class 8 maths chapter 14 exercise 14.2
class 8 maths chapter 14 exercise 14.2 (1)
1 Factorize the following expressions :
(i) a²+8a+16 (ii) p² – 10p+25 (iii) 25m²+30m+9
(iv) 49y²+84yz+36z² (v) 4x² – 8x+4 (vi) 121b² – 88bc +16c²
(vii) (l+m)² – 4lm [Hint: Expand (l+m)² first ]
(viii) a⁴+2a²b²+b⁴
Solution to class 8 maths chapter 14 exercise 14.2 (1)
(i) a²+8a+16=a²+(4+4)a+4×4
Using Identity x²+(a+b)x+ab=(x+a)(x+b)
Here x=a, a=4 and b=4
a²+8a+16= (a+4)(a+4)=(a+4)²
(ii) p² – 10p+25= p² – 2(p)(5) +(5)²
Using Identity a² – 2ab+b² =(a – b)²
Here a=p, b= 5
p² – 10p+25=(p-5)²
(iii) 25m²+30m+9=(5m)²+2×5m×3+(3)²
Using Identity a² + 2ab+b² =(a + b)²
Here a=5m, b=3
25m²+30m+9=(5m+3)²
(iv) 49y²+84yz+36z²=(7y)²+2×7y×6z+(6z)²
Using Identity a² + 2ab+b² =(a + b)² ,
Here a=7y, b=6z
49y²+84yz+36z²= (7y+6z)²
(v) 4x² – 8x+4= (2x)² – 2×2x×2+(2)²
Using Identity a² – 2ab+b² =(a – b)² ,
Here a=2x, b=2
4x² – 8x+4 = (2x – 2)² = (2)²(x – 1)² = 4(x – 1)²
(vi) 121b² – 88bc+16c²= (11b)² – 2×11b×4c+(4c)²
Using Identity a² – 2ab+b² =(a – b)² ,
Here a=11b, b=4c
121b² – 88bc+16c² = (11b – 4c)²
(vii) (l+m)² – 4lm = l²+2×l×m+m² – 4lm [∵ a² + 2ab+b² =(a + b)²]
= l²+2lm+m² – 4lm
= l² – 2lm+m²
= (l – m)² [∵ (a – b)²= a² – 2ab+b² ]
(viii) a⁴+2a²b²+b⁴ = (a²)²+2×a²×b²+(b²)²
=(a²+b²)² [∵ a² + 2ab+b² =(a + b)²]
class 8 maths chapter 14 exercise 14.2 (2)
2 Factorize :
(i) 4p² – 9q² (ii) 63a² – 112b² (iii) 49x² – 36 (iv) 16x⁵ – 144x² (v) (l+m)² – (l –m)²
(vi) 9x²y² – 16 (vii) (x² – 2xy +y²)–z² (viii) 25a² – 4b²+28bc – 49c²
Solution to class 8 maths chapter 14 exercise 14.2 (2)
(i) 4p² – 9q² = (2p)² – (3q)²
= (2p – 3q)(2p + 3q) [ a² – b² =(a-b)(a+b)]
(ii) 63a² – 112b² = 7(9a² – 16b²)= 7[(3a)² – (4b)²]
= 7(3a – 4b)(3a + 4b) [ a² – b² =(a-b)(a+b)]
(iii) 49x² – 36 = (7x)² – (6)²
= (7x – 6)(7x+ 6) [ ∵a² – b² =(a-b)(a+b)]
(iv) 16x⁵ – 144x² =16x³(x² –9)
=16x³[(x)² –(3)²]
=16x³(x –3)(x+3) [ ∵a² – b² =(a-b)(a+b)]
(v) (l+m)² – (l –m)²=[(l+m)+(l –m)][(l+m)–(l –m)] [ ∵a² – b² =(a-b)(a+b)]
=(l+m+l–m)(l+m–l+m)
=(2l)(2m)=4lm
(vi) 9x²y² –16=(3xy)² –(4)²
=(3xy –4)(3xy+4) [ ∵a² – b² =(a-b)(a+b)]
(vii) (x² –2xy+y²) –z²=(x –y)² –z² [∵ (a – b)²= a² – 2ab+b² ]
=(x –y –z)(x –y+z) [ ∵a² – b² =(a-b)(a+b)]
(viii) 25a² –4b²+28bc –49c²
= 25a² –(4b² –28bc+49c²)
