 k-l)^{2}$ Using

(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca

with

a=-3m, b=4k, c=-l

.

(-3m)^2+(4k)^2+(-l)^2+2(-3m)(4k)+2(4k)(-l)+2(-l)(-3m)

= 9m^2+16k^2+l^2-24mk-8kl+6ml

Final Answer:

9m^2+16k^2+l^2-24mk-8kl+6ml

(viii) $\left(x-\frac{1}{3} y\right)^{3}$ Using

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

.

x^3 - 3(x)^2\left(\frac{1}{3}y\right) + 3(x)\left(\frac{1}{3}y\right)^2 - \left(\frac{1}{3}y\right)^3

= x^3 - x^2y + 3x\left(\frac{1}{9}y^2\right) - \frac{1}{27}y^3 = x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3

Final Answer:

x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3

(ix) $\left(\frac{7}{2} k-\frac{2}{3} m\right)^{3}$ Using

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

.

\left(\frac{7}{2}k\right)^3 - 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 - \left(\frac{2}{3}m\right)^3

= \frac{343}{8}k^3 - 3\left(\frac{49}{4}k^2\right)\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{4}{9}m^2\right) - \frac{8}{27}m^3

= \frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3

Final Answer:

\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3

Q2End-of-Chapter Exercises

Find the values using suitable identities:

(i)

17 × 21

(ii)

104 × 96

(iii)

24 × 16

(iv)

$147^{3}$

(v)

$199^{3}$

(vi)

$127^{3}$

(vii)

$(-107)^{3}$

(viii)

$(-299)^{3}$

Solution

(i) 17 × 21 Using

(a-b)(a+b) = a^2-b^2

with

a=19, b=2

.

17 \times 21 = (19-2)(19+2) = 19^2 - 2^2 = 361 - 4 = 357

Final Answer: 357

(ii) 104 × 96 Using

(a+b)(a-b) = a^2-b^2

with

a=100, b=4

.

104 \times 96 = (100+4)(100-4) = 100^2 - 4^2 = 10000 - 16 = 9984

Final Answer: 9984

(iii) 24 × 16 Using

(a+b)(a-b) = a^2-b^2

with

a=20, b=4

.

24 \times 16 = (20+4)(20-4) = 20^2 - 4^2 = 400 - 16 = 384

Final Answer: 384

(iv) $147^{3}$ Using

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

with

a=150, b=3

.

147^3 = (150-3)^3 = 150^3 - 3(150^2)(3) + 3(150)(3^2) - 3^3

= 3375000 - 9(22500) + 9(150) - 27

= 3375000 - 202500 + 1350 - 27 = 3173823

Final Answer: 3173823

(v) $199^{3}$ Using

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

with

a=200, b=1

.

199^3 = (200-1)^3 = 200^3 - 3(200^2)(1) + 3(200)(1^2) - 1^3

= 8000000 - 3(40000) + 600 - 1 = 8000000 - 120000 + 600 - 1 = 7880599

Final Answer: 7880599

(vi) $127^{3}$ Using

(a+b)^3 = a^3+3a^2b+3ab^2+b^3

with

a=100, b=27

is complicated. Using

a=130, b=3

is better with

(a-b)^3

.

127^3 = (130-3)^3 = 130^3 - 3(130^2)(3) + 3(130)(3^2) - 3^3

= 2197000 - 9(16900) + 9(130) - 27

= 2197000 - 152100 + 1170 - 27 = 2046043

Final Answer: 2046043

(vii) $(-107)^{3}$

(-107)^3 = -(107)^3 = -(100+7)^3

Using

(a+b)^3 = a^3+3a^2b+3ab^2+b^3

.

-(100^3 + 3(100^2)(7) + 3(100)(7^2) + 7^3)

-(1000000 + 3(10000)(7) + 300(49) + 343)

-(1000000 + 210000 + 14700 + 343) = -1225043

Final Answer: -1225043

(viii) $(-299)^{3}$

(-299)^3 = -(299)^3 = -(300-1)^3

Using

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

.

-(300^3 - 3(300^2)(1) + 3(300)(1^2) - 1^3)

-(27000000 - 3(90000) + 900 - 1)

-(27000000 - 270000 + 900 - 1) = -(26730899) = -26730899

Final Answer: -26730899

Q3End-of-Chapter Exercises

Factor the following algebraic expressions:

(i)

$4 y^{2}+1+\frac{1}{16 y^{2}}$

(ii)

$9 m^{2}-\frac{1}{25 n^{2}}$

(iii)

$\quad 27 b^{3}-\frac{1}{64 b^{3}}$

(iv)

$x^{2}+\frac{5 x}{6}+\frac{1}{6}$

(v)

$27 u^{3}-\frac{1}{125}-\frac{27 u^{2}}{5}+\frac{9 u}{25}$

(vi)

$64 y^{3}+\frac{1}{125} z^{3}$

(vii)

$p^{3}+27 q^{3}+r^{3}-9 p q r$

(viii)

$9 m^{2}-12 m+4$

(ix)

$9 x^{3}-\frac{8}{3} y^{3}+\frac{z^{3}}{3}+6 x y z$

(x)

$4 x^{2}+9 y^{2}+36 z^{2}+12 x z+36 y z+24 x y$ (xi) $27 u^{3}-\frac{1}{216}-\frac{9 u^{2}}{2}+\frac{u}{4}$

Solution

(i) $4 y^{2}+1+\frac{1}{16 y^{2}}$ This is of the form

a^2+2ab+b^2

. Let

a=2y, b=\frac{1}{4y}

.

(2y)^2+2(2y)\left(\frac{1}{4y}\right)+\left(\frac{1}{4y}\right)^2 = 4y^2+1+\frac{1}{16y^2}

This matches the given expression. So it factors to

(2y+\frac{1}{4y})^2

. Final Answer:

(2y+\frac{1}{4y})^2

(ii) $9 m^{2}-\frac{1}{25 n^{2}}$ Using

a^2-b^2=(a-b)(a+b)

.

(3m)^2 - \left(\frac{1}{5n}\right)^2 = \left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)

Final Answer:

\left(3m-\frac{1}{5n}\right)\left(3m+\frac{1}{5n}\right)

(iii) $27 b^{3}-\frac{1}{64 b^{3}}$ Using

a^3-b^3=(a-b)(a^2+ab+b^2)

.

(3b)^3 - \left(\frac{1}{4b}\right)^3 = \left(3b-\frac{1}{4b}\right)\left((3b)^2+(3b)\left(\frac{1}{4b}\right)+\left(\frac{1}{4b}\right)^2\right)

= \left(3b-\frac{1}{4b}\right)\left(9b^2+\frac{3}{4}+\frac{1}{16b^2}\right)

Final Answer:

\left(3b-\frac{1}{4b}\right)\left(9b^2+\frac{3}{4}+\frac{1}{16b^2}\right)

(iv) $x^{2}+\frac{5 x}{6}+\frac{1}{6}$ We need two numbers with product

\frac{1}{6}

and sum

\frac{5}{6}

. The numbers are

\frac{1}{2}

and

\frac{1}{3}

.

x^2+\frac{1}{2}x+\frac{1}{3}x+\frac{1}{6} = x\left(x+\frac{1}{2}\right)+\frac{1}{3}\left(x+\frac{1}{2}\right) = \left(x+\frac{1}{3}\right)\left(x+\frac{1}{2}\right)

Final Answer:

\left(x+\frac{1}{3}\right)\left(x+\frac{1}{2}\right)

(v) $27 u^{3}-\frac{1}{125}-\frac{27 u^{2}}{5}+\frac{9 u}{25}$ Rearranging:

27u^3 - \frac{27u^2}{5} + \frac{9u}{25} - \frac{1}{125}

. This is of the form

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

with

a=3u, b=\frac{1}{5}

.

(3u)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)^3 = 27u^3 - \frac{27u^2}{5} + \frac{9u}{25} - \frac{1}{125}

Final Answer:

\left(3u-\frac{1}{5}\right)^3

(vi) $64 y^{3}+\frac{1}{125} z^{3}$ Using

a^3+b^3=(a+b)(a^2-ab+b^2)

.

