Question

How do you factor `16x^2 + 80x + 84`?

Answer 1

`16x^2+80x+84=4(2x+3)(2x+7)`

Explanation:

Separate out the common scalar factor `4`, complete the square, then use the difference of squares identity:

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

with `a=(2x+5)` and `b=2` as follows:

`16x^2+80x+84`

`=4(4x^2+20x+21)`

`=4((2x)^2+2(2x)(5)+21)`

`=4((2x+5)^2-25+21)`

`=4((2x+5)^2-4)`

`=4((2x+5)^2-2^2)`

`=4((2x+5)-2)((2x+5)+2)`

`=4(2x+3)(2x+7)`

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Alternatively, we can use an AC method to factor `4x^2+20x+21`:

Look for a pair of factors of `AC=4*21 = 84` with sum `B=20`.

The pair `14, 6` works.

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

`4x^2+20x+21`

`=4x^2+14x+6x+21`

`=(4x^2+14x)+(6x+21)`

`=2x(2x+7)+3(2x+7)`

`=(2x+3)(2x+7)`