EXERCISE 2.6
Page No 48:
Question 1:
Use suitable identities to find the following products:(i) (x+4)(x+10)
(ii) (x+8)(x-10)
(iii) (3x+4)(3x-5)
(iv) $\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$
(v) (3-2x)(3+2x)
Answer:
(i) By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$,$(x+4)(x+10)=x^{2}+(4+10) x+4 \times 10$
$=x^{2}+14 x+40$
(ii) By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$,
$(x+8)(x-10)=x^{2}+(8-10) x+(8)(-10)$
$=x^{2}-2 x-80$
(iii) $(3 x+4)(3 x-5)=9\left(x+\frac{4}{3}\right)\left(x-\frac{5}{3}\right)$
By using the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$,
$9\left(x+\frac{4}{3}\right)\left(x-\frac{5}{3}\right)=9\left[x^{2}+\left(\frac{4}{3}-\frac{5}{3}\right) x+\left(\frac{4}{3}\right)\left(-\frac{5}{3}\right)\right]$
$=9\left[x^{2}-\frac{1}{3} x-\frac{20}{9}\right]$
$=9 x^{2}-3 x-20$
(iv) By using the identity $(x+y)(x-y)=x^{2}-y^{2}$,
$\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}$
$=y^{4}-\frac{9}{4}$
(v) By using the identity $(x+y)(x-y)=x^{2}-y^{2}$,
$(3-2 x)(3+2 x)=(3)^{2}-(2 x)^{2}$
$=9-4 x^{2}$
Question 2:
Evaluate the following products without multiplying directly:(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96
Answer:
(i) 103 × 107 = (100 + 3) (100 + 7)= (100)2 + (3 + 7) 100 + (3) (7)
[By using the identity$(x+a)(x+b)=x^{2}+(a+b) x+a b$, where
x = 100, a = 3, and b = 7]
= 10000 + 1000 + 21
= 11021
(ii) 95 × 96 = (100 − 5) (100 − 4)
= (100)2 + (− 5 − 4) 100 + (− 5) (− 4)
[By using the identity$(x+a)(x+b)=x^{2}+(a+b) x+a b$, where
x = 100, a = −5, and b = −4]
= 10000 − 900 + 20
= 9120
(iii) 104 × 96 = (100 + 4) (100 − 4)
= (100)2 − (4)2 $\left[(x+y)(x-y)=x^{2}-y^{2}\right]$
= 10000 − 16
= 9984
Question 3:
Factorize the following using appropriate identities:(i) 9x2 + 6xy + y2
(ii) $4 y^{2}-4 y+1$
(iii) $x^{2}-\frac{y^{2}}{100}$
Answer:
(i) $9 x^{2}+6 x y+y^{2}=(3 x)^{2}+2(3 x)(y)+(y)^{2}$=(3x+y)(3x+y) $\left[x^{2}+2 x y+y^{2}=(x+y)^{2}\right]$
(ii) $4 y^{2}-4 y+1=(2 y)^{2}-2(2 y)(1)+(1)^{2}$
=(2 y-1)(2 y-1) $\left[x^{2}-2 x y+y^{2}=(x-y)^{2}\right]$
(iii) $x^{2}-\frac{y^{2}}{100}=x^{2}-\left(\frac{y}{10}\right)^{2}$
$=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)$ $\left[x^{2}-y^{2}=(x+y)(x-y)\right]$
Page No 49:
Question 4:
Expand each of the following, using suitable identities:(i) $(x+2 y+4 z)^{2}$
(ii) $(2 x-y+z)^{2}$
(iii) $(-2 x+3 y+2 z)^{2}$
(iv) $(3 a-7 b-c)^{2}$
(v) $(-2 x+5 y-3 z)^{2}$
(vi) $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
Answer:
It is known that,$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$
(i) $(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x)$
$=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z$
(ii) $(2 x-y+z)^{2}=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(2 x)(-y)+2(-y)(z)+2(z)(2 x)$
$=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$
(iii) $(-2 x+3 y+2 z)^{2}$
$=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)+2(3 y)(2 z)+2(2 z)(-2 x)$
$=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z$
(iv) $(3 a-7 b-c)^{2}$
$=(3 a)^{2}+(-7 b)^{2}+(-c)^{2}+2(3 a)(-7 b)+2(-7 b)(-c)+2(-c)(3 a)$
$=9 a^{2}+49 b^{2}+c^{2}-42 a b+14 b c-6 a c$
(v) $(-2 x+5 y-3 z)^{2}$
$=(-2 x)^{2}+(5 y)^{2}+(-3 z)^{2}+2(-2 x)(5 y)+2(5 y)(-3 z)+2(-3 z)(-2 x)$
$=4 x^{2}+25 y^{2}+9 z^{2}-20 x y-30 y z+12 x z$
(vi) $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
$=\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2\left(\frac{1}{4} a\right)(1)$
$=\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$
Question 5:
Factorize:(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
Answer:
It is known that,$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
$=(2 x)^{2}+(3 y)^{2}+(-4 z)^{2}+2(2 x)(3 y)+2(3 y)(-4 z)+2(2 x)(-4 z)$
$=(2 x+3 y-4 z)^{2}$
=(2 x+3 y-4 z)(2 x+3 y-4 z)
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
$=(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)(-\sqrt{2} x+y+2 \sqrt{2} z)$
Question 6:
Write the following cubes in expanded form:(i) $(2 x+1)^{3}$
(ii) $(2 a-3 b)^{3}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}$
(iv) $\left[x-\frac{2}{3} y\right]^{3}$
Answer:
It is known that,$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
(i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$
$=8 x^{3}+1+6 x(2 x+1)$
$=8 x^{3}+1+12 x^{2}+6 x$
$=8 x^{3}+12 x^{2}+6 x+1$
(ii) $(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b)$
$=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)$
$=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2} x$
$=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1$
(vi) $\left[x-\frac{2}{3} y\right]^{3}=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
$=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$
$=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$
Question 7:
Evaluate the following using suitable identities:(i) (99)3 (ii) (102)3 (iii) (998)3
Answer:
It is known that,$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$
(i) (99)3 = (100 − 1)3
= (100)3 − (1)3 − 3(100) (1) (100 − 1)
= 1000000 − 1 − 300(99)
= 1000000 − 1 − 29700
= 970299
(ii) (102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600 (102)
= 1000000 + 8 + 61200
= 1061208
(iii) (998)3= (1000 − 2)3
= (1000)3 − (2)3 − 3(1000) (2) (1000 − 2)
= 1000000000 − 8 − 6000(998)
= 1000000000 − 8 − 5988000
= 1000000000 − 5988008
= 994011992
Question 8:
Factorize each of the following:(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
(iii) $27-125 a^{3}-135 a+225 a^{2}$
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$
Answer:
It is known that,$(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$
and $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
$=(2 a)^{3}+(b)^{3}+3(2 a)^{2} b+3(2 a)(b)^{2}$
$=(2 a+b)^{3}$
=(2 a+b)(2 a+b)(2 a+b)
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
$=(2 a)^{3}-(b)^{3}-3(2 a)^{2} b+3(2 a)(b)^{2}$
$=(2 a-b)^{3}$
=(2 a-b)(2 a-b)(2 a-b)
(iii) $27-125 a^{3}-135 a+225 a^{2}$
$=(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2}$
$=(3-5 a)^{3}$
=(3-5 a)(3-5 a)(3-5 a)
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
$=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$
$=(4 a-3 b)^{3}$
=(4 a-3 b)(4 a-3 b)(4 a-3 b)
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$
$=(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2}\left(\frac{1}{6}\right)+3(3 p)\left(\frac{1}{6}\right)^{2}$
$=\left(3 p-\frac{1}{6}\right)^{3}$
$=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
Question 9:
Verify:(i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
Answer:
(i) It is known that,$(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$
$x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$
$=(x+y)\left[(x+y)^{2}-3 x y\right]$
$=(x+y)\left(x^{2}+y^{2}+2 x y-3 x y\right)$
$=(x+y)\left(x^{2}+y^{2}-x y\right)$
$=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) It is known that,
$(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$
$x^{3}-y^{3}=(x-y)^{3}+3 x y(x-y)$
$=(x-y)\left[(x-y)^{2}+3 x y\right]$
$=(x-y)\left(x^{2}+y^{2}-2 x y+3 x y\right)$
$=(x-y)\left(x^{2}+y^{2}+x y\right)$
$=(x-y)\left(x^{2}+x y+y^{2}\right)$
Question 10:
Factorize each of the following:(i) $27 y^{3}+125 z^{3}$
(ii) $64 m^{3}-343 n^{3}$
[Hint: See question 9.]