(4y)^3 + \left(\frac{1}{5}z\right)^3 = \left(4y+\frac{1}{5}z\right)\left((4y)^2-(4y)\left(\frac{1}{5}z\right)+\left(\frac{1}{5}z\right)^2\right)

= \left(4y+\frac{1}{5}z\right)\left(16y^2-\frac{4}{5}yz+\frac{1}{25}z^2\right)

Final Answer:

\left(4y+\frac{1}{5}z\right)\left(16y^2-\frac{4}{5}yz+\frac{1}{25}z^2\right)

(vii) $p^{3}+27 q^{3}+r^{3}-9 p q r$ Using

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

.

p^3+(3q)^3+r^3-3(p)(3q)(r) = (p+3q+r)(p^2+(3q)^2+r^2-p(3q)-(3q)r-rp)

= (p+3q+r)(p^2+9q^2+r^2-3pq-3qr-rp)

Final Answer:

(p+3q+r)(p^2+9q^2+r^2-3pq-3qr-rp)

(viii) $9 m^{2}-12 m+4$ Using

a^2-2ab+b^2=(a-b)^2

.

(3m)^2 - 2(3m)(2) + 2^2 = (3m-2)^2

Final Answer:

(3m-2)^2

(ix) $9 x^{3}-\frac{8}{3} y^{3}+\frac{z^{3}}{3}+6 x y z$ Taking

\frac{1}{3}

as a common factor:

\frac{1}{3}(27x^3-8y^3+z^3+18xyz)

. This is of the form

a^3+b^3+c^3-3abc

with

a=3x, b=-2y, c=z

.

a^3+b^3+c^3-3abc = (3x)^3+(-2y)^3+z^3-3(3x)(-2y)(z) = 27x^3-8y^3+z^3+18xyz

So, the expression is

\frac{1}{3}(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

.

= \frac{1}{3}(3x-2y+z)((3x)^2+(-2y)^2+z^2-(3x)(-2y)-(-2y)z-z(3x))

= \frac{1}{3}(3x-2y+z)(9x^2+4y^2+z^2+6xy+2yz-3zx)

Final Answer:

\frac{1}{3}(3x-2y+z)(9x^2+4y^2+z^2+6xy+2yz-3zx)

(x) $4 x^{2}+9 y^{2}+36 z^{2}+12 x z+36 y z+24 x y$ Note: There appears to be a typo in the question. The cross-product terms for the identity

(a+b+c)^2

with

a=2x, b=3y, c=6z

should be

2ab=12xy

,

2bc=36yz

, and

2ca=24xz

. The question has the coefficients for

xy

and

xz

swapped. Assuming the question intended to be

4x^2+9y^2+36z^2+12xy+36yz+24xz

, the factorization is:

4x^2+9y^2+36z^2+12xy+36yz+24xz = (2x)^2+(3y)^2+(6z)^2+2(2x)(3y)+2(3y)(6z)+2(6z)(2x)

= (2x+3y+6z)^2

Final Answer (based on corrected expression):

(2x+3y+6z)^2

(xi) $27 u^{3}-\frac{1}{216}-\frac{9 u^{2}}{2}+\frac{u}{4}$ Rearranging:

27u^3 - \frac{9u^2}{2} + \frac{u}{4} - \frac{1}{216}

. This is of the form

(a-b)^3=a^3-3a^2b+3ab^2-b^3

with

a=3u, b=\frac{1}{6}

.

(3u)^3-3(3u)^2\left(\frac{1}{6}\right)+3(3u)\left(\frac{1}{6}\right)^2-\left(\frac{1}{6}\right)^3 = 27u^3 - 3(9u^2)\frac{1}{6} + 9u\frac{1}{36} - \frac{1}{216}

= 27u^3 - \frac{9u^2}{2} + \frac{u}{4} - \frac{1}{216}

Final Answer:

\left(3u-\frac{1}{6}\right)^3

Q4End-of-Chapter Exercises

Simplify the following:

(i)

$\frac{4 x^{2}+4 x+1}{4 x^{2}-1}$

(ii)

$\frac{9\left(3 a^{3}-24 b^{3}\right)}{9 a^{2}-36 b^{2}}$

(iii)

$\frac{s^{3}+125 t^{3}}{s^{2}-2 s t-35 t^{2}}$ Note: Assume that the denominators are not equal to 0.

Solution

(i) $\frac{4 x^{2}+4 x+1}{4 x^{2}-1}$ Factor numerator:

4x^2+4x+1 = (2x+1)^2

. Factor denominator:

4x^2-1 = (2x)^2-1^2 = (2x-1)(2x+1)

.

\frac{(2x+1)^2}{(2x-1)(2x+1)} = \frac{2x+1}{2x-1}

Final Answer:

\frac{2x+1}{2x-1}

(ii) $\frac{9\left(3 a^{3}-24 b^{3}\right)}{9 a^{2}-36 b^{2}}$ Factor numerator:

9(3a^3-24b^3) = 9 \cdot 3(a^3-8b^3) = 27(a-2b)(a^2+2ab+4b^2)

. Factor denominator:

9a^2-36b^2 = 9(a^2-4b^2) = 9(a-2b)(a+2b)

.

\frac{27(a-2b)(a^2+2ab+4b^2)}{9(a-2b)(a+2b)} = \frac{3(a^2+2ab+4b^2)}{a+2b}

Final Answer:

\frac{3(a^2+2ab+4b^2)}{a+2b}

(iii) $\frac{s^{3}+125 t^{3}}{s^{2}-2 s t-35 t^{2}}$ Factor numerator:

s^3+125t^3 = s^3+(5t)^3 = (s+5t)(s^2-5st+25t^2)

. Factor denominator:

s^2-2st-35t^2 = (s-7t)(s+5t)

.

\frac{(s+5t)(s^2-5st+25t^2)}{(s-7t)(s+5t)} = \frac{s^2-5st+25t^2}{s-7t}

Final Answer:

\frac{s^2-5st+25t^2}{s-7t}

Q5End-of-Chapter Exercises

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i)

$25 a^{2}-30 a b+9 b^{2}$

(ii)

$36 s^{2}-49 t^{2}$

Solution

(i) Area = $25 a^{2}-30 a b+9 b^{2}$ Area of a rectangle = length

\times

breadth. We factor the expression for the area.

25a^2-30ab+9b^2 = (5a)^2 - 2(5a)(3b) + (3b)^2 = (5a-3b)^2

Since the factors are identical, the rectangle is a square. Possible length =

5a-3b

units. Possible breadth =

5a-3b

units. Final Answer: Possible expressions are length =

(5a-3b)

and breadth =

(5a-3b)

.

(ii) Area = $36 s^{2}-49 t^{2}$ We factor the expression for the area using the identity

x^2-y^2=(x-y)(x+y)

.

36s^2-49t^2 = (6s)^2 - (7t)^2 = (6s-7t)(6s+7t)

Possible length =

6s+7t

units. Possible breadth =

6s-7t

units. (assuming length > breadth, and s, t are positive) Final Answer: Possible expressions are length =

(6s+7t)

and breadth =

(6s-7t)

.

Q6End-of-Chapter Exercises

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i)

$6 a^{2}-24 b^{2}$

(ii)

$3 p s^{2}-15 p s+12 p$

Solution

(i) Volume = $6 a^{2}-24 b^{2}$ Volume of a cuboid = length

\times

breadth

\times

height. We factor the expression for the volume.

6a^2-24b^2 = 6(a^2-4b^2) = 6(a-2b)(a+2b)

Possible expressions for the dimensions are

6

,

(a-2b)

, and

(a+2b)

. Final Answer: Possible dimensions are

6

,

(a-2b)

, and

(a+2b)

.

(ii) Volume = $3 p s^{2}-15 p s+12 p$ We factor the expression for the volume.

3ps^2-15ps+12p = 3p(s^2-5s+4)

Now, factor the quadratic

s^2-5s+4 = (s-1)(s-4)

. So, Volume =

3p(s-1)(s-4)

. Possible expressions for the dimensions are

3p

,

(s-1)

, and

(s-4)

. Final Answer: Possible dimensions are

3p

,

(s-1)

, and

(s-4)

.

Q7End-of-Chapter Exercises

The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$ .

Solution

Given: Side of the square playground =

40

metres. Width of the path around the playground =

s

metres.

To Find: An expression for the area of the path.

Solution: The playground is a square. The path is around it. So, we have an outer square and an inner square. Inner square side = side of playground =

40

m. Area of inner square =

(40)^2 = 1600

m

^2

.

