Question

Find the value of $a$ if $(x-5)$ is a factor of $(x^3-3x^2+ax-10)$.

Answer 1

The Factor Theorem states that if $m$ is the root of any polynomial $p\left(x\right)$ that is if $p\left(m\right)=0$, then $(x-a)$ is the factor of the polynomial $p\left(x\right)$.

Let $p\left(x\right)=x^3-3x^2+ax-10$ and it is given that $(x-5)$ is a factor of $p\left(x\right)$, therefore, by factor theorem $p\left(5\right)=0$.

Let us first find $p\left(5\right)$ as follows:

$p\left(5\right)=5^3-(3\times 5^2)+(a\times 5)-10$

$=125-(3\times 25)+5a-10$

$=125-75+5a-10$

$=5a+125-85=5a+40$

Now equate $p\left(5\right)=0$ as shown below:

$5a+40=0\Rightarrow 5a=-40\Rightarrow a=-\frac{40}{5}\Rightarrow a=-8$

Hence, $a=-8$.