# Ncert Solutions For Class 9 Maths Ex 2.5

## Ncert Solutions For Class 9 Maths Chapter 2 Ex 2.5

Q.1.Use suitable identities to find the following products:

(i)(x + 4) (x + 10)

Answer: (x+4)(x+10)=x2+(4+10)x+(4×10)=x2+14x+40$(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 \times 10)\\ = x^{2} + 14x + 40$
(ii)(x + 8) (x – 10)

Answer: (x+8)(x10)=x2+(8+(10))x+(8×(10))=x2+(810)x80=x22x80$(x + 8) (x -10 ) = x^{2} + (8+(- 10))x + (8\times(-10))\\ = x^{2} +(8-10)x – 80\\ =x^{2}-2x-80$

(iii)(3x + 4) (3x – 5)

Answer: (3x+4)(3x5)=(3x)2+4+(5)3x+4×(5)=9x2+3x(45)20=9x23x20$(3x + 4)(3x -5) = (3x) ^{2} + {4 + (-5)}3x + {4×(-5)}\\ = 9x^{2} + 3x(4 – 5) – 20\\ = 9x^{2} – 3x – 20$
(iv)(y2+32)(y232)$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2})$

Answer: (y2+32)(y232)=(y2)2(32)2=y494$(y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2}) = (y^{2})^{2} – (\frac{3}{2})^{2}\\ = y^{4} – \frac{9}{4}$

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   103×107=(100+3)×(100+7)=(100)2+(3+7)(100+(3×7))=10000+1000+21=11021$103 \times 107 = (100 + 3)\times (100 + 7)\\ = (100)^{2} + (3 + 7)(100 + (3 \times 7))\\ = 10000 + 1000 + 21 \\ = 11021$

(ii) 95 × 96

Answer: 95×96=(90+5)×(90+6)=(90)2+90(5+6)+(5×6)=8100+990+30=9120$95 \times 96 = (90 + 5) \times (90 + 6)\\ = (90)^{2} + 90(5 + 6) + (5 \times 6) \\ = 8100 + 990 + 30 = 9120$

(iii) 104 × 96
Answer: 104×96=(100+4)×(1004)=(100)2(4)2=1000016=9984$104 \times 96 = (100 + 4) \times (100 – 4)\\ = (100)^{2} – (4)^{2}\\ = 10000 – 16 \\ = 9984$

Q.3. Factorize the following using appropriate identities:

(i)9x2+6xy+y2$9x^{2} + 6xy + y^{2}$

Answer: 9x2+6xy+y2=(3x)2+(2×3x×y)+y2=(3x+y)2=(3x+y)(3x+y)$9x^{2} + 6xy + y^{2} \\ = (3x) ^{2} + (2\times 3x\times y) + y^{2}\\ = (3x + y)^{2} \\ = (3x + y) (3x + y)$

(ii) 4y24y+1$4y^{2}- 4y + 1$

Answer: 4y24y+1=(2y)2(2×2y×1)+12=(2y1)2=(2y1)(2y1)$4y^{2}- 4y + 1 \\ =(2y)^{2} – (2\times 2y\times 1) + 1^{2}\\ = (2y – 1)^{2}\\ = (2y – 1) (2y – 1)$

(iii) x2y2100$x2 – \frac{y^{2}}{100}$

Answer: x2(y10)2=(xy10)(x+y10)$x^{2} – (\frac{y}{10})^{2} = (x – \frac{y}{10}) (x + \frac{y}{10})$

Q.4. Expand each of the following, using suitable identities:

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

(ii) (2xy+z)2$(2x- y + z)^{2}$

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

(iv) [14a12b+1]2$[\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}$

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

Using identity,

(x+2y+4z)2=x2+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x)=x2+4y2+16z2+4xy+16yz+8xz$(x + 2y + 4z)^{2} \\ = x^{2} + (2y)^{2} + (4z)^{2} + (2\times x\times 2y) + (2\times 2y\times 4z) + (2\times 4z\times x)\\ = x^{2} + 4y^{2} + 16z2 + 4xy + 16yz + 8xz$
(ii)  (2xy+z)2$(2x- y + z)^{2}$
Using identity,

