Question

How do you factor the trinomial `20x² + 47x − 24`?

Answer 1

`20x^2+47x-24 = 20(x+47/40-sqrt(4129)/40)(x+47/40+sqrt(4129)/40)`

Explanation:

Use the quadratic formula to find the zeros and hence the linear factors...

`20x^2+47x-24`

is in the form:

`ax^2+bx+c`

with `a=20`, `b=47`, `c=-24`

Let us take a quick look at the discriminant to decide how we should solve this:

`Delta = b^2-4ac = 47^2-4(20)(-24) = 2209+1920 = 4129`

Since `Delta > 0` this quadratic has two distinct Real zeros and will factor as a product of two linear terms with Real coefficients.

Note that `4129` is not a perfect square - it is a prime number - so the zeros are irrational. That means that we will not find any factors using an AC method or similar. We could use completing the square, but the coefficients `20` and `47` make that a little painful.

So let us use the quadratic formula...

The zeros given are by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (-47+-sqrt(4129))/40`

`color(white)(x) = -47/40+-sqrt(4129)/40`

Hence we can factor the given quadratic as:

`20x^2+47x-24 = 20(x+47/40-sqrt(4129)/40)(x+47/40+sqrt(4129)/40)`