How do you factor y=x^3-4x^2-17x+60 ?

Feb 9, 2016

The factors are $y$ =$\left(x - 3\right)$ * $\left(x - 5\right)$ * $\left(x + 4\right)$

Explanation:

Assume that the factors are $\left(x - a\right)$, $\left(x - b\right)$ and $\left(x - c\right)$. Two things arise from this.

(i) if $x$ is either equal to $a$ or $b$ or $c$, $y = 0$ and

(ii) $a \cdot b \cdot c = - 60$, which means $a$ or $b$ and $c$ are factors of 60.

So, which factor of 60 (such as 2, 3, 5, 6, 10 etc.) makes $y = 0$.

Hit and trial will show that $x = 3$ returns $y = 0$

and hence $\left(x - 3\right)$ is a factor of $y$.

When we divide $y$ by $\left(x - 3\right)$, we get ${x}^{2} - x - 20$

Similarly $x = 5$ makes ${x}^{2} - x - 20$ equal to zero

and hence next factor is $\left(x - 5\right)$

Dividing ${x}^{2} - x - 20$ by $\left(x - 5\right)$, we get $\left(x + 4\right)$

Hence the factors of $y = {x}^{3} - 4 {x}^{2} - 17 x + 60$ are

$y$ =$\left(x - 3\right)$ * $\left(x - 5\right)$ * $\left(x + 4\right)$