Question

factorise 2x²+x³-14x²-19x-6

Answer 1

Sol:
Let f(x) = 2x4 + x3 - 14x2 - 19x – 6

f(-1) = 2(-1)4 + (-1)3 - 14(-1)2 - 19(-1) - 6

= 2 - 1 - 14 + 19 - 6

= 0

So, -1 is the zero of the polynomial and (x + 1) is the factor.

f(-2) = 2(-2)4 + (-2)3 - 14(-2)2 - 19(-2) - 6

= 32 - 8 - 56 + 38 - 6

= 0

So, -2 is the zero of the polynomial and (x + 2) is the factor.

(x + 1)(x + 2) = (x2 + 3x + 2)




The quotient when (2x4 + x3 - 14x2 - 19x – 6) is divided by (x2 + 3x + 2) is (2x2 - 5 x - 3).

2x2 - 5 x - 3 = 2x2 - 6x + x - 3
= 2x(x - 3) + 1(x - 3)
= (2x + 1)(x - 3)
Therefore, the factors of the polynomial are (x + 1)(x + 2)(2x + 1)(x - 3).