Given that an expression $2x^3+px^2-8x+q$ is exactly divisible by $2x^2-7x+6$, determine the value of $p$ and $q$.

Question

Given that an expression $2x^3+px^2-8x+q$ is exactly divisible by $2x^2-7x+6$, determine the value of $p$ and $q$.

Answer 1

Since $\;2x^2-7x+6=2(x-2)\left(x-\frac32\right)\;$, we get that both $\;2,\,\frac32\;$ are roots of the cubic, so:

$$\begin{cases}2\cdot2^3+p\cdot2^2-8\cdot2+q=0\\{}\\2\cdot\left(\frac32\right)^3+p\cdot\left(\frac32\right)^2-8\cdot\frac32\cdot+q=0\end{cases}$$

Now solve the easy linear non-homogeneous system above.

Answer 2

Observe that $2x^2-7x+6 = (2x-3)(x-2)$

If $2x^3+px^2-8x+q$ is exactly divisible by $2x^2-7x+6 = (2x-3)(x-2)$ , then $(2x-3)$ and $(x-2)$ must be its factors.

So when $x=2$ and $x=\frac{3}{2}$ , $2x^3+px^2-8x+q$ should be zero.

Answer 3

For some $a, b \in \mathbb Z$, we know that: \begin{align*} 2x^3 + px^2 - 8x + q &= (ax + b)(2x^2 - 7x + 6) \\ &= (2a)x^3 + (-7a + 2b)x^2 + (6a - 7b)x + (6b) \end{align*} Comparing the $x^3$ coefficients, we see that $2a = 2 \iff a = 1$. Comparing the $x$ coefficients, we see that $6 - 7b = -8 \iff b = 2$. Substituting these values and comparing the remaining coefficients, we conclude that $p = -3$ and $q = 12$.