Is it possible to factor $y = 3 x^{2} + 2 x - 8$ $y=3x^2+2x-8$ ? If so, what are the factors?

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Answer 1 · 2018-04-19T16:37:23

$y = (x + 2) (3 x - 4)$ $y=(x+2)(3x-4)$

$y = 3 x^{2} + 2 x - 8$ $y=3x^2+2x-8$

to begin the factorising process

we look for factors of $3 \times - 8 = - 24$ $3xx-8=-24$

that sum to the middle term $2$ $2$

by inspection we will have $+ 6, - 4$ $+6, -4$

#we now split teh middle term with these factors and factorise by grouping

$y = 3 x^{2} + 6 x - 4 x - 8$ $y=3x^2+6x-4x-8$

$y = (3 x^{2} + 6 x) - (4 x + 8)$ $y=(3x^2+6x)-(4x+8)$

$y = 3 x (x + 2) - 4 . (x + 2)$ $y=3x(x+2)-4.(x+2)$

$y = (x + 2) (3 x - 4)$ $y=(x+2)(3x-4)$

Is it possible to factor y=3x2+2x−8y=3x^2+2x-8 ? If so, what are the factors?