NCERT Solutions
Class 9 Maths
Polynomials
Ex.2.4 Q.5
Factorise:
(1) x3 – 2x2 – x + 2
(2) x3 – 3x2 – 9x – 5
(3) x3 + 13x2 + 32x + 20
(4) 2y3 + y2 – 2y – 1
(1) Let p(x) = x3 – 2x2 – x + 2
Put x = 1, we get
p (1) = 13 – 2 × 12 – 1 + 2
= 1 – 2 – 1 + 2
= 3 – 3
= 0
Hence, x = 1 i.e. (x – 1) is a factor of p(x).
Now, divide x3 – 2x2 – x + 2 by x – 1, we get
Now, x3 – 2x2 – x + 2 = (x2 – 3x + 2) (x + 1)
= (x2 – x – 2x + 2) (x + 1)
= {x (x - 1) – 2(x - 1)} (x + 1)
= (x - 1) (x - 2) (x + 1)
(2) Let p(x) = x3 – 3x2 – 9x – 5
Put x = 1, we get
p (1) = 13 – 3 × 12 – 9 × 1 – 5
= 1 – 3 – 9 – 5
= 1 – 17
= -16 ≠ 0
Again, put x = -1, we get
p (1) = (-1)3 – 3 × (-1)2 – 9 × (-1) – 5
= -1 – 3 + 9 – 5
= -9 + 9
= 0
Hence, x = -1 i.e. (x + 1) is a factor of p(x)
Now, divide x3 – 3x2 – 9x – 5 by (x + 1), we get
Now, x3 – 3x2 – 9x – 5 = (x + 1) (x2 – 4x - 5)
= (x + 1) (x2 – 5x + x - 5)
= (x + 1) {x (x – 5) + 1(x - 5)}
= (x + 1) (x + 1) (x - 5)
= (x + 1)2 (x - 5)
(3) Let p(x) = x3 + 13x2 + 32x + 20
Put x = -1, we get
p (1) = (-1)3 + 13 × (-1)2 + 32 × (-1) + 20
= -1 + 13 - 32 + 20
= -33 + 33
= 0
Hence, x = -1 i.e. (x + 1) is a factor of p(x)
Now, divide x3 + 13x2 + 32x + 20 by (x + 1), we get
Now, x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20)
= (x + 1) (x2 + 2x + 10x + 20)
= (x + 1) {x (x + 2) + 10(x + 2)}
= (x + 1) (x + 2) (x + 10)
(4) Let p(y) = 2y3 + y2 – 2y – 1
Put y = 1, we get
p (1) = 2 × (1)3 - (1)2 - 2 × (-1) - 1
= 2 - 1 + 2 - 1
= -3 + 3
= 0
Hence, y = 1 i.e. (y - 1) is a factor of p(y)
Now, divide (2y3 + y2 – 2y – 1) by (y - 1), we get
Again, 2y3 + y2 – 2y – 1 = (y - 1) (2y2 + 3y + 1)
= (y - 1) (2y2 + 2y + y + 1)
= (y - 1) {2y (y + 1) + 1(y + 1)}
= (y - 1) (y + 1) (2y + 1)