## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 11 Factorisation Check Your Progress

Question 1.
Find the HCF of the given polynomials:
(i) 14pq, 28p2q2
(ii) 8abc, 24ab2, 12a2b
Solution:
(i) 14pq, 28p2q2 (HCF of 14, 28 = 14)
HCF of 1 4pq, 28p2q2 = 14pq
(ii) 8abc, 24ab2, 12a2b
HCF of 8, 24, 12 = 4
HCF of 8abc, 24ab2, 12a2b = 4ab

Question 2.
Factorise the following:
(i) 10x2 – 18x3 + 14x4
(ii) 5x2y + 10xyz + 15xy2
(iii) p2x2 + c2x2 – ac2 – ap2
(iv) 15(x + y)2 – 5x – 5y
(v) (ax + by)2 + (ay – bx)2
(vi) ax + by + cx + bx + cy + ay
(vii) 49x2 – 70xy + 25y2
(viii) 4a2 + 12ab + 9b2
(ix) 49p2 – 36q2
(x) 100x3 – 25xy2
(xi) x2 – 2xy + y2 – z2
(xii) x8 – y8
(xiii) 12x3 – 14x2 – 10x
(xiv) p2 – 10p + 21
(xv) 2x2 – x – 6
(xvi) 6x2 – 5xy – 6y2
(xvii) x2 + 2xy – 99y2
Solution:
(i) 10x2 – 18x3 + 14x4
HCF of 10, 18, 14 = 2
∴ 10x2 – 18x3/sup> + 14 x 4
= 2x2 (5 – 9x + 7x2)

(ii) 5x2y + 10xyz + 15xy2
HCF of 5, 10, 15 = 5
∴ 5x2y + 10xyz + 15xy2
= 5xy(x + 2z + 3y)

(iii) p2x2 + c2x2 – ac2 – ap
= p2x2 – ap2 + c2x2 – ac2
p2(x2 – a) + c2(x2 – a)
= (x2 – a) (p2 + c2)

(iv) 15(x + y)2 – 5x – 5y
= 15(x + y)2 – 5(x + y)
= 5(x + y) [3(x + y) – 1]
= 5(x + y) (3x + 3y – 1)

(v) (ax + by)2 + (ay – bx)2
a2x2 + b2y2 + 2abxy + a2y2 + b2x2 – 2abxy
= a2x2 + a2y2 + b2x2 + b2y2
= a2(x2 + y2) + b2(x2 + y2)
= (x2 + y2) (a2 + b2)

(vi) ax + by + cx + bx + cy + ay
= ax + bx + cx + ay + by + cy (grouping)
= x(a + b + c) + y(a + b + c)
= (a + b + c) (x + y)

(vii) 49x2 – 70xy + 25y2
= (7x)2 – 2 × 7x × 5y + (5y)2
{∵ (a – b)2 = a2 – 2ab + b2}
= (7x – 5y)2

(viii) 4a2 + 12ab + 9b2
= (2a)22 + 2 × 2a × 3b + (3b)2
{∵ (a + b)2 = a2 + 2ab + b2}
= (2a + 3b)2

(ix) 49p2 – 36q2
= (7p)2 – (6q)2
= (7p + 6q) (7p – 6q)
{∵ a2 – b2 = (a + b) (a – b)}

(x) 100x3 – 25xy2
= 25x(x2 – y2) = 25x{(x)2 – (y)2}
= 25x(x + y) (x – y)

(xi) x2 – 2xy + y2 – z2
= (x – y)2 – (z)2
{∵ a2 -2ab + b2 = (a – b)2)
a2 – b2 = (a + b)(a – b)}
= (x – y + z)(x – y – z)

(xii) x8 – y8
= (x4)2 – (y4)2
{∵ a2 – b2 = (a + b)(a- b)}
= (x4 + y4)(x4 – y4)
= (x4 + y4) {(x2)2 – (y2)1}
= (x4 + y4) (x2 + y2) (x2 – y2)
= (x4 + y4 (x2 + y2) (x + y) (x – y)

(xiii) 12x3 – 14x2 – 10x
= 2x(6x2 – 7x – 5)
{∵ 6 × (-5) = -30
∴ -30 = -10 × 3
-7 = -10 + 3}
= 2x{6x2 + 3x – 10x – 5}
= 2x{3x(2x + 1) – 5(2x + 1)}
= 2x(2x + 1) (3x – 5)

(xiv) p2 – 10p + 21
= p2 – 3p – 7p + 21
{∵ 21 =-3 × (-7)
-10 = -3 – 7}
= p(p – 3) – 7(p – 3)
= (p – 3)(p – 7)

(xv) 2x2 – x – 6
= 2x2 – 4x + 3x – 6
{ ∵ -6 × 2 = -12
∴ -12 = -4 × 3
-1 = -4 + 3}
= 2x(x – 2) + 3(x – 2)
= (x – 2) (2x + 3)

(xvi) 6x2 – 5xy – 6y2
= 6x2 – 9xy + 4xy – 6y2
{∵ 6 × (-6) = -36
∴ – 36 = -9 × 4
– 5 = -9 + 4}
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y) (3x + 2y)

(xvii) x2 + 2xy – 99y2
= x2 + 11xy – 9xy – 99y2
{∵ -99 = -11 × 9
-2 = -11 + 9 }
= x(x + 11y) – 9y(x + 11y)
= (x + 11y) (x – 9y)

Question 3.
Divide as directed:
(i) 15(y + 3)(y2 – 16) ÷ 5(y2 – y – 12)
(ii) (3x3 – 6x2 – 24x) ÷ (x – 4) (x + 2)
(iii) (x4 – 81) ÷ (x3 + 3x2 + 9x + 27)
Solution:
(i) 15(y + 3)(y2 – 16) ÷ 5(y2 – y – 12)
y2 – 16 = (y)2 – (4)2
= (y + 4)(y – 4)
y2 – y – 12 = y2 – 4y + 3y – 12
= y(y – 4) + 3(y – 4)
= (y – 4)(y + 3)
Now, $$\frac{15(y+3)\left(y^{2}-16\right)}{5\left(y^{2}-y-12\right)}$$
$$\frac{15 \times(y+3)(y+4)(y-4)}{5(y-4)(y+3)}$$
= 3(y + 4)

(ii) (3x3 – 6x2 – 24x) ÷ (x – 4) (x + 2)
3x3 – 6x2 – 24x = 3x(x2 – 2x – 8)
= 3x{x2 – 4x + 2x – 8}
= 3x{x(x – 4) + 2(x – 4)}
= 3x(x – 4) (x + 2)
∴ $$\frac{3 x^{3}-6 x^{2}-24 x}{(x-4)(x+2)}=\frac{3 x(x-4)(x+2)}{(x-4)(x+2)}=3 x$$

(iii) (x4 – 81) ÷ (x3 + 3x2 + 9x + 27)
x4 – 81 = (x2)2 – (9)2 = (x2 + 9) (x2 – 9)
= (x2 + 9) {(x)2 – (3)2}
= (x2 + 9) (x + 3) (x – 3)
x3 + 3x2 + 9x + 27
= (x)2 + (x + 3) + 9 (x + 3)
= (x2 + 9) (x + 3)
Now, $$\frac{x^{4}-81}{x^{3}+3 x^{2}+9 x+27}$$
= $$\frac{\left(x^{2}+9\right)(x+3)(x-3)}{\left(x^{2}+9\right)(x+3)}$$
= x = -3