Question

How do you factor `18r ^ { 2} + 51r + 35= 0`?

Answer 1

`0 = 18r^2+51r+35 = (6r+7)(3r+5)`

has roots: `" "r = -7/6" "` and `" "r = -5/3`

Explanation:

Complete the square, premultiplying by `8` first to avoid fractions...

`8(18r^2+51r+35) = 144r^2+408r+280`

`color(white)(8(18r^2+51r+35)) = (12r)^2+2(12r)(17)+(17)^2-9`

`color(white)(8(18r^2+51r+35)) = (12r+17)^2-3^2`

`color(white)(8(18r^2+51r+35)) = ((12r+17)-3)((12r+17)+3)`

`color(white)(8(18r^2+51r+35)) = (12r+14)(12r+20)`

`color(white)(8(18r^2+51r+35)) = (2(6r+7))(4(3r+5))`

`color(white)(8(18r^2+51r+35)) = 8(6r+7)(3r+5)`

Dividing both ends by `8`, we then find:

`18r^2+51r+35 = (6r+7)(3r+5)`

which has zeros: `" "r=-7/6" "` and `" "r=-5/3`

Answer 2

`0 = 18r^2+30r+21r+35 = (6r+7)(3r+5)`

Hence roots:

`r = -7/6" "` or `" "r = -5/3`

Explanation:

Given:

`18r^2+51r+35 = 0`

Use an AC method to factorise:

Look for a pair of factors of `AC = 18*35 = 630` with sum `B = 51`

Note that `51` is divisible by `3` and there are two prime factors `3` in `630`. So `51` must be a sum of two factors both divisible by `3`.

Putting those factors to one side for a moment, let us find a pair of factors of `630/9 = 70` with sum `51/3 = 17`.

The pair `10, 7` works. So multiplying both by `3` will give us the original pair of factors we wanted: `30`, `21`.

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

`0 = 18r^2+51r+35`

`color(white)(0) = 18r^2+30r+21r+35`

`color(white)(0) = (18r^2+30r)+(21r+35)`

`color(white)(0) = 6r(3r+5)+7(3r+5)`

`color(white)(0) = (6r+7)(3r+5)`