KC Sinha Mathematics Solution Class 9 Chapter 4 Algebraic identities exercise 4.2

Page 4.7

Exercise 4.2

Q1 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Type 1

Question 1

Factorize the following :
(i) x2+4x+4
Sol:
Using identity:
a2+b2+2ab=(a+b)2=(a+b)(a+b)
⇒x2+22+2(x)(2)
⇒(x+2)2
⇒(x+2)(x+2)
(ii) x2-18x+81
Sol:
Using identity:
a2+b2-2ab=(a-b)2=(a-b)(a-b)
⇒x2+92-2(x)(9)
⇒(x-9)2
⇒(x-9)(x-9)
(iii) 4x2+4x+1
Sol:
Using identity:
a2+b2+2ab=(a+b)2=(a+b)(a+b)
⇒(2x)2+12+2(2x)(1)
⇒(2x+1)2
⇒(2x+1)(2x+1)
(iv) $\dfrac{4}{9}a^2+b^2+\dfrac{4}{3}ab$
Sol:
Using identity:
a2+b2+2ab=(a+b)2=(a+b)(a+b)
⇒$\left(\dfrac{2}{3}a\right)^2+b^2+2\left(\dfrac{2}{3}a\right)(b)$
⇒$\left(\dfrac{2}{3}a+b\right)^2$
⇒$\left(\dfrac{2}{3}a+b\right)$ $\left(\dfrac{2}{3}a+b\right)$
(v) 25a2-60ab+36b2
Sol:
Using identity:
a2+b2-2ab=(a-b)2=(a-b)(a-b)
⇒(5a)2+(6b)2-2(5a)(6b)
⇒(5a-6b)2
⇒(5a-6b)(5a-6b)
(vi) 9a2-30ab+25b2
Sol:
Using identity:
a2+b2-2ab=(a-b)2=(a-b)(a-b)
⇒(3a)2+(5b)2-2(3a)(5b)
⇒(3a-5b)2
⇒(3a-5b)(3a-5b)
(vii) 4(x-y)2-12(x-y)(x+y)+9(x+y)2
Sol:
Using identity:
(a2-b2)=(a+b)(a-b)
⇒4(x2+y2-2xy)-12(x2-y2)+9(x2+y2+2xy)
⇒4x2+4y2-8xy-12x2+12y2+9x2+9y2+18xy
⇒4x2-12x2+9x2+4y2+12y2+9y2-8xy+18xy
⇒13x2-12x2+13y2+12y2+10xy
⇒x2+25y2+10xy
⇒x2+(5y)2+2(x)(5y)
Using identity:
a2+b2+2ab=(a+b)2
⇒(x+5y)2
⇒(x+5y)(x+5y)

Q2 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 2

Factorize the following :
(i) p2+q2+9r2+2pq+6pr+6qr
Sol:
Using identity:
a2+b2+c2+2ab+2bc+2ca=(a+b+c)2
⇒p2+q2+(3r)2+2pq+2(q)(3r)+2(3r)(p)
⇒(p+q+3r)2
⇒(p+q+3r)(p+q+3r)
(ii) 4a2+9b2+c2+12ab+4ac+6bc
Sol:
Using identity:
a2+b2+c2+2ab+2bc+2ca=(a+b+c)2
⇒(2a)2+(3b)2+c2+2(2a)(3b)+2(3b)(c)+2(c)(2a)
⇒(2a+3b+c)2
⇒(2a+3b+c)(2a+3b+c)
(iii) x2+4+9z2+4x-6xz-12z
Sol:
Using identity:
a2+b2+c2+2ab+2bc+2ca=(a+b+c)2
⇒(x)2+(2)2+(-3z)2+2(x)(2)+2(2)(-3z)+2(-3z)(x)
⇒(x+2-3z)2
⇒(x+2-3z)(x+2-3z)
(iv) 4x2+9y2+z2-12xy-4xz+6yz
Sol:
Using identity:
a2+b2+c2+2ab+2bc+2ca=(a+b+c)2
⇒(-2x)2+(3y)2+(z)2+2(-2x)(3y)+2(3y)(z)+2(z)(-2x)
⇒(-2x+3y+z)2
[Taking common (-)]
⇒(2x-3y-z)(2x-3y-z)