The path of width

s

is on all sides. So the side of the outer square will be the side of the inner square plus the width of the path on both sides. Outer square side =

40 + s + s = 40 + 2s

m. Area of outer square =

(40+2s)^2

m

^2

.

Area of the path = Area of outer square - Area of inner square. Area of path =

(40+2s)^2 - 40^2

.

Using the identity

a^2-b^2 = (a-b)(a+b)

: Area of path =

((40+2s) - 40)((40+2s) + 40)

= (2s)(80+2s)

= 160s + 4s^2

Final Answer: The area of the path is

(160s + 4s^2)

square metres.

Q8End-of-Chapter Exercises

If a number plus its reciprocal equals $\frac{10}{3}$ , find the number.

Solution

Let: The number be

x

. Its reciprocal is

\frac{1}{x}

.

Given: A number plus its reciprocal equals

\frac{10}{3}

.

x + \frac{1}{x} = \frac{10}{3}

Solution: To solve for

x

, we first form a quadratic equation.

\frac{x^2+1}{x} = \frac{10}{3}

3(x^2+1) = 10x

3x^2+3 = 10x

3x^2 - 10x + 3 = 0

We can factor this quadratic equation by splitting the middle term. We need two numbers whose product is

3 \times 3 = 9

and whose sum is

-10

. These numbers are

-9

and

-1

.

3x^2 - 9x - x + 3 = 0

3x(x-3) - 1(x-3) = 0

(3x-1)(x-3) = 0

This gives two possible solutions:

3x-1=0 \implies x = \frac{1}{3}

x-3=0 \implies x = 3

Both numbers satisfy the condition. If the number is 3, its reciprocal is 1/3, and their sum is

3+1/3 = 10/3

. If the number is 1/3, its reciprocal is 3, and their sum is

1/3+3 = 10/3

.

Final Answer: The number is 3 or

\frac{1}{3}

.

Q9End-of-Chapter Exercises

A rectangular pool has area $2 x^{2}+7 x+3$ square hastas. If its width is $2 x+1$ hastas, find its length. Hasta was a unit used to measure length.

Solution

Given: Area of the rectangular pool =

(2x^2+7x+3)

square hastas. Width of the pool =

(2x+1)

hastas.

To Find: The length of the pool.

Formula: Area of a rectangle = Length

\times

Width So, Length =

\frac{\text{Area}}{\text{Width}}

Solution: We need to factor the expression for the area,

2x^2+7x+3

. We look for two numbers whose product is

2 \times 3 = 6

and whose sum is

7

. These numbers are

6

and

1

.

2x^2+7x+3 = 2x^2+6x+x+3

= 2x(x+3) + 1(x+3)

= (2x+1)(x+3)

Now we can find the length:

\text{Length} = \frac{(2x+1)(x+3)}{2x+1}

\text{Length} = x+3

Final Answer: The length of the pool is

(x+3)

hastas.

Q10End-of-Chapter Exercises

*10. If both $x-2$ and $x-\frac{1}{2}$ are factors of $p x^{2}+5 x+r$ , show that $p=r$ .

Solution

Given:

x-2

and

x-\frac{1}{2}

are factors of the polynomial

P(x) = px^2+5x+r

.

To Show:

p=r

.

Proof: According to the Factor Theorem, if

(x-a)

is a factor of a polynomial

P(x)

, then

P(a)=0

.

Since

(x-2)

is a factor,

P(2)=0

.

p(2)^2 + 5(2) + r = 0

4p + 10 + r = 0 \quad \dots(1)

Since

(x-\frac{1}{2})

is a factor,

P(\frac{1}{2})=0

.

p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r = 0

p\left(\frac{1}{4}\right) + \frac{5}{2} + r = 0

Multiplying the entire equation by 4 to eliminate fractions:

p + 10 + 4r = 0 \quad \dots(2)

Now we have a system of two linear equations with two variables,

p

and

r

. From equation (1), we have

4p+r = -10

. From equation (2), we have

p+4r = -10

.

Since both expressions are equal to

-10

, we can set them equal to each other:

4p+r = p+4r

4p - p = 4r - r

3p = 3r

p = r

Hence Proved.

Q11End-of-Chapter Exercises

*11. If $a+b+c=5$ and $a b+b c+c a=10$ , then prove that $a^{3}+b^{3}+c^{3}-3 a b c=-25$ .

Solution

Given:

a+b+c=5

ab+bc+ca=10

To Prove:

a^3+b^3+c^3-3abc = -25

.

Proof: We use the identity:

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2 - (ab+bc+ca))

First, we need to find the value of

a^2+b^2+c^2

. We use the identity:

(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)

Substitute the given values:

(5)^2 = a^2+b^2+c^2 + 2(10)

25 = a^2+b^2+c^2 + 20

a^2+b^2+c^2 = 25 - 20 = 5

Now substitute all known values into the main identity:

a^3+b^3+c^3-3abc = (5)(5 - 10)

a^3+b^3+c^3-3abc = (5)(-5)

a^3+b^3+c^3-3abc = -25

Hence Proved.

Q12End-of-Chapter Exercises

*12. By factoring the expression, check that $n^{3}-n$ is always divisible by 6 for all natural numbers $n$ . Give reasons.

Solution

To Check:

n^3-n

is always divisible by 6 for any natural number

n

.

Solution: First, let's factor the expression

n^3-n

.

n^3-n = n(n^2-1)

Using the identity

a^2-b^2=(a-b)(a+b)

, we get:

n(n-1)(n+1)

Rearranging the terms, we get:

(n-1)n(n+1)

Reasoning: The expression

(n-1)n(n+1)

represents the product of three consecutive integers.

Divisibility by 2: In any set of three consecutive integers, at least one integer must be even (a multiple of 2). Therefore, their product is always divisible by 2.
Divisibility by 3: In any set of three consecutive integers, exactly one integer must be a multiple of 3. Therefore, their product is always divisible by 3.

Since the product is divisible by both 2 and 3, and 2 and 3 are coprime, the product must be divisible by their product, which is

2 \times 3 = 6

.

Thus,

n^3-n

is always divisible by 6 for all natural numbers

n

.

Example:

If $n=1$ , $1^3-1=0$ , which is divisible by 6.
If $n=2$ , $2^3-2=6$ , which is divisible by 6.
If $n=3$ , $3^3-3=24$ , which is divisible by 6.
If $n=4$ , $4^3-4=60$ , which is divisible by 6.

Q13End-of-Chapter Exercises

*13. Find the value of

(i)

$x^{3}+y^{3}-12 x y+64$ , when $x+y=-4$

(ii)

$x^{3}-8 y^{3}-36 x y-216$ , when $x=2 y+6$

Solution

(i) $x^{3}+y^{3}-12 x y+64$ , when $x+y=-4$

Given: Expression:

x^3+y^3-12xy+64

Condition:

x+y = -4

Solution: Let's rewrite the expression to match the identity

a^3+b^3+c^3-3abc

.

x^3+y^3+64-12xy = x^3+y^3+4^3-3(x)(y)(4)

This matches the identity with

a=x, b=y, c=4

.

The identity states:

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

So,

x^3+y^3+4^3-3(x)(y)(4) = (x+y+4)(x^2+y^2+4^2-xy-4y-4x)

.

We are given that

x+y=-4

. Let's find the value of the first factor

(x+y+4)

.

x+y+4 = (-4)+4 = 0

Since one of the factors is 0, the entire product is 0.

x^3+y^3-12xy+64 = 0 \times (x^2+y^2+16-xy-4y-4x) = 0

Final Answer: 0

(ii) $x^{3}-8 y^{3}-36 x y-216$ , when $x=2 y+6$

Given: Expression:

x^3-8y^3-36xy-216

Condition:

x=2y+6

Solution: Let's rewrite the expression to match the identity

a^3+b^3+c^3-3abc

.

x^3-8y^3-216-36xy = x^3+(-2y)^3+(-6)^3-3(x)(-2y)(-6)

Let's check the last term:

-3(x)(-2y)(-6) = -36xy

. This matches the given expression. This matches the identity with

a=x, b=-2y, c=-6

.

The identity states:

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

So,

x^3+(-2y)^3+(-6)^3-3(x)(-2y)(-6) = (x-2y-6)(x^2+(-2y)^2+(-6)^2 - x(-2y) - (-2y)(-6) - (-6)x)

.

