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

Page 4.7

Exercise 4.2

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)


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-6rz-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)


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

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)


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)


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

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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