(2xy+z)2=(2x)2+(y)2+z2+(2×2x×y)+(2×y×z)+(2×z×2x)=4x2+y2+z24xy2yz+4xz$(2x-y + z)^{2}\\ = (2x)^{2} + (-y)^{2}+ z^{2} + (2\times 2x\times -y) + (2\times -y\times z) + (2\times z\times 2x) \\ = 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz$

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

Using identity,

(2x+3y+2z)2=(2x)2+(3y)2+(2z)2+(2×2x×3y)+(2×3y×2z)+(2×2z×2x)=4x2+9y2+4z212xy+12yz8xz$(-2x+ 3y + 2z)^{2} \\ =(-2x)^{2} + (3y)^{2} + (2z)^{2} + (2\times -2x\times 3y) + (2\times 3y\times 2z) + (2\times 2z\times -2x)\\ = 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8xz$
(iv) [14a12b+1]2$[\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}$
Using identity,
[14a12b+1]2=(14a)2+(12b)2+12+(2×14a×12b)+(2×12b×1)+(2×1×14a)=116a2+14b2+114abb+12a$[\frac{1}{4} a -\frac{ 1}{2} b + 1]^{2} = (\frac{1}{4} a)^{2} + (-\frac{1}{2} b)^{2} + 1^{2} + (2\times \frac{1}{4} a\times -\frac{1}{2} b) + (2\times -\frac{1}{2} b\times 1) + (2\times 1\times \frac{1}{4} a) \\ = \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 – \frac{1}{4} ab – b + \frac{1}{2} a$

Q.5. Factorize:
(i) 4x2+9y2+16z2+12xy24yz16xz$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$

(ii) 2x2+y2+8z222xy+42yz8xz$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$

(i) 4x2+9y2+16z2+12xy24yz16xz$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$

4x2+9y2+16z2+12xy24yz16xz$4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz$

=(2x)2+(3y)2+(4z)2+(2×2x×3y)+(2×3y×4z)+(2×4z×2x)=(2x+3y4z)2=(2x+3y4z)(2x+3y4z)$= (2x)^{2} + (3y)^{2} + (-4z)^{2} + (2\times 2x\times 3y) + (2\times 3y\times -4z) + (2\times -4z\times 2x)\\ = (2x + 3y – 4z)^{2}\\ = (2x + 3y – 4z) (2x + 3y – 4z)$
(ii) 2x2+y2+8z222xy+42yz8xz$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$

2x2+y2+8z222xy+42yz8xz$2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz$
=(2x)2+(y)2+(22z)2+(2×2x×y)+(2×y×22z)+(2×22z×2x)=(2x+y+22z)2=(2x+y+22z)(2x+y+22z)$= (-\sqrt{2}x)^{2} + (y)^{2} + (2\sqrt{2}z)^{2} + (2\times -\sqrt{2}x\times y) + (2\times y\times 2\sqrt{2}z) + (2\times 2\sqrt{2}z\times -\sqrt{2}x)\\ = (-\sqrt{2}x + y + 2\sqrt{2}z)^{2}\\ = (-\sqrt{2}x + y + 2\sqrt{2}z) (-\sqrt{2}x + y + 2\sqrt{2}z)$

Q.6. Write the following cubes in expanded form:

(i)(2x+1)3$(2x + 1)^{3}$

(ii) (2a3b)3$(2a -3b)^{3}$

(iii) [32x+1]3$[\frac{3}{2} x + 1]^{3}$

(iv) [x23y]3$[ x -\frac{2}{3} y]^{3}$

(i)(2x+1)3$(2x + 1)^{3}$

(2x+1)3=(2x)3+13+(3×2x×1)(2x+1)=8x3+1+6x(2x+1)=8x3+12x3+6x+1$(2x + 1)^{3} = (2x)^{3} + 1^{3} + (3\times 2x\times 1)(2x + 1)\\ = 8x^{3} + 1 + 6x(2x + 1)\\ = 8x^{3} + 12x^{3} + 6x + 1$

(ii) (2a3b)3$(2a -3b)^{3}$

(2a3b)3=(2a)3(3b)3(3×2a×3b)(2a3b)=8a327b318ab(2a3b)=8a327b336a2b+54ab2$(2a-3b)^{3}=(2a)^{3}-(3b)^{3} – (3\times 2a\times 3b)(2a – 3b)\\ = 8a^{3} – 27b^{3} – 18ab(2a – 3b)\\ = 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}$

