Question

Factorise : (i) x³+13x²+32x+20.​

Answer 1

X³+13X²+32X+20.

X³+13X²+32X+20.= x³ + 13x² + 32x + 20

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + + 20 (x + 1)

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + 1) + 20 (x + 1)= (x – 1) (x² + 12x + 20)

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + 1) + 20 (x + 1)= (x – 1) (x² + 12x + 20)= (x – 1) (x² + 10x + 2x + 20)

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + 1) + 20 (x + 1)= (x – 1) (x² + 12x + 20)= (x – 1) (x² + 10x + 2x + 20)= (x – 1) [x (x + 10) + 2 (x + 10)]

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + 1) + 20 (x + 1)= (x – 1) (x² + 12x + 20)= (x – 1) (x² + 10x + 2x + 20)= (x – 1) [x (x + 10) + 2 (x + 10)]= (x – 1) (x + 2) (x + 10)

X³+13X²+32X+20.= x³ + 13x² + 32x + 20= x³+ x² + 12x² + 12x + 20x + 20= x² (x + 1) + 12x (x + 1) + 20 (x + 1)= (x – 1) (x² + 12x + 20)= (x – 1) (x² + 10x + 2x + 20)= (x – 1) [x (x + 10) + 2 (x + 10)]= (x – 1) (x + 2) (x + 10)Therefore, the value of x³ + 13x² + 32x + 20 is (x – 1) (x + 2) (x + 10)