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The Factor Theorem

The Factor Theorem

The linear expression $ax+b$ is a factor of the polynomial $P(x)$ if and only if $P\left(-\frac{b}{a}\right) = 0$

The factor theorem 1

Determine the value of $a$ for which $(3x - 4)$ is a factor of the polynomial $P(x) = 2x^3 + ax + 12$
solution - press button to display
The factor theorem say $ax + b$ is a factor of $P(x)$ if and only if $P\left(\frac{-b}{a}\right) = 0$ Therefore let us solve $P\left(\frac{4}{3}\right) =0$ $$P \left(\frac{4}{3}\right) = 2\left(\frac{4}{3}\right)^3 + a\left(\frac{4}{3}\right) + 12 = \frac{128}{27} + \frac{4a}{3} + 12$$ $$ \frac{452}{27} + \frac{4a}{3} = 0 \Leftrightarrow a = -\frac{113}{9} $$ Therefore the polynomial $P(x) = 2x^3 - \frac{113}{9}x + 12$ has $3x-4$ as a factor.

The factor theorem 2

Given that $(x-2)$ is a factor of both $f(x)$ and $f'(x)$. Determine the value of $a$ if $f(x)$ can be expressed as $$ f(x) = x^3 - (5+a)x^2 + (6+5a)x - 6a $$

Hence factorise $f(x)$ completely.

solution - press button to display
$$ f'(x) = 3x^2 - 2(5+a)x + (6+5a) $$ Since $f'(2) =0$, we get $$ \begin{align} 0 &= 12 - 20 - 4a + 6 + 5a \\ &= -2 + a\\ a &= 2 \end{align} $$

It follows that $f(x) = x^3 - 7x^2 + 16x - 12$, but since $(x-2)$ is a factor of both $f(x)$ and $f'(x)$, $x=2$ must be a repeated root of $f(x) = 0$, so $f(x) = (x-2)^2(x-k)$. It follows that $(-2)^2(-k) = -12$ hence $k = 3$. This gives $$ f(x)= (x-2)^2(x-3) $$