(iii) [32x+1]3$[\frac{3}{2} x + 1]^{3}$

[32x+1]3=(32x)3+13+(3×32x×1)(32x+1)=278x3+1+92x(32x+1)=278x3+1+2742+92x=278x3+2742+92x+1$[\frac{3}{2} x + 1]^{3} = (\frac{3}{2} x)^{3} + 1^{3} + (3\times\frac{ 3}{2} x\times 1)(\frac{3}{2} x + 1)\\ = \frac{27}{8} x^{3}+ 1 + \frac{9}{2}x (\frac{3}{2} x + 1)\\ = \frac{27}{8} x3 + 1 + \frac{27}{4}^{2} + \frac{9}{2} x\\ = \frac{27}{8} x3 + \frac{27}{4} ^{2} + \frac{9}{2} x + 1$
(iv) [x23y]3$[ x -\frac{2}{3} y]^{3}$
[x23y]3=(x)3(23y)3(3×x×23y)(x23y)=(x)3827y32xy(x23y)=(x)3827y32x2y+43xy2$[x – \frac{2}{3} y]^{3} = (x)^{3} – (\frac{2}{3} y)^{3} – (3\times x\times \frac{2}{3} y)(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y^{3} – 2xy(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y3 – 2x^{2}y + \frac{4}{3}xy^{2}$

Q.7. Evaluate the following using suitable identities:

(i)(99)3$(99)^{3}$

(ii)(102)3$(102)^{3}$

(iii) (998)3$(998)^{3}$

(i) (99)3=(1001)3$(99)^{3} = (100 – 1)^{3}$

(1001)3=(100)313(3×100×1)(1001)$(100 – 1)^{3} = (100)^{3} – 1^{3} – (3\times 100\times 1)(100 – 1)$

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300 = 970299

(ii) (102)3=(100+2)3$(102)^{3} = (100 + 2)^{3}$
(100+2)3=(100)3+23+(3×100×2)(100+2)$(100 + 2)^{3} = (100)^{3} + 2^{3}+ (3\times 100\times 2)(100 + 2)$
= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200= 1061208

(iii) (998)3=(10002)3$(998)^3= (1000 – 2)^{3}$

=(1000)323(3×1000×2)(10002)$= (1000)^{3} – 2^{3} – (3\times 1000\times 2)(1000 – 2)$

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000 = 994011992

Q.8. Factorise each of the following:
(i)
8a3+b3+12a2b+6ab2$8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}$

(ii) 8a3b312a2b+6ab2$8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}$

(iii)27 – 125a3 – 135a + 225a2

(iv) 64a327b3144a2b+108ab2$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}$

(v) 27p3121692p2+14p$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$

(i) 8a3+b3+12a2b+6ab2$8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}$

8a3+b3+12a2b+6ab2=(2a)3+b3+3(2a)2b+3(2a)(b)2=(2a+b)3=(2a+b)(2a+b)(2a+b)$8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}\\ = (2a)^{3} + b^{3} + 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a + b)^{3}\\ = (2a + b)(2a + b)(2a + b)$
(ii) 8a3b312a2b+6ab2$8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}$

8a3b312a2b+6ab2=(2a)3b33(2a)2b+3(2a)(b)2=(2ab)3=(2ab)(2ab)(2ab)$8a^{3} – b^{3}- 12a^{2}b + 6ab^{2}\\ = (2a)^{3} – b^{3} – 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a – b)^{3}\\ = (2a – b)(2a – b)(2a – b)$
(iii)27 – 125a3 – 135a + 225a2

27125a3135a+225a2=33(5a)33(3)2(5a)+3(3)(5a)2=(35a)3=(35a)(35a)(35a)$27 – 125a^{3} – 135a + 225a^{2}\\ = 3^{3} – (5a)^{3} – 3(3)^{2}(5a) + 3(3)(5a)^{2}\\ = (3 – 5a)^{3}\\ = (3 – 5a)(3 – 5a)(3 – 5a)$

(iv) 64a327b3144a2b+108ab2$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}$

64a327b3144a2b+108ab2=(4a)3(3b)33(4a)2(3b)+3(4a)(3b)2=(4a3b)3=(4a3b)(4a3b)(4a3b)$64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}\\ = (4a)^{3} – (3b)^{3} – 3(4a)^{2}(3b) + 3(4a)(3b)^{2}\\ = (4a – 3b)^{3}\\ = (4a – 3b)(4a – 3b)(4a – 3b)$
(v) 27p3121692p2+14p$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$

