Factor Theorem (Finding values of a and b)

Question

Question: The polynomial $p(x) = 2x^3 - ax^2 + bx + 48$ has $(x-4)$ as a repeated factor, find the values of $a$ and $b$.

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What I have attempted

if $x-4$ is a factor then $x = 4$ is a root

then $$ p(4) = 2(4)^3 -a(4)^2 + b(4) + 48 = 0 $$

then $$ -16a + 4b = -176 $$

then I have 2 equations which are the same thing so I cannot find a and b?

My teacher said I could find the derivative of $p(x)$ and then plug in $x=4$ but why does that work? I'm confused..

Answer 1

The OP asked why a repeated root is a factor of the derivative. Let $p$ be a polynomial with root $r$. By the factor theorem $(x-r)^2 \mid p$ so there is some polynomial $q$ with $p = q (x-r)$ and $x-r \mid q$. Then $$\frac{dp}{dx} = \frac{d}{dx} q (x-r) = (x-r)\frac{dq}{dx} + q\frac{d}{dx} (x-r).$$ As $x-r \mid q$, the rightmost expression is clearly divisible by $x-r$.

In fact, we didn't even use that $x-r$ and $p$ are polynomials. If $f$ is any function of $x$ and $p$ is any function of $x$ divisible by $f^2$ then $f \mid \frac{dp}{dx}$ by the same argument.

Answer 2

$p'(x) = 6x^2 - 2ax + b$.

Since $(x-4)$ is a repeated factor, we know that $(x-4)$ is a factor of the derivative. Hence $p'(4)= 0$.

Thus $p'(4) = 96 - 8a + b = 0$.

The rest is just algebra.