Question

How do you factor `8x ^ { 3} + 7x ^ { 2} - 32x + 28` completely?

Answer 1

Cannot be factored. (See explanation for alternate problem which was possibly the question intended.)

Explanation:

The problem as stated is irreducible and cannot be factored as written. However, it's incredibly close to a problem which CAN indeed be factored, assuming there's a typo and the problem mean for either the `-32x` term to be positive, or the `28` term to be negative. I'm going to assume the `28` term should have been negative arbitrarily and provide a solution to this problem:

Factor `8x^3+7x^2-32x-28`

When you come across factoring problems like this, there are many methods you can use. One thing to immediately look for is to see if there's any pattern in the coefficients than can be exploited in factoring.

For instance, the coefficients are 8, 7, -32, and -28. Can those be grouped in some manner where a number and a multiple of that number are together? If so, can you group them so that each pair has the same multiplier?

If we rearrange the terms such that we put the `8x^3` and the `-32x` together, you can immediately see than -32 is a multiple of 8. (It's equal to `8*(-4)`). At the same time, if we put the `7x^2` and the `-28` together, we can see that -28 is also a multiple of 7 - by -4 as well.

That suggests we can use a practice known as factoring by grouping. Let's rewrite the terms to group them together as we talked about:

`8x ^ { 3} + 7x ^ { 2} - 32x - 28`

`= 8x^3 - 32x + 7x^2 - 28`

`= 8x(x^2 - 4) + 7(x^2 - 4)`

Notice how both groupings have factored in a way that leaves behind `(x^2-4)`? That means we can "switch" our perspective and factor out `(x^2-4)` from the last line we wrote above:

`= (x^2-4)(8x + 7)`

Lastly, we can recognize that `(x^2-4)` is the "difference of two squares" common expression, which has a well-known factoring:

`= (x+2)(x-2)(8x+7)`

Special note

Had I chosen to show the problem with a positive `32x` term as opposed to the `-28` term, we could have proceeded mostly the same, except we would have ended up with an irreducible quadratic term at the front.