Answer:
(i) $27 y^{3}+125 z^{3}$$=(3 y)^{3}+(5 z)^{3}$
$=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$
$=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]$
(ii) $64 m^{3}-343 n^{3}$
$=(4 m)^{3}-(7 n)^{3}$
$=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right]$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$
$=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]$
Question 11:
Factorize: $27 x^{3}+y^{3}+z^{3}-9 x y z$Answer:
It is known that,$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$\therefore 27 x^{3}+y^{3}+z^{3}-9 x y z$
$=(3 x)^{3}+(y)^{3}+(z)^{3}-3(3 x)(y)(z)$
$=(3 x+y+z)\left[(3 x)^{2}+y^{2}+z^{2}-(3 x)(y)-(y)(z)-z(3 x)\right]$
$=(3 x+y+z)\left[9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z\right]$
Question 12:
Verify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$Answer:
It is known that,$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$=\frac{1}{2}(x+y+z)\left[2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right]$
$=\frac{1}{2}(x+y+z)\left[\left(x^{2}+y^{2}-2 x y\right)+\left(y^{2}+z^{2}-2 y z\right)+\left(x^{2}+z^{2}-2 z x\right)\right]$
$=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$
Question 13:
If x + y + z = 0, show that$x^3 + y^3 + z^3 = 3xyz.$
Answer:
It is known that,$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
Put x + y + z = 0,
$x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x^{3}+y^{3}+z^{3}-3 x y z=0$
$x^{3}+y^{3}+z^{3}=3 x y z$
Question 14:
Without actually calculating the cubes, find the value of each of the following:(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
Answer:
(i) $(-12)^{3}+(7)^{3}+(5)^{3}$Let x = −12, y = 7, and z = 5
It can be observed that,
x + y + z = − 12 + 7 + 5 = 0
It is known that if x + y + z = 0, then
$x^{3}+y^{3}+z^{3}=3 x y z$
∴ $(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$
= −1260
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
Let x = 28, y = −15, and z = −13
It can be observed that,
x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0
It is known that if x + y + z = 0, then
$x^{3}+y^{3}+z^{3}=3 x y z$
$\therefore(28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)$
=16380
Question 15:
Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given:
Answer:
Area = Length × BreadthThe expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.
(i) $25 a^{2}-35 a+12=25 a^{2}-15 a-20 a+12$
=5a(5a-3)-4(5a-3)
=(5a-3)(5a-4)
Therefore, possible length = 5a − 3
And, possible breadth = 5a − 4
(ii) $35 y^{2}+13 y-12=35 y^{2}+28 y-15 y-12$
=7 y(5 y+4)-3(5 y+4)
=(5 y+4)(7 y-3)
Therefore, possible length = 5y + 4
And, possible breadth = 7y − 3
Page No 50:
Question 16:
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Answer:
Volume of cuboid = Length × Breadth × HeightThe expression given for the volume of the cuboid has to be factorised. One of its factors will be its length, one will be its breadth, and one will be its height.
(i) $3 x^{2}-12 x=3 x(x-4)$
One of the possible solutions is as follows.
Length = 3, Breadth = x, Height = x − 4
(ii) $12 k y^{2}+8 k y-20 k=4 k\left(3 y^{2}+2 y-5\right)$
$=4 k\left[3 y^{2}+5 y-3 y-5\right]$
=4 k[y(3 y+5)-1(3 y+5)]
=4 k(3 y+5)(y-1)
One of the possible solutions is as follows.
Length = 4k, Breadth = 3y + 5, Height = y − 1
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