Q3 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 3

Factorize:
4x2+y2+z2-4xy-2yz+4zx
Sol:
Using identity:
a2+b2+c2+2ab+2bc+2ca=(a+b+c)2
⇒(2x)2-y2+z2+2(2x)(-y)+2(-y)(z)+2(z)(2x)
⇒(2x-y+z)2
⇒(2x-y+z)(2x-y+z)

Type 2
Q4 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 4

Factorize the following :
(i) x2-1
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒x2-12
⇒(x-1)(x+1)
(ii) x2-4
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒x2-22
⇒(x-2)(x+2)
(iii) 49-64x2
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒(7)2-(8x)2
⇒(7-8x)(7+8x)
(iv) 4x2-81
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒(2x)2-(9)2
⇒(2x-9)(2x+9)
(v) 4a2-(2b-c)2
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒(2a)2-(2b-c)2
⇒[2a+2b-c][2a-(2b-c)]
⇒(2a+2b-c)(2a-2b+c)
(vi) 16(2x-1)2-25z2
Sol:
Using identity:
a2-b2=(a+b)(a-b)
⇒[4(2x-1)]2-(5z)2
⇒[4(2x-1)+5z][4(2x-1)-5z]
⇒[8x-4+5z][8x-4-5z]
⇒(8a-5z-4)(8x+5z-4)

Q5 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 5

Factorize the following :
(i) x2+y2+2xy-z2
Sol:
Using identity:
a2+b2+2ab=(a+b)2
⇒(x+y)2-(z)2
Using identity:
a2-b2=(a+b)(a-b)
⇒(x+y+z)(x+y-z)
(ii) x2+y2-z2-2xy
Sol:
Using identity:
a2+b2-2ab=(a-b)2
⇒x2+y2-2xy-z2
⇒(x-y)2-(z)2
Using identity:
a2-b2=(a+b)(a-b)
⇒[(x-y)+(z)][(x-y)-(z)]
⇒(x-y+z)(x-y-z)
(iii) a2-b2-c2-2bc
Sol:
⇒(a)2-(b2+c2+2bc)
Using identity:
x2+y2+2xy=(x+y)2
⇒(a)2-(b+c)2
Using identity:
x2-y2=(x+y)(x-y)
⇒[(a)+(b+c)][(a)-(b+c)]
⇒(a+b+c)(a-b-c)
(iv) x2-1-2a-a2
Sol:
⇒x2-[a2+12+2(a)(1)]
Using identity:
a2+b2+2ab=(a+b)2
⇒(x)2-(a+1)2
Using identity:
a2-b2=(a+b)(a-b)
⇒[x+a+1][x-(a+1)]
⇒(x+a+1)(x-a-1)
(v) x4-14x2y2+y4
Sol:
⇒x4+y4-14x2y2
⇒[(x2)2+(y2)2+2(x2)(y2)]-[16(x2)(y2)]
Using identity:
a2-b2=(a+b)(a-b)
⇒[(x2+y2)2]-[16(x2)(y2)]
⇒(x2+y2)2-(4xy)2
⇒(x2+y2-4xy)(x2+y2+4xy)
(vi) x4-7x2y2+y4
Sol:
⇒x4+y4-7x2y2
⇒[(x2)2+(y2)2+2(x2)(y2)]-[9x2y2]
Using identity:
a2+b2+2ab=(a+b)2
⇒[(x2+y2)2]-[9x2y2]
⇒(x2+y2)2-(3xy)2
Using identity:
a2-b2=(a+b)(a-b)
⇒[(x2+y2)+(3xy)][(x2+y2)-(3xy)]
⇒(x2+y2+3xy)(x2+y2-3xy)

