Question

How do you factor the expression `2 x^2 + 5 x +12`?

Answer 1

`2x^2+5x+12 = 1/8(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)`

Explanation:

The difference of squares identity can be written:

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

We use this below with `a=(4x+5)` and `b=sqrt(71)i`.

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Multiply by `8`, complete the square, then divide by `8`...

`8(2x^2+5x+12)`

`=16x^2+40x+96`

`=(4x+5)^2-25+96`

`=(4x+5)^2+71`

`=(4x+5)^2-(sqrt(71)i)^2`

`=((4x+5)-sqrt(71)i)((4x+5)+sqrt(71)i)`

`=(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)`

So:

`2x^2+5x+12 = 1/8(4x+5-sqrt(71)i)(4x+5+sqrt(71)i)`