We are given the condition

x=2y+6

. Let's rearrange this to find the value of the first factor

(x-2y-6)

.

x-2y-6 = 0

Since one of the factors is 0, the entire product is 0.

x^3-8y^3-36xy-216 = 0 \times (\text{some expression}) = 0

Final Answer: 0

Q1Exercise Set 4.1

Using the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$ , expand the following:

(i)

$(7 x+4 y)^{2}$

(ii)

$\left(\frac{7}{5} x+\frac{3}{2} y\right)^{2}$

(iii)

$(2.5 p+1.5 q)^{2}$

(iv)

$\left(\frac{3}{4} s+8 t\right)^{2}$

(v)

$\left(x+\frac{1}{2 y}\right)^{2}$

(vi)

$\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$

Solution

Identity Used:

(a+b)^{2}=a^{2}+2 a b+b^{2}

(i) $(7 x+4 y)^{2}$ Here,

a = 7x

and

b = 4y

.

(7x+4y)^2 = (7x)^2 + 2(7x)(4y) + (4y)^2

= 49x^2 + 56xy + 16y^2

Final Answer:

49x^2 + 56xy + 16y^2

(ii) $\left(\frac{7}{5} x+\frac{3}{2} y\right)^{2}$ Here,

a = \frac{7}{5}x

and

b = \frac{3}{2}y

.

\left(\frac{7}{5}x + \frac{3}{2}y\right)^2 = \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right) + \left(\frac{3}{2}y\right)^2

= \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2

Final Answer:

\frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2

(iii) $(2.5 p+1.5 q)^{2}$ Here,

a = 2.5p

and

b = 1.5q

.

(2.5p + 1.5q)^2 = (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2

= 6.25p^2 + 7.5pq + 2.25q^2

Final Answer:

6.25p^2 + 7.5pq + 2.25q^2

(iv) $\left(\frac{3}{4} s+8 t\right)^{2}$ Here,

a = \frac{3}{4}s

and

b = 8t

.

\left(\frac{3}{4}s + 8t\right)^2 = \left(\frac{3}{4}s\right)^2 + 2\left(\frac{3}{4}s\right)(8t) + (8t)^2

= \frac{9}{16}s^2 + 12st + 64t^2

Final Answer:

\frac{9}{16}s^2 + 12st + 64t^2

(v) $\left(x+\frac{1}{2 y}\right)^{2}$ Here,

a = x

and

b = \frac{1}{2y}

.

\left(x + \frac{1}{2y}\right)^2 = x^2 + 2(x)\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2

= x^2 + \frac{x}{y} + \frac{1}{4y^2}

Final Answer:

x^2 + \frac{x}{y} + \frac{1}{4y^2}

(vi) $\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$ Here,

a = \frac{1}{x}

and

b = \frac{1}{y}

.

\left(\frac{1}{x} + \frac{1}{y}\right)^2 = \left(\frac{1}{x}\right)^2 + 2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right) + \left(\frac{1}{y}\right)^2

= \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}

Final Answer:

\frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}

Q2Exercise Set 4.1

Using the same identity, find the values of the following:

(i)

$(64)^{2}$

(ii)

$(105)^{2}$

(iii)

$(205)^{2}$

Solution

Identity Used:

(a+b)^{2}=a^{2}+2 a b+b^{2}

(i) $(64)^{2}$ We can write

64

as

(60+4)

.

(64)^2 = (60+4)^2

= (60)^2 + 2(60)(4) + (4)^2

= 3600 + 480 + 16

= 4096

Final Answer:

4096

(ii) $(105)^{2}$ We can write

105

as

(100+5)

.

(105)^2 = (100+5)^2

= (100)^2 + 2(100)(5) + (5)^2

= 10000 + 1000 + 25

= 11025

Final Answer:

11025

(iii) $(205)^{2}$ We can write

205

as

(200+5)

.

(205)^2 = (200+5)^2

= (200)^2 + 2(200)(5) + (5)^2

= 40000 + 2000 + 25

= 42025

Final Answer:

42025

Q1Exercise Set 4.2

Factor completely:

(i)

$9 x^{2}+24 x y+16 y^{2}$

(ii)

$4 s^{2}+20 s t+25 t^{2}$

(iii)

$49 x^{2}+28 x y+4 y^{2}$

(iv)

$64 p^{2}+\frac{32}{3} p q+\frac{4}{9} q^{2}$ (v) $\quad 3 a^{2}+4 a b+\frac{4}{3} b^{2}$ (vi) $\frac{9}{5} s^{2}+6 s v+5 v^{2}$ (Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)

Solution

Identity Used:

a^{2}+2 a b+b^{2}=(a+b)^{2}

(i) $9 x^{2}+24 x y+16 y^{2}$ We can write the expression as:

= (3x)^2 + 2(3x)(4y) + (4y)^2

This is in the form

A^2 + 2AB + B^2

where

A=3x

and

B=4y

.

= (3x+4y)^2

Final Answer:

(3x+4y)^2

(ii) $4 s^{2}+20 s t+25 t^{2}$ We can write the expression as:

= (2s)^2 + 2(2s)(5t) + (5t)^2

This is in the form

A^2 + 2AB + B^2

where

A=2s

and

B=5t

.

= (2s+5t)^2

Final Answer:

(2s+5t)^2

(iii) $49 x^{2}+28 x y+4 y^{2}$ We can write the expression as:

= (7x)^2 + 2(7x)(2y) + (2y)^2

This is in the form

A^2 + 2AB + B^2

where

A=7x

and

B=2y

.

= (7x+2y)^2

Final Answer:

(7x+2y)^2

(iv) $64 p^{2}+\frac{32}{3} p q+\frac{4}{9} q^{2}$ We can write the expression as:

= (8p)^2 + 2(8p)\left(\frac{2}{3}q\right) + \left(\frac{2}{3}q\right)^2

This is in the form

A^2 + 2AB + B^2

where

A=8p

and

B=\frac{2}{3}q

.

= \left(8p+\frac{2}{3}q\right)^2

Final Answer:

\left(8p+\frac{2}{3}q\right)^2

(v) $3 a^{2}+4 a b+\frac{4}{3} b^{2}$ We can write the expression as:

= (\sqrt{3}a)^2 + 2(\sqrt{3}a)\left(\frac{2}{\sqrt{3}}b\right) + \left(\frac{2}{\sqrt{3}}b\right)^2

Checking the middle term:

2(\sqrt{3}a)(\frac{2}{\sqrt{3}}b) = 4ab

. This is correct. This is in the form

A^2 + 2AB + B^2

where

A=\sqrt{3}a

and

B=\frac{2}{\sqrt{3}}b

.

= \left(\sqrt{3}a+\frac{2}{\sqrt{3}}b\right)^2

Final Answer:

\left(\sqrt{3}a+\frac{2}{\sqrt{3}}b\right)^2

(vi) $\frac{9}{5} s^{2}+6 s v+5 v^{2}$ We can write the expression as:

= \left(\frac{3}{\sqrt{5}}s\right)^2 + 2\left(\frac{3}{\sqrt{5}}s\right)(\sqrt{5}v) + (\sqrt{5}v)^2

Checking the middle term:

2(\frac{3}{\sqrt{5}}s)(\sqrt{5}v) = 6sv

. This is correct. This is in the form

A^2 + 2AB + B^2

where

A=\frac{3}{\sqrt{5}}s

and

B=\sqrt{5}v

.

= \left(\frac{3}{\sqrt{5}}s+\sqrt{5}v\right)^2

Final Answer:

\left(\frac{3}{\sqrt{5}}s+\sqrt{5}v\right)^2

Q2Exercise Set 4.2

Find the values of the following using the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$ .

(i)

$(79)^{2}$

(ii)

$(193)^{2}$

(iii)

$(299)^{2}$

Solution

Identity Used:

(a-b)^{2}=a^{2}-2 a b+b^{2}

(i) $(79)^{2}$ We can write

79

as

(80-1)

.

(79)^2 = (80-1)^2

= (80)^2 - 2(80)(1) + (1)^2

= 6400 - 160 + 1

= 6241

Final Answer:

6241

(ii) $(193)^{2}$ We can write

193

as

(200-7)

.

(193)^2 = (200-7)^2

= (200)^2 - 2(200)(7) + (7)^2

= 40000 - 2800 + 49

= 37249

Final Answer:

37249

(iii) $(299)^{2}$ We can write

299

as

(300-1)

.

