How do you solve $6x^2 + 5x - 4 = 0$ by factoring?

Question

17120 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-17T06:35:15

The solutions are
$color(blue)(x=-4/3$

$color(blue)(x=1/2$

$6x^2+5x−4=0$

We can Split the Middle Term of this expression to factorise it and thereby find solutions.

In this technique, if we have to factorise an expression like $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 6*-4 = -24$
and
$N_1 +N_2 = b = 5$

After trying out a few numbers we get $N_1 = 8$ and $N_2 =-3$
$8*-3 = -24$ , and $8+(-3)= 5$

$6x^2+5x−4=6x^2-3x+8x−4$

$=3x(2x-1) +4(2x-1)=0$

$=(3x+4)(2x-1) =0$

Now we equate the factors to zero.
$=3x+4 =0, color(blue)(x=-4/3$

$=2x-1 =0, color(blue)(x=1/2$

How do you solve 6x^2 + 5x - 4 = 06x^2 + 5x - 4 = 0 by factoring?