Question

How do you write a four term polynomial that can be factored by grouping?

Answer 1

e.g.: `x^3+3x^2-4x-12 = (x-2)(x+2)(x+3)`

Explanation:

We can construct such a polynomial by reversing the factorisation.

For example, to get an example cubic, pick any numbers `a, b, c, d` and multiply out:

`(ax+b)(cx^2+d) = acx^3+bcx^2+adx+bd`

So picking `a=1`, `b=3`, `c=1`, `d=-4`, we would get:

`x^3+3x^2-4x-12 = (x^3+3x^2)-(4x+12)`

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

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

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

Note that in this example, I chose `c=1` and `d=-4=-2^2`, so the quadratic factor factored cleanly.

If we want clean factorisations like this, we could use the alternative pattern:

`(ax+b)(cx-d)(cx+d) = (ax+b)(c^2x^2-d^2)`

`color(white)((ax+b)(cx-d)(cx+d)) = ac^2x^3+bc^2x^2-ad^2x-bd^2`

Answer 2

Here is one way I've done it when writing exam questions.

Explanation:

Make sure that the first two terms have a common factor. Not a constant common factor, but something involving the variable.

Warning: This may lead to a polynomial that cannot be factored completely over the integers.

For example we might start with

`14x^5-49x^3`

There is a common factor of `7x^3`.

The other factor is `2x^2-7`

For our third and fourth terms we need a multiple of `2x^2-7` In order to avoid a common factor of `x` use a constant.

We could use the following:

`5` to get `14x^5-49x^3 + 5(2x^2-7) = 14x^5-49x^3+10x^2-35`
Which factors as `(7x^3+5)(2x^2-7)`

Neither of these factor can be factored over the integers.