(299)^2 = (300-1)^2

= (300)^2 - 2(300)(1) + (1)^2

= 90000 - 600 + 1

= 89401

Final Answer:

89401

Q1Exercise Set 4.3

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i)

$117^{2}$

(ii)

$78^{2}$

(iii)

$198^{2}$

(iv)

$214^{2}$

(v)

$\quad 1104^{2}$

(vi)

$1120{ }^{2}$

Solution

Identities:

$(a+b)^2 = a^2 + 2ab + b^2$
$(a-b)^2 = a^2 - 2ab + b^2$
$(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$

(i) $117^{2}$ Using

(a+b)^2

is easier. Let

117 = 100+17

.

117^2 = (100+17)^2 = 100^2 + 2(100)(17) + 17^2

= 10000 + 3400 + 289 = 13689

Alternatively, using

(a-b)^2

:

117 = 120-3

.

117^2 = (120-3)^2 = 120^2 - 2(120)(3) + 3^2 = 14400 - 720 + 9 = 13689

Final Answer:

13689

(ii) $78^{2}$ Using

(a-b)^2

is easier. Let

78 = 80-2

.

78^2 = (80-2)^2 = 80^2 - 2(80)(2) + 2^2

= 6400 - 320 + 4 = 6084

Final Answer:

6084

(iii) $198^{2}$ Using

(a-b)^2

is easier. Let

198 = 200-2

.

198^2 = (200-2)^2 = 200^2 - 2(200)(2) + 2^2

= 40000 - 800 + 4 = 39204

Final Answer:

39204

(iv) $214^{2}$ Using

(a+b)^2

is easier. Let

214 = 200+14

.

214^2 = (200+14)^2 = 200^2 + 2(200)(14) + 14^2

= 40000 + 5600 + 196 = 45796

Final Answer:

45796

(v) $1104^{2}$ Using

(a+b)^2

is easier. Let

1104 = 1100+4

.

1104^2 = (1100+4)^2 = 1100^2 + 2(1100)(4) + 4^2

= 1210000 + 8800 + 16 = 1218816

Final Answer:

1218816

(vi) $1120^{2}$ Using

(a+b)^2

is easier. Let

1120 = 1100+20

.

1120^2 = (1100+20)^2 = 1100^2 + 2(1100)(20) + 20^2

= 1210000 + 44000 + 400 = 1254400

Final Answer:

1254400

Q2Exercise Set 4.3

Factor using suitable identities:

(i)

$16 y^{2}-24 y+9$

(ii)

$\frac{9}{4} s^{2}+6 s t+4 t^{2}$

(iii)

$\frac{m^{2}}{9}+\frac{m k}{3}+\frac{k^{2}}{4}+3 n k+2 m n+9 n^{2}$

(iv)

$\frac{p^{2}}{16}-2+\frac{16}{p^{2}}$

(v)

$9 a^{2}+4 b^{2}+c^{2}-12 a b+6 a c-4 b c$

Solution

(i) $16 y^{2}-24 y+9$ This expression is of the form

a^2-2ab+b^2

.

16y^2-24y+9 = (4y)^2 - 2(4y)(3) + (3)^2

Using the identity

a^2-2ab+b^2=(a-b)^2

, with

a=4y

and

b=3

.

= (4y-3)^2

Final Answer:

(4y-3)^2

(ii) $\frac{9}{4} s^{2}+6 s t+4 t^{2}$ This expression is of the form

a^2+2ab+b^2

.

\frac{9}{4}s^2+6st+4t^2 = \left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2

Using the identity

a^2+2ab+b^2=(a+b)^2

, with

a=\frac{3}{2}s

and

b=2t

.

= \left(\frac{3}{2}s+2t\right)^2

Final Answer:

\left(\frac{3}{2}s+2t\right)^2

(iii) $\frac{m^{2}}{9}+\frac{m k}{3}+\frac{k^{2}}{4}+3 n k+2 m n+9 n^{2}$ This expression is of the form

a^2+b^2+c^2+2ab+2bc+2ca

. Rearranging the terms:

= \frac{m^2}{9} + \frac{k^2}{4} + 9n^2 + 2mn + 3nk + \frac{mk}{3}

= \left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)(3n) + 2\left(\frac{k}{2}\right)(3n) + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right)

This matches the expansion of

(a+b+c)^2

with

a=\frac{m}{3}

,

b=\frac{k}{2}

, and

c=3n

. No, let's check the terms again. The question has

2mn, 3nk, mk/3

. My expansion gives

2mn, 3nk, mk/3

. It matches. So the order of terms in my expansion should be

a=\frac{m}{3}, c=3n, b=\frac{k}{2}

. Let's rewrite it properly.

= \left(\frac{m}{3}\right)^2 + (3n)^2 + \left(\frac{k}{2}\right)^2 + 2\left(\frac{m}{3}\right)(3n) + 2(3n)\left(\frac{k}{2}\right) + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) = \frac{m^2}{9}+9n^2+\frac{k^2}{4}+2mn+3nk+\frac{mk}{3}

So it factors to

\left(\frac{m}{3}+3n+\frac{k}{2}\right)^2

. Final Answer:

\left(\frac{m}{3}+\frac{k}{2}+3n\right)^2

(iv) $\frac{p^{2}}{16}-2+\frac{16}{p^{2}}$ This is of the form

a^2-2ab+b^2

.

\frac{p^2}{16}-2+\frac{16}{p^2} = \left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2

Using the identity

a^2-2ab+b^2=(a-b)^2

, with

a=\frac{p}{4}

and

b=\frac{4}{p}

.

= \left(\frac{p}{4}-\frac{4}{p}\right)^2

Final Answer:

\left(\frac{p}{4}-\frac{4}{p}\right)^2

(v) $9 a^{2}+4 b^{2}+c^{2}-12 a b+6 a c-4 b c$ This is of the form

(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx

. The squared terms are

(3a)^2, (2b)^2, c^2

. So let's try

x=3a, y=2b, z=c

. The negative signs are in the terms

-12ab

and

-4bc

. The common variable in these terms is

b

. This suggests that the term with

b

is negative. Let's try

x=3a, y=-2b, z=c

.

(3a-2b+c)^2 = (3a)^2+(-2b)^2+c^2+2(3a)(-2b)+2(-2b)(c)+2(c)(3a)

= 9a^2+4b^2+c^2-12ab-4bc+6ac

This matches the given expression. Final Answer:

(3a-2b+c)^2

Q3Exercise Set 4.3

Expand the following using the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a:$

(i)

$(p+3 q+7 r)^{2}$

(ii)

$(3 x-2 y+4 z)^{2}$

Solution

Identity Used:

(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a

(i) $(p+3 q+7 r)^{2}$ Here,

a=p, b=3q, c=7r

.

(p+3q+7r)^2 = p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)

= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

Final Answer:

p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp

(ii) $(3 x-2 y+4 z)^{2}$ Here,

a=3x, b=-2y, c=4z

.

(3x-2y+4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)

= 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx

Final Answer:

9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx

Q4Exercise Set 4.3

Is this an identity? $(a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=2 a^{2}+2 b^{2}+2 c^{2}.$

Solution

To Verify: Whether

(a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=2 a^{2}+2 b^{2}+2 c^{2}

is an identity.

An equation is an identity if it is true for all values of the variables. Let's expand the Left Hand Side (LHS).

LHS:

(a+b-c)^{2} = a^2+b^2+(-c)^2+2ab+2b(-c)+2(-c)a = a^2+b^2+c^2+2ab-2bc-2ca

(a-b+c)^{2} = a^2+(-b)^2+c^2+2a(-b)+2(-b)c+2ca = a^2+b^2+c^2-2ab-2bc+2ca

(a-b-c)^{2} = a^2+(-b)^2+(-c)^2+2a(-b)+2(-b)(-c)+2(-c)a = a^2+b^2+c^2-2ab+2bc-2ca

Now, let's add these three expansions: LHS =

(a^2+b^2+c^2+2ab-2bc-2ca) + (a^2+b^2+c^2-2ab-2bc+2ca) + (a^2+b^2+c^2-2ab+2bc-2ca)

Combine like terms:

a^2

terms:

a^2+a^2+a^2 = 3a^2

b^2

terms:

b^2+b^2+b^2 = 3b^2

c^2

terms:

c^2+c^2+c^2 = 3c^2

ab

terms:

2ab-2ab-2ab = -2ab

bc

terms:

-2bc-2bc+2bc = -2bc

ca

terms:

-2ca+2ca-2ca = -2ca

So, LHS =

3a^2+3b^2+3c^2-2ab-2bc-2ca

.

