Question

If (x + 3) completely divides (3x2 + kx – 6), then the value of k is​

Answer 1

Step-by-step explanation:

hope this helps you and look the answer once

Answer 2

Given:

(x+3) divides (3x^2+kx-6)  

Find:

We have to calculate the value of k

step-by-step explanation:

Since (x+3) divides (3x^2+kx-6) completely, the remainder must be zero.

∴ $\displaystyle\frac{3x^2+kx-6}{x+3}=0$

multiply N & D by (x-3)

∴  $\displaystyle\frac{3x^2+kx-6}{x+3}\ast\frac{x-3}{x-3}=0$

∴ $\displaystyle\frac{\left ( 3x^2+kx-6 \right ) \ast \left ( x+3 \right )}{x^2-9}=0$

∴ $x\neq \pm 3$

Now, let x= 0

∴ (x+3)= (0+3) = 3 solve equation (3x^2+kx-6) completely.

∴ 3x^2+kx-6 = 0

∴ 3(3)^2+k(3)-6 = 0

∴ 27+3k-6 = 0

∴ 3k = -21

∴ k = -7