# Can y=x^3-3x^2-10x  be factored? If so what are the factors ?

Jul 11, 2017

$y = x \left(x - 5\right) \left(x + 2\right)$

#### Explanation:

Take x out as a factor first to give:

$y = x \left({x}^{2} - 3 x - 10\right)$

Then factorise the quadratic in the brackets to give the answer above.

Jul 11, 2017

$y = x \left(x - 5\right) \left(x + 2\right)$

#### Explanation:

Inspection reveals that all three terms have $x$ common in the given polynomial
$y = {x}^{3} - 3 {x}^{2} - 10 x$ .....(1)

LHS can be written as
$x \left({x}^{2} - 3 x - 10\right)$

The quadratic in the brackets can be factorized by splitting the middle term
$\left({x}^{2} - 3 x - 10\right)$
$\implies \left({x}^{2} - 5 x + 2 x - 10\right)$
$\implies \left(x \left[x - 5\right] + 2 \left[x - 5\right]\right)$
$\implies \left(\left[x - 5\right] \left[x + 2\right]\right)$

Hence factors are

$y = x \left(x - 5\right) \left(x + 2\right)$

Jul 12, 2017

$y = x \left(x - 5\right) \left(x + 2\right)$

#### Explanation:

Take out the common factor of $x$ first:

$y = {x}^{3} - 3 {x}^{2} - 10 x$

$y = x \left({x}^{2} - 3 x - 10\right)$

To factorise the quadratic trinomial, find factors of $10$ which subtract to give $3$

$15 \times 2 = 10 \mathmr{and} 5 - 2 = 3$ so these are the factors we need.

Their signs must be different, but there must be more negatives.
This gives:

$y = x \left(x - 5\right) \left(x + 2\right)$