RHS: RHS =

2a^2+2b^2+2c^2

.

Since LHS

\neq

RHS, the given equation is not true for all values of

a, b, c

. For example, let

a=1, b=1, c=1

. LHS =

(1+1-1)^2 + (1-1+1)^2 + (1-1-1)^2 = 1^2+1^2+(-1)^2 = 1+1+1=3

. RHS =

2(1)^2+2(1)^2+2(1)^2 = 2+2+2=6

. Since

3 \neq 6

, the equation is not an identity.

Final Answer: No, this is not an identity.

Q1Exercise Set 4.4

Fill in the blanks to complete the following identities:

(i)

$s^{2}-11 s+24=( \_\_\_\_ ) ( \_\_\_\_ )$

(ii)

( $)(x+1)=\left(3 x^{2}-4 x-7\right)$

(iii)

$10 x^{2}-11 x-6=(2 x- \_\_\_\_ ( \_\_\_\_ + 2)$

(iv)

$6 x^{2}+7 x+2=( \_\_\_\_ () \_\_\_\_$

Solution

(i) $s^{2}-11 s+24=( \_\_\_\_ ) ( \_\_\_\_ )$ We need to find two numbers whose sum is

-11

and product is

24

. These numbers are

-3

and

-8

. So,

s^2-11s+24 = (s-3)(s-8)

. Answer:

(s-3)(s-8)

(ii) ( $)(x+1)=\left(3 x^{2}-4 x-7\right)$ We need to factor the quadratic expression

3x^2-4x-7

. We can split the middle term. We need two numbers whose product is

3 \times (-7) = -21

and sum is

-4

. These numbers are

-7

and

3

.

3x^2-4x-7 = 3x^2-7x+3x-7 = x(3x-7)+1(3x-7) = (x+1)(3x-7)

. So,

(3x-7)(x+1) = 3x^2-4x-7

. Answer:

(3x-7)

(iii) $10 x^{2}-11 x-6=(2 x- \_\_\_\_ ) ( \_\_\_\_ + 2)$ Let the expression be

(2x-A)(Bx+2)

. Expanding this gives

2Bx^2 + (4-AB)x - 2A

. Comparing with

10x^2-11x-6

:

Coefficient of $x^2$ : $2B = 10 \implies B=5$ .
Constant term: $-2A = -6 \implies A=3$ .
Coefficient of $x$ : $4-AB = 4-(3)(5) = 4-15 = -11$ . This matches. So the factorization is $(2x-3)(5x+2)$ . Answer: $(2x-3)(5x+2)$

(iv) $6 x^{2}+7 x+2=( \_\_\_\_ ) ( \_\_\_\_ )$ We need to factor

6x^2+7x+2

. We need two numbers whose product is

6 \times 2 = 12

and sum is

7

. These numbers are

3

and

4

.

6x^2+7x+2 = 6x^2+3x+4x+2 = 3x(2x+1)+2(2x+1) = (3x+2)(2x+1)

. Answer:

(3x+2)(2x+1)

Q2Exercise Set 4.4

Select and use the identity that will help you to find the following products without multiplying directly:

(i)

$(41)^{2}$

(ii)

$(27)^{2}$

(iii)

$(23 \times 17)$

(iv)

$(135)^{2}$

(v)

$(97)^{2}$

(vi)

$(18 \times 29)$

(vii)

$(34 \times 43)$

(viii)

$(205)^{2}$

Solution

(i) $(41)^{2}$ Using

(a+b)^2 = a^2+2ab+b^2

.

41^2 = (40+1)^2 = 40^2 + 2(40)(1) + 1^2 = 1600+80+1 = 1681

Final Answer:

1681

(ii) $(27)^{2}$ Using

(a-b)^2 = a^2-2ab+b^2

.

27^2 = (30-3)^2 = 30^2 - 2(30)(3) + 3^2 = 900-180+9 = 729

Final Answer:

729

(iii) $(23 \times 17)$ Using

(a+b)(a-b) = a^2-b^2

.

23 \times 17 = (20+3)(20-3) = 20^2 - 3^2 = 400-9 = 391

Final Answer:

391

(iv) $(135)^{2}$ Using

(a+b)^2 = a^2+2ab+b^2

.

135^2 = (130+5)^2 = 130^2 + 2(130)(5) + 5^2 = 16900 + 1300 + 25 = 18225

Final Answer:

18225

(v) $(97)^{2}$ Using

(a-b)^2 = a^2-2ab+b^2

.

97^2 = (100-3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = 9409

Final Answer:

9409

(vi) $(18 \times 29)$ Using the distributive property or

(x+a)(x+b) = x^2+(a+b)x+ab

.

18 \times 29 = (20-2)(30-1) = 20(30-1) - 2(30-1) = 600 - 20 - 60 + 2 = 522

Final Answer:

522

(vii) $(34 \times 43)$ Using

(x+a)(x+b) = x^2+(a+b)x+ab

.

34 \times 43 = (40-6)(40+3) = 40^2 + (-6+3)40 + (-6)(3) = 1600 - 3(40) - 18 = 1600 - 120 - 18 = 1462

Final Answer:

1462

(viii) $(205)^{2}$ Using

(a+b)^2 = a^2+2ab+b^2

.

205^2 = (200+5)^2 = 200^2 + 2(200)(5) + 5^2 = 40000 + 2000 + 25 = 42025

Final Answer:

42025

Q3Exercise Set 4.4

Factor the following:

(i)

$9 a^{2}+b^{2}+4 c^{2}-6 a b+12 a c-4 b c$

(ii)

$16 s^{2}+25 t^{2}-40 s t$

(iii)

$r^{2}-r-42$

(iv)

$49 g^{2}+14 g h+h^{2}$

(v)

$64 u^{2}+121 v^{2}+4 w^{2}-176 u v-32 u w+44 v w$

Solution

(i) $9 a^{2}+b^{2}+4 c^{2}-6 a b+12 a c-4 b c$ This seems to be of the form

(x+y+z)^2

. The squared terms are

(3a)^2, b^2, (2c)^2

. The negative cross-product terms are

-6ab

and

-4bc

. The common variable is

b

. This suggests the term with

b

is negative. Let's try

(3a-b+2c)^2

.

(3a-b+2c)^2 = (3a)^2+(-b)^2+(2c)^2+2(3a)(-b)+2(-b)(2c)+2(2c)(3a) = 9a^2+b^2+4c^2-6ab-4bc+12ac

. The question has

+12ac

, not

-12ac

. Wait, let me re-read the question:

-6ab+12ac-4bc

. My expansion gives

-6ab+12ac-4bc

. It matches. So the factorization is correct. Final Answer:

(3a-b+2c)^2

(ii) $16 s^{2}+25 t^{2}-40 s t$ Rearranging gives

16s^2 - 40st + 25t^2

. This is of the form

a^2-2ab+b^2

.

= (4s)^2 - 2(4s)(5t) + (5t)^2 = (4s-5t)^2

Final Answer:

(4s-5t)^2

(iii) $r^{2}-r-42$ We need two numbers that multiply to

-42

and add to

-1

. These are

-7

and

6

.

r^2-r-42 = (r-7)(r+6)

Final Answer:

(r-7)(r+6)

(iv) $49 g^{2}+14 g h+h^{2}$ This is of the form

a^2+2ab+b^2

.

= (7g)^2 + 2(7g)(h) + h^2 = (7g+h)^2

Final Answer:

(7g+h)^2

(v) $64 u^{2}+121 v^{2}+4 w^{2}-176 u v-32 u w+44 v w$ This seems to be of the form

(x+y+z)^2

. The squared terms are

(8u)^2, (11v)^2, (2w)^2

. The negative cross-product terms are

-176uv

and

-32uw

. The common variable is

u

. Let's try

(-8u+11v+2w)^2

.

