Question

How do you factor `x^ { 2} + 6 x - 27= 0`?

Answer 1

`(x - 3) ( x + 9)`

Explanation:

To factor `x^2 + 6x - 27`

Find to numbers whose product is -27 and whose sum is 6. For this it is -3 and 9.

`-3 xx 9 = - 27` and `-3 + 9 = 6`.

Rewrite the equation as:

`x^2 - 3x + 9x -27`

Factor the first 2 terms:

`x(x - 3)`

Factor the last 2 terms:

`9(x - 3)`

So now we have:

`x(x -3) + 9(x -3)`

Notice we can bracket off this and factor out the expressions in parenthesis:

`[x(x-3) + 9(x-3)]`

`(x-3)[ x + 9 ]`

These are our required factors:

`(x - 3) ( x + 9)`

Answer 2

`(x+9)(x-3) = 0`

Explanation:

Given:

`x^2+6x-27`

Here are a couple of methods:

Method 1 - Fishing for factors

Find a pair of factors of `27` with difference `6`.

The pair `9, 3` works in that `9*3 = 27` and `9-3=6`

So we find:

`x^2+6x-27 = (x+9)(x-3)`

Method 2 - Completing the square

`x^2+6x-27 = x^2+6x+9-36`

`color(white)(x^2+6x-27) = (x+3)^2-6^2`

`color(white)(x^2+6x-27) = ((x+3)-6)((x+3)+6)`

`color(white)(x^2+6x-27) = (x-3)(x+9)`