27p3121692p2+14p$27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p$

=(3p)3(16)33(3p)2(16)+3(3p)(16)2=(3p16)3=(3p16)(3p16)(3p16)$= (3p)^{3} – (\frac{1}{6})^{3} – 3(3p)^{2}(\frac{1}{6}) + 3(3p)(\frac{1}{6})^{2}\\ = (3p – \frac{1}{6})^{3}\\ = (3p – \frac{1}{6})(3p – \frac{1}{6})(3p – \frac{1}{6})$

Q.9. Verify:

(i) x3+y3=(x+y)(x2xy+y2)$x^{3} + y^{3}\\ = (x + y) (x^{2} – xy + y^{2})$

(ii) x3y3=(xy)(x2+xy+y2)$x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})$

(i) x3+y3=(x+y)(x2xy+y2)Weknowthat,(x+y)3=x3+y3+3xy(x+y)x3+y3=(x+y)33xy(x+y)x3+y3=(x+y)[(x+y)23xy]Taking(x+y)commonx3+y3=(x+y)[(x2+y2+2xy)3xy]x3+y3=(x+y)(x2+y2xy)$x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})\\ We \: know\: that, \\ (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)^{3} – 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)[(x + y)^{2} – 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} +y^{3}= (x + y)[(x^{2} + y^{2} + 2xy) – 3xy] \\ \Rightarrow x^{3} +y^{3} = (x + y)(x^{2} + y^{2} – xy)$
(ii) x3y3=(xy)(x2+xy+y2)$x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})$
x3y3=(xy)(x2+xy+y2)Weknowthat,(xy)3=x3y33xy(xy)x3y3=(xy)3+3xy(xy)x3y3=(xy)[(xy)2+3xy]Taking(x+y)commonx3y3=(xy)[(x2+y22xy)+3xy]x3+y3=(xy)(x2+y2+xy)$x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})\\ We \: know\: that, \\ (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)^{3} + 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)[(x – y)^{2} + 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} -y^{3}= (x – y)[(x^{2} + y^{2} – 2xy) + 3xy] \\ \Rightarrow x^{3} +y^{3} = (x – y)(x^{2} + y^{2} + xy)$

Q.10. Factorize each of the following:

(i) 27y3+125z3$27y^{3} + 125z^{3}$

(ii) 64m3343n3$64m^{3} – 343n^{3}$

(i) 27y3+125z3$27y^{3} + 125z^{3}$

27y3+125z3=(3y)3+(5z)3=(3y+5z)[(3y)2(3y)(5z)+(5z)2]=(3y+5z)(9y215yz+25z)2$27y^{3} + 125z^{3}\\ = (3y)^{3} + (5z)^{3}\\ = (3y + 5z) [{(3y)^{2} – (3y)(5z) + (5z)^{2}}]\\ = (3y + 5z) (9y^{2} – 15yz + 25z)^{2}$
(ii) 64m3343n3$64m^{3} – 343n^{3}$

64m3343n3=(4m)3(7n)3=(4m+7n)[(4m)2+(4m)(7n)+(7n)2]=(4m+7n)(16m2+28mn+49n)2$64m^{3} – 343n^{3} \\ = (4m)^{3} – (7n)^{3}\\ = (4m + 7n) [{(4m)^{2} + (4m)(7n) + (7n)^{2}}]\\ = (4m + 7n) (16m^{2} + 28mn + 49n)^{2}$

Q.11. Factorise : 27x3+y3+z39xyz$27x^{3} + y^{3} + z^{3} – 9xyz$

27x3+y3+z39xyz=(3x)3+y3+z33(3x)(y)(z)=(3x+y+z)(3x)2+y2+z23xyyz3xz=(3x+y+z)(9x2+y2+z23xyyz3xz)$27x^{3} + y^{3} + z^{3} – 9xyz \\ = (3x)^{3} + y^{3} + z^{3} – 3 (3x)(y)(z)\\ = (3x + y + z) {(3x)^{2} + y^{2} + z^{2} – 3xy – yz – 3xz}\\ = (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)$
Q.12. Verify that:

x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]$x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]$