(-8u+11v+2w)^2 = (-8u)^2+(11v)^2+(2w)^2+2(-8u)(11v)+2(11v)(2w)+2(2w)(-8u)$$ $$= 64u^2+121v^2+4w^2-176uv+44vw-32uw$$ This matches the given expression. Another possibility is

(8u-11v-2w)^2

. Let's check:

(8u-11v-2w)^2 = (8u)^2+(-11v)^2+(-2w)^2+2(8u)(-11v)+2(-11v)(-2w)+2(-2w)(8u)

= 64u^2+121v^2+4w^2-176uv+44vw-32uw$$ This also matches. Both are valid factorizations. Final Answer:

(-8u+11v+2w)^2

or

(8u-11v-2w)^2

Q1Exercise Set 4.5

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i)

$\frac{3 p^{2}-3 p q-18 q^{2}}{p^{2}+3 p q-10 q^{2}}$

(ii)

$\frac{n^{3}-3 n^{2} m+3 n m^{2}-m^{3}}{5 m^{2}-10 m n+5 n^{2}}$

(iii)

$\frac{w^{3}-v^{3}+x^{3}+3 w v x}{w^{2}+v^{2}+x^{2}-2 w v-2 v x+2 w x}$

(iv)

$\frac{4 y^{2}-20 y z+25 z^{2}}{\left(25 z^{2}-4 y^{2}\right)}$

(v)

$\frac{\left(x^{2}+x-6\right)\left(x^{2}-7 x+12\right)}{\left(x^{2}-6 x+8\right)\left(x^{2}-9\right)}$

(vi)

$\frac{p^{4}-16}{p^{2}-4 p+4}$

Solution

(i) $\frac{3 p^{2}-3 p q-18 q^{2}}{p^{2}+3 p q-10 q^{2}}$ Factor the numerator:

3p^2-3pq-18q^2 = 3(p^2-pq-6q^2) = 3(p-3q)(p+2q)

. Factor the denominator:

p^2+3pq-10q^2 = (p+5q)(p-2q)

. The expression becomes

\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}

. There are no common factors to cancel. Final Answer: The expression cannot be simplified further. It is

\frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}

.

(ii) $\frac{n^{3}-3 n^{2} m+3 n m^{2}-m^{3}}{5 m^{2}-10 m n+5 n^{2}}$ Numerator is the expansion of

(n-m)^3

. Denominator:

5m^2-10mn+5n^2 = 5(m^2-2mn+n^2) = 5(m-n)^2

. So, the expression is

\frac{(n-m)^3}{5(m-n)^2} = \frac{(-(m-n))^3}{5(m-n)^2} = \frac{-(m-n)^3}{5(m-n)^2} = -\frac{m-n}{5}

or

\frac{n-m}{5}

. Final Answer:

-\frac{m-n}{5}

or

\frac{n-m}{5}

(iii) $\frac{w^{3}-v^{3}+x^{3}+3 w v x}{w^{2}+v^{2}+x^{2}-2 w v-2 v x+2 w x}$ Numerator:

w^3+(-v)^3+x^3-3(w)(-v)(x)

. This is of the form

a^3+b^3+c^3-3abc

with

a=w, b=-v, c=x

. It factors to

(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (w-v+x)(w^2+v^2+x^2+wv+vx-wx)

. Denominator:

w^2+v^2+x^2-2wv-2vx+2wx

. This is the expansion of

(w-v+x)^2

. So the expression is

\frac{(w-v+x)(w^2+v^2+x^2+wv+vx-wx)}{(w-v+x)^2}

.

= \frac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}

Final Answer:

\frac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}

(iv) $\frac{4 y^{2}-20 y z+25 z^{2}}{\left(25 z^{2}-4 y^{2}\right)}$ Numerator:

4y^2-20yz+25z^2 = (2y-5z)^2 = (5z-2y)^2

. Denominator:

25z^2-4y^2 = (5z)^2-(2y)^2 = (5z-2y)(5z+2y)

. The expression is

\frac{(5z-2y)^2}{(5z-2y)(5z+2y)} = \frac{5z-2y}{5z+2y}

. Final Answer:

\frac{5z-2y}{5z+2y}

(v) $\frac{\left(x^{2}+x-6\right)\left(x^{2}-7 x+12\right)}{\left(x^{2}-6 x+8\right)\left(x^{2}-9\right)}$ Factor each quadratic:

x^2+x-6 = (x+3)(x-2)

x^2-7x+12 = (x-3)(x-4)

x^2-6x+8 = (x-2)(x-4)

x^2-9 = (x-3)(x+3)

The expression is

\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}

. Cancelling all common factors, we get

1

. Final Answer:

1

(vi) $\frac{p^{4}-16}{p^{2}-4 p+4}$ Numerator:

p^4-16 = (p^2)^2-4^2 = (p^2-4)(p^2+4) = (p-2)(p+2)(p^2+4)

. Denominator:

p^2-4p+4 = (p-2)^2

. The expression is

\frac{(p-2)(p+2)(p^2+4)}{(p-2)^2} = \frac{(p+2)(p^2+4)}{p-2}

. Final Answer:

\frac{(p+2)(p^2+4)}{p-2}

Exploring Algebraic IdentitiesClass 9 Mathematics NCERT Solutions

Use suitable identities to find the following products:

(−3x+4)2(-3 x+4)^{2}(−3x+4)2

(2s+7)(2s−7)(2 s+7)(2 s-7)(2s+7)(2s−7)

(p2+12)(p2−12)\left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)(p2+21​)(p2−21​)

(2n+7)(2n−7)(2 n+7)(2 n-7)(2n+7)(2n−7)

(s−2t)(s2+2st+4t2)(s-2 t)\left(s^{2}+2 s t+4 t^{2}\right)(s−2t)(s2+2st+4t2)

(12r−4r)2\left(\frac{1}{2 r}-4 r\right)^{2}(2r1​−4r)2

(−3m+4k−l)2(-3 m+4 k-l)^{2}(−3m+4k−l)2

(x−13y)3\left(x-\frac{1}{3} y\right)^{3}(x−31​y)3

(72k−23m)3\left(\frac{7}{2} k-\frac{2}{3} m\right)^{3}(27​k−32​m)3

Solution

Find the values using suitable identities:

17 × 21

104 × 96

24 × 16

1473147^{3}1473

1993199^{3}1993

1273127^{3}1273

(−107)3(-107)^{3}(−107)3

(−299)3(-299)^{3}(−299)3

Solution

Factor the following algebraic expressions:

4y2+1+116y24 y^{2}+1+\frac{1}{16 y^{2}}4y2+1+16y21​

9m2−125n29 m^{2}-\frac{1}{25 n^{2}}9m2−25n21​

27b3−164b3\quad 27 b^{3}-\frac{1}{64 b^{3}}27b3−64b31​

x2+5x6+16x^{2}+\frac{5 x}{6}+\frac{1}{6}x2+65x​+61​

27u3−1125−27u25+9u2527 u^{3}-\frac{1}{125}-\frac{27 u^{2}}{5}+\frac{9 u}{25}27u3−1251​−527u2​+259u​

64y3+1125z364 y^{3}+\frac{1}{125} z^{3}64y3+1251​z3

p3+27q3+r3−9pqrp^{3}+27 q^{3}+r^{3}-9 p q rp3+27q3+r3−9pqr

9m2−12m+49 m^{2}-12 m+49m2−12m+4

9x3−83y3+z33+6xyz9 x^{3}-\frac{8}{3} y^{3}+\frac{z^{3}}{3}+6 x y z9x3−38​y3+3z3​+6xyz

4x2+9y2+36z2+12xz+36yz+24xy4 x^{2}+9 y^{2}+36 z^{2}+12 x z+36 y z+24 x y4x2+9y2+36z2+12xz+36yz+24xy (xi) 27u3−1216−9u22+u427 u^{3}-\frac{1}{216}-\frac{9 u^{2}}{2}+\frac{u}{4}27u3−2161​−29u2​+4u​

Solution

Simplify the following:

4x2+4x+14x2−1\frac{4 x^{2}+4 x+1}{4 x^{2}-1}4x2−14x2+4x+1​

9(3a3−24b3)9a2−36b2\frac{9\left(3 a^{3}-24 b^{3}\right)}{9 a^{2}-36 b^{2}}9a2−36b29(3a3−24b3)​

s3+125t3s2−2st−35t2\frac{s^{3}+125 t^{3}}{s^{2}-2 s t-35 t^{2}}s2−2st−35t2s3+125t3​ Note: Assume that the denominators are not equal to 0.

