Question

`2x^3-7x^2-2x+7`? How to factor it completely?

factor completely

Answer 1

`p(x)=(x+1)(x-1)(2x-7)`

Explanation:

We take,

`p(x)=2x^3-7x^2-2x+7`

`=2x^3-2x-7x^2+7`

`=2xcolor(red)((x^2-1))-7color(red)((x^2-1))`

`=(x^2-1)(2x-7)`

`=(x+1)(x-1)(2x-7)`

Answer 2

`2x^3-7x^2-2x+7 = (x+1)(x-1)(2x-7)`

Explanation:

First, use trial and error to find an `x` value that will make the equation equal to zero. From the coefficients, you can tell that `x=1` will make the equation equal to zero.

`2(1)^3 -7(1)^2 -2(1) +7 = 0`

Then you do a polynomial divsion of `(2x^3-7x^2-2x+7)/(x+1)`

(Don't know how to format this so I wrote it out and took a picture of it)

`2x^3-7x^2-2x+7 = (x+1)(2x^2-9x+7)`

(The explaining of polynomial long division)

Now factorise the quadratic.

`2x^2 - 9x+7 = x^2-9x+14`

Which factors add up to nine and their product is 14? The factors are 2 and 7.

`x^2-9x+14=(x-2)(x-7)`

Then put the coefficient of `x` back in and simplify where necessary.

`(2x-2)(2x-7) = (x-1)(x-7)`

`2x^2 -9x +7 = (x-1)(x-7)`

`A= (x+1)(x-1)(2x-7)`