Factoring the Cubic Equation $2x^3 + x^2 + kx + 6$

Question

Find the value of $k$ such that $x + 2$ is a factor of $2x^3 + x^2 + k x + 6$ then factorise completely.

Answer 1

Hint: Let $2x^3+x^2+kx+6 = (x+3)P(x)$ for some quadratic $P(x)$. Now, plug in $x = -3$.

Answer 2

If $(x+3)$ is a factor, this function should evaluate to $0$ for $x = -3$.

Fill in $-3$ for $x$ and have $-54 + 9 - 3k + 6 = 0$, and find $k = 13$

Did I say $k = 13$, that was a bad copying to or from my notes?

$-54 + 9 + 6 = -39 = 3k $

so $k$ should have been $-13$.

Answer 3

Since, $x+3$ is a factor of the equation, then $f(-3)=0$

submit $x=-3$ to get $k$.

Answer 4

By long division we can obtain polynomials $q(x)$ and $r(x)$ such that : $2x^3+x^2+kx+6 = (x+3)q(x) + r(x)$

$r(x) = 0$ since $(x+3)$ is a factor :

$2x^3+x^2+kx+6 = (x+3)q(x) + 0$

This is an equality and must work for ALL values of x, including x = -3 :

$2(-3)^3+(-3)^2+k(-3)+6 = (-3+3)q(-3)$

$2(-3)^3+(-3)^2+k(-3)+6 = 0$

$k$ can be solved

You may refer factor theorem for more info