Question

How do you factor `80x^2+68x+12`?

Answer 1

`80x^2+68x+12 = 4(4x+1)(5x+3)`

Explanation:

Given:

`80x^2+68x+12`

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this with `a=(40x+17)` and `b=7`

First let us multiply the whole expression by `20`, complete the square, then divide by `20` at the end:

`20(80x^2+68x+12) = 1600x^2+1360x+240`

`color(white)(20(80x^2+68x+12)) = (40x)^2+2(40x)(17)+17^2-49`

`color(white)(20(80x^2+68x+12)) = (40x+17)^2-7^2`

`color(white)(20(80x^2+68x+12)) = ((40x+17)-7)((40x+17)+7)`

`color(white)(20(80x^2+68x+12)) = (40x+10)(40x+24)`

`color(white)(20(80x^2+68x+12)) = 10(4x+1)(8)(5x+3)`

`color(white)(20(80x^2+68x+12)) = 20*4(4x+1)(5x+3)`

Then dividing both ends by `20` we get:

`80x^2+68x+12 = 4(4x+1)(5x+3)`

`color(white)()`
Why did I choose to multiply by `20`?

I wanted a multiplier that would make the leading term into a perfect square and give the middle term enough powers of `2` so that we would not end up with fractions when divided by the square root of resulting leading term and by `2`.

The multiplier had to be a multiple of `5`, to match the existing factor `5` in the given coefficient of `x^2`. The additional `xx 2^2` helps with the middle coefficient, which would otherwise be `2(20x)(17/2)`. We could have just used `5` as a multiplier, but our arithmetic would involve fractions.