Maiclubk

2021-02-02

Solve by factorization

$6{x}^{2}+x-2=0$

grbavit

Skilled2021-02-03Added 109 answers

Step 1

factor the following$6{x}^{2}+x-2=0$

Step 2

Factor the quadratic$6{x}^{2}+x-2$

The coefficient of$x}^{2$ is 6 and the constant term is −2.

The product of 6 and −2 is −12. The factors of −12 which sum to

1 are −3 and 4

So$6{x}^{2}+x-2=6{x}^{2}+4x-3x-2=2(2x-1)+3x(2x-1)$ :

2(2x−1)+3x(2x−1):

Factor 2x−1 from 2(2x−1)+3x(2x−1)=(2x−1)(3x+2),

=(2x−1)(3x+2)

Now set (2x−1)(3x+2)=0

(2x−1)=0 and (3x+2)=0

$x=\frac{1}{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}x=-\frac{2}{3}$

factor the following

Step 2

Factor the quadratic

The coefficient of

The product of 6 and −2 is −12. The factors of −12 which sum to

1 are −3 and 4

So

2(2x−1)+3x(2x−1):

Factor 2x−1 from 2(2x−1)+3x(2x−1)=(2x−1)(3x+2),

=(2x−1)(3x+2)

Now set (2x−1)(3x+2)=0

(2x−1)=0 and (3x+2)=0

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