Show that $a = 1$ if $(x - 3)$ is a factor of $x^3 - 8x^2 + ax + 42$ and $a$ is an integer

Question

$(x - 3)$ is a factor of $x^3 - 8x^2 + ax + 42$ where $a$ is an integer

Show that $a = 1$

I don't know where to start with this... Any insight?

Answer 1

As $x - 3$ is a factor of the given equation, and since $x - 3 = 0$ when $x = 3$, put $x = 3$ in the equation:

$(3)^3 - 8(3)^2 + a(3) + 42 = 0$

$27 - 72 + 3a + 42 = 0$

$3a - 3 = 0$

$3a = 3$

$a = 1$

Answer 2

Hint: If $x-3$ is a factor of $x^3-8x^2+ax+42$, then $x^3-8x^2+ax+42$ can be written as $(x-3)$ times something. Importantly, this means that when $x=3$, $x^3-8x^2+ax+42=0$. Use this to find $a$.

Answer 3

Hint:

The property ‘$x-3$’ is a factor of the polynomial $p(x)$ simply means $p(3)=0$, since the remainder of the division of $p(x)$ by $x-\alpha\;$ is equal to $p(\alpha)$.

Answer 4

You may recall in the past that you would find roots of a polynomial to help factor it; e.g. $3$ is a root of the polynomial $f(x)$ if and only if $(x-3)$ is one of the factors of $f(x)$.

What you are overlooking is that you can use this same fact in other ways. Rather than using roots to find a factor, use the factor to find a root.

If $(x-3)$ is a factor of $x^3 - 8x^2 + ax + 42$, then $3$ is a root of $x^3 - 8x^2 + ax + 42$.

If $3$ is a root of $x^3 - 8x^2 + ax + 42$, then $3^3 - 8 \cdot 3^2 + a \cdot 3 + 42 = 0$.

This is an equation you can solve for $a$.