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

Jul 27, 2017

e.g.: ${x}^{3} + 3 {x}^{2} - 4 x - 12 = \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

#### 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:

$\left(a x + b\right) \left(c {x}^{2} + d\right) = a c {x}^{3} + b c {x}^{2} + a \mathrm{dx} + b d$

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

${x}^{3} + 3 {x}^{2} - 4 x - 12 = \left({x}^{3} + 3 {x}^{2}\right) - \left(4 x + 12\right)$

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

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

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

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:

$\left(a x + b\right) \left(c x - d\right) \left(c x + d\right) = \left(a x + b\right) \left({c}^{2} {x}^{2} - {d}^{2}\right)$

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

Jul 28, 2017

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.

$14 {x}^{5} - 49 {x}^{3}$

There is a common factor of $7 {x}^{3}$.

The other factor is $2 {x}^{2} - 7$

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

We could use the following:

$5$ to get $14 {x}^{5} - 49 {x}^{3} + 5 \left(2 {x}^{2} - 7\right) = 14 {x}^{5} - 49 {x}^{3} + 10 {x}^{2} - 35$
Which factors as $\left(7 {x}^{3} + 5\right) \left(2 {x}^{2} - 7\right)$

Neither of these factor can be factored over the integers.