We know that,
x3+y3+z33xyz=(x+y+z)(x2+y2+z2xyyzxz)x3+y3+z33xyz=12×(x+y+z)[2(x2+y2+z2xyyzxz)]=12(x+y+z)(2x2+2y2+2z22xy2yz2xz)=12(x+y+z)[(x2+y22xy)+(y2+z22yz)+(x2+z22xz)]=12(x+y+z)[(xy)2+(yz)2+(zx)2]$x^{3} + y^{3} + z^{3} -3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)\\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}\times (x + y + z) [2(x^{2} + y^{2} + z^{2} – xy – yz – xz)]\\ = \frac{1}{2}(x + y + z) (2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2xz) \\ = \frac{1}{2}(x + y + z) [(x^{2} + y^{2} -2xy) + (y^{2} + z^{2} – 2yz) + (x^{2} + z^{2} – 2xz)] \\ = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]$

Q.13. If x + y + z = 0, show that x3+y3+z3=3xyz.$x^{3} + y^{3} + z^{3} = 3xyz.$

We know that,
x3+y3+z3=3xyz$x^{3} + y^{3} + z^{3} = 3xyz$= (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Now put (x + y + z) = 0,
x3+y3+z3=3xyz=(0)(x2+y2+z2xyyzxz)x3+y3+z33xyz=0$x^{3} + y^{3} + z^{3} = 3xyz= (0)(x^{2} + y^{2}+ z^{2} – xy – yz – xz) \\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = 0$

Q.14. Without actually calculating the cubes, find the value of each of the following:
(i) (12)3+(7)3+(5)3$(-12)^{3} + (7)^{3} + (5)^{3}$

(ii) (28)3+(15)3+(13)3$(28)^{3} +(-15)^{3}+(-13)^{3}$

(i) (12)3+(7)3+(5)3$(-12)^{3} + (7)^{3} + (5)^{3}$
(12)3+(7)3+(5)3=0+3(12)(7)(5)(Since.(12)+(7)+(5)=0)UsingIdentity=1260$(-12)^{3} + (7)^{3} + (5)^{3}= 0+3(-12)(7)(5)………………(Since. (-12)+(7)+(5)=0)Using\; Identity\\ = -1260$
(ii) (28)3+(15)3+(13)3$(28)^{3} +(-15)^{3}+(-13)^{3}$
x + y + z = 28 – 15 – 13 = 0

(28)3+(15)3+(13)3=0+3(28)(15)(13)=16380$(28)^{3}+ (-15)^{3} + (-13)^{3}=0+ 3(28)(-15)(-13) = 16380$

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a235a+12$25a^{2} – 35a + 12$

(ii) Area : 35y2+13y12$35 y^{2} + 13y – 12$

(i) Area : 25a235a+12$25a^{2} – 35a + 12$
25a235a+12=25a215a20a+12=5a(5a3)4(5a3)=(5a4)(5a3)$25a^{2} – 35a + 12\\ = 25a^{2} – 15a -20a + 12\\ = 5a(5a – 3) – 4(5a – 3)\\ = (5a – 4)(5a – 3)$

Possible expression for length = 5a – 4
Possible expression for breadth = 5a – 3

(ii) Area : 35y2+13y12$35 y^{2} + 13y – 12$

35y2+13y12=35y215y+28y12=5y(7y3)+4(7y3)=(5y+4)(7y3)$35 y^{2} + 13y – 12\\ = 35y^{2} – 15y + 28y – 12\\ = 5y(7y – 3) + 4(7y – 3)\\ = (5y + 4)(7y – 3)$

Possible expression for length = (5y + 4)
Possible expression for breadth = (7y – 3)

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

(i) Volume : 3x212x$3x^{2} – 12x$
(ii) Volume : 12ky2+8ky20k$12ky^{2} + 8ky – 20k$

(i) Volume : 3x212x$3x^{2} – 12x$
3x212x$3x^{2} – 12x$
= 3x(x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

(ii) Volume : 12ky2+8ky20k$12ky^{2} + 8ky – 20k$

12ky2+8ky20k=4k(3y2+2y5)=4k(3y2+5y3y5)=4k[y(3y+5)1(3y+5)]=4k(3y+5)(y1)$12ky^{2} + 8ky – 20k\\ = 4k(3y^{2} + 2y – 5)\\ = 4k(3y^{2} +5y – 3y – 5)\\ = 4k[y(3y +5) – 1(3y + 5)]\\ = 4k (3y +5) (y – 1)$

Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)