Distributive Factoring: 2x^2 − 6x − 8 = 2(x − 4)(x + 1)

[iceasson]

Could someone help me out with this one

2x² − 6x − 8=2(x − 4)(x + 1),


The Expression is
2x^2 -6x -8

These are the steps i have done

(X+A) (X+B)
X^2+(A+B)x+AB
A+B= -6
A.B=-8

Factors of 8 are 1,2 and 4
so i went with -2+-4 to get A+B=-6
but -2.-4= Possitive 8

so, i factored everything by 2

2(x^2-3-4)
which equal 2x² − 6x − 8

Then why 2(x − 4)(x + 1) is the correct answer?

 

[Otis]

Could someone help me out with this one

2x^2 − 6x − 8=2(x − 4)(x + 1),

Could someone point me out on this one

The Expression is 2x^2 - 6x - 8

These are the steps i have done

(X+A) (X+B)
X^2+(A+B)x+AB
A+B= -6
A.B=-8

Factors of 8 are 1,2 and 4
so i went with -2+-4 to get A+B=-6
but -2.-4= Positive 8

so, i factored everything by 2

2(x^2 - 3x - 4)
which equal 2x2 − 6x − 8

Then why 2(x − 4)(x + 1) is the correct answer?

Note: I fixed your typo (missing x) above, in red.

The posted answer is correct because it is the complete factorization.

You correctly factored out a 2, but you ought to have done that first.

2x^2 - 6x - 8 = 2(x^2 - 3x - 4)

Now, the polynomial inside parentheses can itself be factored.

Look for two numbers whose sum is -3 and whose product is -4. Then you can write the complete factorization of the original polynomial.

:idea: Remember: the factoring rule that you showed (highlighted in blue above) works only when the coefficient on the x^2 term is 1. You tried to apply that rule to a polynomial whose coefficient on the x^2 term is 2, and that's why it didn't work out. By factoring out a 2 first, you get a new polynomial whose leading coefficient is 1, so then the rule in blue applies.