Question

How do you factor `w^2 - 3w + 28`?

Answer 1

You can only factor this expression using complex roots found from the quadratic formula.

Explanation:

When you try to factor this expression, it becomes clear that you cannot easily break it into a statement like

`(w+?)(w+?)`

Usually, you would experiment with numbers using FOIL in reverse. The F in First, gives you `wxxw=w^2`. The O and I for Outer and Inner would have to be added together to give you the `-3` and multiplied together to give you the `28`. But no matter how many combinations you try, it doesn't work, because the signs don't work out nicely.

You can prove an "easy" answer doesn't work in two ways. First, you could graph it and quickly see the expression does not hit the horizontal `x`-axis, and therefore does not have any real roots to factor. That graph looks like this:

graph{y=x^2-3x+28 [-20,20,-10,160]}

Second, you can set the expression equal to zero and use the quadratic formula to solve for `w` and get the roots of the equation.

`w^2-3w+28=0`
`w=(-(-3)+-sqrt((-3)^2-4*1*28))/(2*1)`
`=(3+-sqrt(9-112))/2`
`=(3+-sqrt(-103))/2` [note that `sqrt(-1)=i`]
`=3/2+-(sqrt(103)/2)i`

So `w=3/2+(sqrt(103)/2)i` and `w=3/2-(sqrt(103)/2)i` are two separate, complex roots to the equation. But this means you can manipulate the roots and multiply them together. The factorization would be:

`w^2-3w+28=(w-3/2-sqrt(103)/2i)(w-3/2+sqrt(103)/2i)`.

It's a bit of a nightmare, but you could multiply these two factors to prove to yourself they do indeed equal the original expression. Finis!