# How do you factor y=-2x^3-6x^2+8x+24 ?

Jun 2, 2018

$y = - 2 \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

#### Explanation:

Given:

$y = - 2 {x}^{3} - 6 {x}^{2} + 8 x + 24$

Note that the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic will factor by grouping. We can first separate out the common factor $- 2$ rather than tidying it up later...

$y = - 2 {x}^{3} - 6 {x}^{2} + 8 x + 24$

$\textcolor{w h i t e}{y} = - 2 \left({x}^{3} + 3 {x}^{2} - 4 x - 12\right)$

$\textcolor{w h i t e}{y} = - 2 \left(\left({x}^{3} + 3 {x}^{2}\right) - \left(4 x + 12\right)\right)$

$\textcolor{w h i t e}{y} = - 2 \left({x}^{2} \left(x + 3\right) - 4 \left(x + 3\right)\right)$

$\textcolor{w h i t e}{y} = - 2 \left({x}^{2} - 4\right) \left(x + 3\right)$

$\textcolor{w h i t e}{y} = - 2 \left({x}^{2} - {2}^{2}\right) \left(x + 3\right)$

$\textcolor{w h i t e}{y} = - 2 \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

Jun 2, 2018

Inspection of coefficients followed by informed trial-and-error

#### Explanation:

First ask whether the coefficients have any common numerical factors. In this case they do - a factor of 2. In fact, we'll take out a factor of -2 because it is almost always useful to make the leading term positive. So $- \frac{1}{2} y = {x}^{3} + 3 {x}^{2} - 4 x - 12$. We could have left this in, but it makes life slightly simpler to take it out.

This is a cubic, and so we know that it can be expressed as three linear factors corresponding to the roots of the equation $y = 0$:
$- \frac{1}{2} y = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$. Now, there is no guarantee that these roots will be at all 'nice' (in general, it is quite likely that two of them won't in fact be real numbers). However, the phrasing of the question suggests that we will have three real 'nice' roots to find here.

It is possible to solve the cubic by radicals to reliably obtain an answer, but this is a long and messy process, one far beyond the scope of the question as asked. If we can find one root, then we can factor that out of the cubic to make a quadratic, which we know how to solve relatively easily to obtain the other two roots.

In this situation we proceed with a mix of intuition and trial-and-error; it's something one gets better at the more one tackles such problems. Let's try some simple trial solutions for $x$... Is 0 a root? No - the RHS equals -12 when $x = 0$.
If we look at the constant in the equation, this gives us clues as to what numbers to look at - what numbers divide into 12? It must be formed from multiplying the roots ${x}_{1}$, ${x}_{2}$, ${x}_{3}$ together. Candidate numbers: -12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 12 - so we potentially have a few to check, but not that many. How about 1? No, -12 RHS again. But now we find it: 2 is a root: ${2}^{3} + 3 \cdot {2}^{2} - 4 \cdot 2 - 12 = 8 + 12 - 8 - 12 = 0$. So $x - 2$ is a factor of our cubic.

We then factor out $x - 2$ from the cubic by long division. To long divide one algebraic term by another, divide the leading term of one by another and then subtract off the result multiplied by the whole dividing factor in order to get the remainder. Then repeat with the remainder as many times as necessary to reach zero. I'll show for this example:

To divide $x - 2$ into ${x}^{3} + 3 {x}^{2} - 4 x - 12$:

Divide one leading term by another: ${x}^{3} / x = {x}^{2}$.
Subtract from the cubic ${x}^{2} \left(x - 2\right) = {x}^{3} - 2 {x}^{2}$ (NB this removes the current leading term):
$\left({x}^{3} + 3 {x}^{2} - 4 x - 12\right) - \left({x}^{3} - 2 {x}^{2}\right) = 5 {x}^{2} - 4 x - 12$

This is our remainder, which now has no ${x}^{3}$ term. Repeat with the quadratic term to get: $6 x - 12$. We can now see immediately (or by another long division step if we don't spot it) that $x - 2$ divides into this exactly: 6. We collect up the results from each long division step to get
$- \frac{1}{2} y = \left(x - 2\right) \left({x}^{2} + 5 x + 6\right)$
(Multiply this back out to check that we get the original cubic back - it's very easy to make mistakes)

Now we are nearly done. It is a much easier job to deduce factorisation of a quadratic from its coefficients than it is for a cubic. Again look at the constant term, this time in the quadratic: 6. This is the product of two roots (as discussed above), and there are only two candidate pairs: $\left(1 , 6\right)$ and $\left(2 , 3\right)$, neglecting positive or negative signs. We also need these two to sum to the $x$ coefficient of 5 - and now we are done - it must be 2 and 3.

So
$- \frac{1}{2} y = \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$
or rather
$y = - 2 \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

The cubic $y$ has roots at $x = - 3 , - 2 , + 2$.