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

##### 2 Answers

#### Answer:

#### Explanation:

Given:

#y=-2x^3-6x^2+8x+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

#y = -2x^3-6x^2+8x+24#

#color(white)(y) = -2(x^3+3x^2-4x-12)#

#color(white)(y) = -2((x^3+3x^2)-(4x+12))#

#color(white)(y) = -2(x^2(x+3)-4(x+3))#

#color(white)(y) = -2(x^2-4)(x+3)#

#color(white)(y) = -2(x^2-2^2)(x+3)#

#color(white)(y) = -2(x-2)(x+2)(x+3)#

#### Answer:

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

This is a cubic, and so we know that it can be expressed as three linear factors corresponding to the roots of the equation

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

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

We then factor out

To divide

Divide one leading term by another:

Subtract from the cubic

This is our remainder, which now has no

(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:

So

or rather

The cubic