Question

How do you factor `2x^2-11x+5`?

Answer 1

f(x) = 2x^2 - 11x + 5 = (x - p)(x - q)

I use the new AC Method.

Converted function: f'(x) = x^2 - 11x + 10 = (x - p')(x - q').

Compose factor pairs of ac = 10: (1, 10). This sum is 11 = -b.
Then p' = -1 and q' = -10. We get: `p = (p')/a = -1/2, and q = (q')/a = -10/2 = -3.`

Factored form: `f(x) = (x - 1/2)(x - 5) = (2x - 1)(x - 5)`

Check by developing: `f(x) = 2x^2 - 10x - x + 5 = 2x^2 - 11x + 5.`OK

Answer 2

`2x^2-11x+5`

`2xx5 = 10`.

Find two numbers whose product is `10` and whose sum is `-11`

A little thought should convince us that the numbers must both be negative.

`-1xx-10` works!, that's good!.

Now split the middle term using these numbers: `-11x = -1x-10x`

`2x^2-11x+5 = 2x^2-1x-10x+5`

Now factor by grouping:

`2x^2-11x+5 = (2x^2-1x)+(-10x+5)`

`color(white)"sssssssssssssss"` `=x(2x-1)-5(2x-1)`

`color(white)"sssssssssssssss"` `=(x-5)(2x-1)`

Check your answer by multiplying.