Question

Factorise, using remainder theorem:
2x^3 + 9x^2 + 7x - 6
​

Answer 1

Answer:

hope this helps you

Step-by-step explanation:

Answer 2

$put \: x = - 2$

$2 {( - 2)}^{3} + 9 {( - 2)}^{2} + 7( - 2) - 6$

$2 \times ( - 8) + 9 \times 4 - 14 - 6$

$- 16 + 36 - 20$

$0$

$so \:( x + 2) \: is \: a \: factor.$

$\frac{2 {x}^{3} + 9 {x}^{2} + 7x - 6 }{x + 2} = 2x {}^{2} + 5x - 3$

$2 {x}^{2} + 5x - 3$

$2 {x}^{2} + 6x - x - 3$

$2x(x + 3) - 1(x + 3)$

$(x + 3)(2x - 1)$

$2 {x}^{3} + 9 {x}^{2} + 7x - 6 = (x + 2)(x + 3)(2x - 1)$