How do you solve $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by factoring?

Question

How do you solve $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by factoring?

1 Answer

Impact of this question

2252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T21:39:28

$(x - 3) (2 x + 1)$ $(x-3)(2x+1)$

Explanation:

$2 x 3 = 6$ $2x3=6$ so then you must find two numbers (a negative and a positive since there are two negatives in the original problem) that will add up to $5$ $5$ and multiply to $6$ $6$ .

The numbers are then $- 6$ $-6$ and $+ 1$ $+1$ which gives you:

$2 x^{2} + 1 x - 6 x - 3$ $2x^2+1x-6x-3$

Split down the middle and factor out an $x$ $x$ and a $- 3$ $-3$

$x (2 x + 1) + - 3 (2 x + 1)$ $x(2x+1) + -3(2x+1)$

Take the leading coefficients and you'll finish with

$(x - 3) (2 x + 1)$ $(x-3)(2x+1)$

Hope this helps!

Science

Math

Humanities

... and beyond

How do you solve $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by factoring?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2x2−5x−32x^2-5x-3 by factoring?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2 x^{2} - 5 x - 3$ $2x^2-5x-3$ by factoring?