=25a² –[(2b)² –2 ×2b ×7c +(7c)²]
=25a² –(2b –7c)² [∵ (a – b)²= a² – 2ab+b² ]
=(5a)² –(2b –7c)²
=[5a –(2b –7c)][5a+(2b –7c)]. [ ∵a² – b² = (a-b)(a+b)]
class 8 maths chapter 14 exercise 14.2 (3)
3 Factorize the expression:
(i) ax²+bx (ii) 7p²+21q² (iii)2x³+2xy²+2xz²
(iv) am²+bm²+bn²+an² (v) (lm+l)+m+l
(vi) y(y+z)+9(y+z) (vii) 5y² –20y –8z+2yz
(viii) 10ab+4a+5b+2 (ix) 6xy –4y+6 –9x
Solution to class 8 maths chapter 14 exercise 14.2 (3)
(i) ax²+bx=x(ax+b)
(ii) 7p²+21q²=7(p²+3q²)
(iii) 2x³+2xy²+2xz²=2x(x²+y²+z²)
(iv) am²+bm²+bn²+an² =m²(a+b)+n²(a+b)
=(a+b)(m²+n²)
(v) (lm+l)+m+1=l(m+1)+1(m+1)=(m+1)(l+1)
(vi) y(y+z)+9(y+z)=(y+z)(y+9)
(vii) 5y² –20y –8z+2yz =5y² –20y+2yz –8z
=5y(y –4)+2z(y –4)
=(y –4)(5y+2z)
(viii) 10ab+4a+5b+2=2a(5b+2)+1(5b+2)
=(5b+2)(2a+1)
(ix) 6xy –4y+6 –9x =6xy –9x –4y+6
=3x(2y –3) –2(2y –3)
=(2y –3)(3x –2)
class 8 maths chapter 14 exercise 14.2 (4)
4 Factorize
(i) a⁴ –b⁴ (ii) p⁴ –81 (iii) x⁴ –(y+z)⁴
(iv) x⁴ –(x –z)⁴ (v) a⁴ –2a²b²+b⁴
Solution to class 8 maths chapter 14 exercise 14.2 (4)
(i) a⁴ –b⁴ =(a²)² – (b²)²
=(a² –b²)(a²+b²) [ ∵a² – b² = (a-b)(a+b)]
=(a –b)(a+b)(a²+b²) [ ∵a² – b² = (a-b)(a+b)]
(ii) p⁴ – 81= (p²)² – (9)²
=(p² –9)(p²+9) [ ∵a² – b² = (a-b)(a+b)]
=(p² –3²)(p²+9)
=(p –3)(p+3)(p²+9) [ ∵a² – b² = (a-b)(a+b)]
(iii) x⁴ –(y+z)⁴ =(x²)² –[(y+z)²]²
=[x² –(y+z)²][x²+(y+z)²] [ ∵a² – b² = (a-b)(a+b)]
=[x –(y+z)][x+(y+z)][x²+(y+z)²]
[ ∵a² – b² = (a-b)(a+b)]
=(x –y –z)(x+y+z)[x²+(y+z)²]
(iv) x⁴ – (x –z)⁴=(x²)² –[(x –z)²]²
=[x²–(x –z)²][x²+(x –z)²] [ ∵a² – b² = (a-b)(a+b)]
=[x–(x –z)][x+(x –z)][x²+(x –z)²] [ ∵a² – b² = (a-b)(a+b)]
= [x –x+z][x+x –z][x²+x² –2xz+z²]
[∵ (a – b)²= a² – 2ab+b² ]
=z[2x –z][2x² –2xz+z²]
(v) a⁴ –2a²b²+b⁴=(a²)² –2a²b²+(b²)²
=(a² – b²)² [∵ (a – b)²= a² – 2ab+b² ]
=[(a –b)(a+b)]² [ ∵a² – b² = (a-b)(a+b)]
=(a –b)²(a+b)² [∵(xy)ᵐ = xᵐ yᵐ]
class 8 maths chapter 14 exercise 14.2 (5)
5 Factorize the following expressions:
(i) p² – 6p+8 (ii) q² –10q+21 (iii) p²+6p –16
Solution to class 8 maths chapter 14 exercise 14.2 (5)
(i) p²+6p+8 = p²+(4+2)p+(4×2)
=p²+4p+2p+(4×2)
=p(p+4)+2(p+4)
=(p+4)(p+2)
(ii) q² –10q+21= q² –(7+3)q+(7×3)
=q² –7q –3q +(7×3)
=q(q –7) –3(q –7)
=(q –7)(q –3)
(iii) p²+6p –16=p²+(8 –2)p –(8 ×2)
=p²+8p –2p –8 ×2
=p(p+8) –2(p+8)
=(p+8)(p –2)