Solution

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

25a2−30ab+9b225 a^{2}-30 a b+9 b^{2}25a2−30ab+9b2

36s2−49t236 s^{2}-49 t^{2}36s2−49t2

Solution

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

6a2−24b26 a^{2}-24 b^{2}6a2−24b2

3ps2−15ps+12p3 p s^{2}-15 p s+12 p3ps2−15ps+12p

Solution

The village playground is shaped as a square of side 40 metres. A path of width sss metres is created around the playground for people to walk. Find an expression for the area of the path in terms of sss.

Solution

If a number plus its reciprocal equals 103\frac{10}{3}310​, find the number.

Solution

A rectangular pool has area 2x2+7x+32 x^{2}+7 x+32x2+7x+3 square hastas. If its width is 2x+12 x+12x+1 hastas, find its length. Hasta was a unit used to measure length.

Solution

*10. If both x−2x-2x−2 and x−12x-\frac{1}{2}x−21​ are factors of px2+5x+rp x^{2}+5 x+rpx2+5x+r, show that p=rp=rp=r.

Solution

*11. If a+b+c=5a+b+c=5a+b+c=5 and ab+bc+ca=10a b+b c+c a=10ab+bc+ca=10, then prove that a3+b3+c3−3abc=−25a^{3}+b^{3}+c^{3}-3 a b c=-25a3+b3+c3−3abc=−25.

Solution

*12. By factoring the expression, check that n3−nn^{3}-nn3−n is always divisible by 6 for all natural numbers nnn. Give reasons.

Solution

*13. Find the value of

x3+y3−12xy+64x^{3}+y^{3}-12 x y+64x3+y3−12xy+64, when x+y=−4x+y=-4x+y=−4

x3−8y3−36xy−216x^{3}-8 y^{3}-36 x y-216x3−8y3−36xy−216, when x=2y+6x=2 y+6x=2y+6

Solution

Using the identity (a+b)2=a2+2ab+b2(a+b)^{2}=a^{2}+2 a b+b^{2}(a+b)2=a2+2ab+b2, expand the following:

(7x+4y)2(7 x+4 y)^{2}(7x+4y)2

(75x+32y)2\left(\frac{7}{5} x+\frac{3}{2} y\right)^{2}(57​x+23​y)2

(2.5p+1.5q)2(2.5 p+1.5 q)^{2}(2.5p+1.5q)2

(34s+8t)2\left(\frac{3}{4} s+8 t\right)^{2}(43​s+8t)2

(x+12y)2\left(x+\frac{1}{2 y}\right)^{2}(x+2y1​)2

(1x+1y)2\left(\frac{1}{x}+\frac{1}{y}\right)^{2}(x1​+y1​)2

Solution

Using the same identity, find the values of the following:

(64)2(64)^{2}(64)2

(105)2(105)^{2}(105)2

(205)2(205)^{2}(205)2

Solution

Factor completely:

9x2+24xy+16y29 x^{2}+24 x y+16 y^{2}9x2+24xy+16y2

4s2+20st+25t24 s^{2}+20 s t+25 t^{2}4s2+20st+25t2

49x2+28xy+4y249 x^{2}+28 x y+4 y^{2}49x2+28xy+4y2

$(-3 x+4)^{2}$

$(2 s+7)(2 s-7)$

$\left(p^{2}+\frac{1}{2}\right)\left(p^{2}-\frac{1}{2}\right)$

$(2 n+7)(2 n-7)$

$(s-2 t)\left(s^{2}+2 s t+4 t^{2}\right)$

$\left(\frac{1}{2 r}-4 r\right)^{2}$

$(-3 m+4 k-l)^{2}$

$\left(x-\frac{1}{3} y\right)^{3}$

$\left(\frac{7}{2} k-\frac{2}{3} m\right)^{3}$

$147^{3}$

$199^{3}$

$127^{3}$

$(-107)^{3}$

$(-299)^{3}$

$4 y^{2}+1+\frac{1}{16 y^{2}}$

$9 m^{2}-\frac{1}{25 n^{2}}$

$\quad 27 b^{3}-\frac{1}{64 b^{3}}$

$x^{2}+\frac{5 x}{6}+\frac{1}{6}$

$27 u^{3}-\frac{1}{125}-\frac{27 u^{2}}{5}+\frac{9 u}{25}$

$64 y^{3}+\frac{1}{125} z^{3}$

$p^{3}+27 q^{3}+r^{3}-9 p q r$

$9 m^{2}-12 m+4$

$9 x^{3}-\frac{8}{3} y^{3}+\frac{z^{3}}{3}+6 x y z$

$4 x^{2}+9 y^{2}+36 z^{2}+12 x z+36 y z+24 x y$ (xi) $27 u^{3}-\frac{1}{216}-\frac{9 u^{2}}{2}+\frac{u}{4}$

$\frac{4 x^{2}+4 x+1}{4 x^{2}-1}$

$\frac{9\left(3 a^{3}-24 b^{3}\right)}{9 a^{2}-36 b^{2}}$

$\frac{s^{3}+125 t^{3}}{s^{2}-2 s t-35 t^{2}}$ Note: Assume that the denominators are not equal to 0.

$25 a^{2}-30 a b+9 b^{2}$

$36 s^{2}-49 t^{2}$

$6 a^{2}-24 b^{2}$

$3 p s^{2}-15 p s+12 p$

The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$ .

If a number plus its reciprocal equals $\frac{10}{3}$ , find the number.

A rectangular pool has area $2 x^{2}+7 x+3$ square hastas. If its width is $2 x+1$ hastas, find its length. Hasta was a unit used to measure length.

*10. If both $x-2$ and $x-\frac{1}{2}$ are factors of $p x^{2}+5 x+r$ , show that $p=r$ .

*11. If $a+b+c=5$ and $a b+b c+c a=10$ , then prove that $a^{3}+b^{3}+c^{3}-3 a b c=-25$ .

*12. By factoring the expression, check that $n^{3}-n$ is always divisible by 6 for all natural numbers $n$ . Give reasons.

$x^{3}+y^{3}-12 x y+64$ , when $x+y=-4$

$x^{3}-8 y^{3}-36 x y-216$ , when $x=2 y+6$

Using the identity $(a+b)^{2}=a^{2}+2 a b+b^{2}$ , expand the following:

$(7 x+4 y)^{2}$

$\left(\frac{7}{5} x+\frac{3}{2} y\right)^{2}$

$(2.5 p+1.5 q)^{2}$

$\left(\frac{3}{4} s+8 t\right)^{2}$

$\left(x+\frac{1}{2 y}\right)^{2}$

$\left(\frac{1}{x}+\frac{1}{y}\right)^{2}$

$(64)^{2}$

$(105)^{2}$

$(205)^{2}$

$9 x^{2}+24 x y+16 y^{2}$

$4 s^{2}+20 s t+25 t^{2}$

$49 x^{2}+28 x y+4 y^{2}$

Find the values of the following using the identity $(a-b)^{2}=a^{2}-2 a b+b^{2}$ .

$(79)^{2}$

$(193)^{2}$

$(299)^{2}$

$117^{2}$

$78^{2}$

$198^{2}$

$214^{2}$

$\quad 1104^{2}$

$1120{ }^{2}$

$16 y^{2}-24 y+9$

$\frac{9}{4} s^{2}+6 s t+4 t^{2}$

$\frac{m^{2}}{9}+\frac{m k}{3}+\frac{k^{2}}{4}+3 n k+2 m n+9 n^{2}$

$\frac{p^{2}}{16}-2+\frac{16}{p^{2}}$

$9 a^{2}+4 b^{2}+c^{2}-12 a b+6 a c-4 b c$

Expand the following using the identity $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a:$

$(p+3 q+7 r)^{2}$

$(3 x-2 y+4 z)^{2}$

Is this an identity? $(a+b-c)^{2}+(a-b+c)^{2}+(a-b-c)^{2}=2 a^{2}+2 b^{2}+2 c^{2}.$

$s^{2}-11 s+24=( \_\_\_\_ ) ( \_\_\_\_ )$

( $)(x+1)=\left(3 x^{2}-4 x-7\right)$

$10 x^{2}-11 x-6=(2 x- \_\_\_\_ ( \_\_\_\_ + 2)$

$6 x^{2}+7 x+2=( \_\_\_\_ () \_\_\_\_$

$(41)^{2}$

$(27)^{2}$

$(23 \times 17)$

$(135)^{2}$

$(97)^{2}$