Question

factor 2x^2-x=3

could someone please explain with steps I'm really bad at factoring

Answer 1

@shamil98

Answer 2

WJHAT

Answer 3

SOMEONE HELP ME PLEASE

Answer 4

Dang

Answer 5

I typed out this long, elaborate way to do this, and I accidently clicked on another person's question qq

Answer 6

Well, bear me for a moment again, please.

Answer 7

LOL TY ANYWAYS

Answer 8

2x^2-x=3
2x^2 - x - 3 = 0
2x^2 - 3x + 2x - 3 = 0
x(2x - 3) + 1(2x - 3) = 0
(2x - 3)(x + 1) = 0

Answer 9

First off, you must set the equation equal to 0:\[2x^2-x-3=0\]
From then, upon seeing the -3, you know that the factors of 3 (which are 1 and 3), upon multiplying them, must be negative, and the only way to get a negative value from multiplying, is for one of the values to be positive and the other negative.
Secondly, since you see the -1 (-x has an invisible -1 in front of it), you must conclude that the negative value must be greater. However, you must keep the "2" in front of the x^2 in mind.
Now, as we said before, the values of 3 are 1 and 3, you may think, what value, multiplied by 2 makes -1? We see that -3+2=-1;\[(2x-3)(x+1)\]
If you're not too sure about the answer, distribute. I hope this helped, pst me if you need more assistance with this or any other problem. :)

Answer 10

my answer key's answer that is closest to that is 1,3 is this correct

Answer 11

@JonnyVonny

Answer 12

Mmk, strange, because if you plug in 1 into the original equation, it does not equal 0, which it is suppose to.