Question

How do you factor `3x^2-4x-32`?

Answer 1

`3x^2-4x-32=(3x+8)(x-4)`

Explanation:

Use an AC method:

Find a pair of factors of `AC = 3*32 = 96` which differ by `B=4`.

The pair `12, 8` works.

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

`3x^2-4x-32`

`=3x^2-12x+8x-32`

`=(3x^2-12x)+(8x-32)`

`=3x(x-4)+8(x-4)`

`=(3x+8)(x-4)`

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Alternative method

Complete the square, then use the difference of squares identity:

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

with `a=(3x-2)` and `b=10` as follows:

First multiply by `3` to make the leading term into a perfect square and simplify the arithmetic.

`3(3x^2-4x-32)`

`=9x^2-12x-96`

`=(3x)^2-2(2)(3x)-96`

`=(3x-2)^2-2^2-96`

`=(3x-2)^2-10^2`

`=((3x-2)-10)((3x-2)+10)`

`=(3x-12)(3x+8)`

`=3(x-4)(3x+8)`

Dividing both ends by `3`, we find:

`3x^2-4x-32 = (x-4)(3x+8)`