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

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Answer 1 · 2016-02-09T04:56:10

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

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

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

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

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

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

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

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

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

and hence next factor is $(x - 5)$ $(x-5)$

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

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

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

How do you factor y=x3−4x2−17x+60y=x^3-4x^2-17x+60 ?