Question

How do you factor `9x^2 +18x + 8`?

Answer 1

`9x^2+18x+8 = (3x+2)(3x+4)`

Explanation:

The quadratic:

`9x^2+18x+8`

is in the form:

`ax^2+bx+c`

with `a=9`, `b=18` and `c=8`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = color(blue)(18)^2-4(color(blue)(9))(color(blue)(8)) = 324 - 288 = 36 = 6^2`

Since this is positive and a perfect square, this quadratic will factor with integer coefficients.

Let's use an AC method:

Find a pair of factors of `AC = 9*8 = 72` with sum `B=18`. (We look for a sum rather than a difference since the sign of `c` is positive).

The pair `12, 6` works in that `12*6=72` and `12+6=18`.

Use this pair to split the middle term and factor by grouping:

`9x^2+18x+8 = (9x^2+12x)+(6x+8)`

`color(white)(9x^2+18x+8) = 3x(3x+4)+2(3x+4)`

`color(white)(9x^2+18x+8) = (3x+2)(3x+4)`

`color(white)()`
We can find the same factorisation by completing the square:

Note that `9x^2 = (3x)^2` and `18x = 2(3x)(3)`.

Hence we find:

`9x^2+18x+8 = (3x)^2+2(3x)(3)+3^2-1`

`color(white)(9x^2+18x+8) = (3x+3)^2-1^2`

`color(white)(9x^2+18x+8) = ((3x+3)-1)((3x+3)+1)`

`color(white)(9x^2+18x+8) = (3x+2)(3x+4)`