Q6 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 6

Factorize the following :
(i) (1-x2)(1-y2)+4xy
Sol:
⇒(1-x2)×(1-y2)+4xy
⇒1×(1-y2)-x2(1-y2)+4xy
⇒1-y2-x2+x2y2+4xy
⇒(1+2xy+x2y2)-(x2+y2-2xy)
Using identities:
a2+b2+2ab=(a+b)2
a2+b2-2ab=(a-b)2
⇒(1+xy)2-(x-y)2
⇒[(1+xy)+(x-y)] × [( 1+xy)- (x-y)]
⇒[(1+x-y+xy) (1-x+y+xy)]

(ii) c2+2ab-(a2+b2)
Sol:
⇒c2+2ab-a2-b2
⇒(c)2-(a2+b2-2ab)
Using identity:
a2+b2-2ab=(a-b)2
⇒c2-(a-b)2
Using identity:
a2-b2=(a+b)(a-b)
⇒(c+a-b)(c-a+b)
(iii) p-2q-p2+4q2
Sol:
⇒p-2q-p2+4q2
⇒p-2q-(p²-4q²)
Using identity:
a²-b²=(a - b)(a + b)
⇒(p-2q)-(p-2q)(p+2q)
[Taking common (p-2q)]
⇒(p-2q)(1-p-2q)
(iv) b2+c2+2(ab+bc+ca)
Sol:
adding and subtracting a2
⇒a2+b2+c2+2(ab+bc+ca)-a2
⇒a2+b2+c2+2ab+2bc+2ca-a2
Using identity:
x2+y2+z2+2xy+2yz+2zx=(x+y+z)2
⇒(a+b+c)2-a2
Using identity:
a2-b2=(a+b)(a-b)
⇒[(a+b+c)+a][(a+b+c)-a]
⇒(b+c+2a)(b+c)
Page 4.16

Type 3

Q7 | Ex-4.2 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

Question 7

Factorize the following :
(i) 27x3+54x2y+36xy2+8y3
Sol:
⇒27x3+8y3+54x2y+36xy2
⇒(3x)3+(2y)3+3(3x)2(y)+3(3x)(y)2
Using identity:
a3+b3+3a2b+3ab2=(a+b)3
⇒(3x+2y)3
⇒(3x+2y)(3x+2y)(3x+2y)
(ii) 8x3+y3+12x2y+6xy2
Sol:
⇒8x3+y3+12x2y+6xy2
⇒(2x)3+y3+3(2x)2(y)+3(2x)(y)2
Using identity:
a3+b3+3a2b+3ab2=(a+b)3
⇒(2x+y)3
⇒(2x+y)(2x+y)(2x+y)
(iii) a3x3-3a2bx2+3ab2x-b3
Sol:
⇒a3x3-b3-3a2bx2+3ab2x
⇒(ax)3-b3-3(ax)2(b)+3(ax)(b)2
Using identity:
a3-b3-3a2b+3ab2=(a-b)3
⇒(ax-b)3
⇒(ax-b)(ax-b)(ax-b)
(iv) x3-12x(x-4)-64
Sol:
⇒x3-43-12x(x-4)
⇒x3-43-12x2+48x
⇒x3-43-3(x)2(4)+3(x)(4)2
Using identity:
a3-b3-3a2b+3ab2=(a-b)3
⇒(x-4)3
⇒(x-4)(x-4)(x-4)
(v) $a^3+\dfrac{3}{2}a^2+\dfrac{3}{4}a+\dfrac{1}{8}$
Sol:
⇒$a^3+\dfrac{1}{8}+\dfrac{3}{2}a^2+\dfrac{3}{4}a$
⇒$a^3+\left(\dfrac{1}{2}\right)^3+3(a)^2\left(\dfrac{1}{2}\right)+3(a)\left(\dfrac{1}{2}\right)^2$
Using identity:
a3+b3+3a2b+3ab2=(a+b)3
⇒$\left(a+\dfrac{1}{2}\right)^3$
⇒$\left(a+\dfrac{1}{2}\right)\left(a+\dfrac{1}{2}\right)\left(a+\dfrac{1}{2}\right)$

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