Question

FACTOR 2x^2 + x - 28 = 0

There should be 2 x's. I am stuck at 2(x^2+x-14)=0

Answer 1

@haleyelizabeth2017 Lists steps

Answer 2

Wait, it was originally \(2x^2+2x-28?\)

Answer 3

What you have is the simplest it can go.

Answer 4

Originally 2x^2+x-28=0

Answer 5

Okay, then what you have so far is not correct.

Answer 6

oh yea i made a mistake, i thought the GCF was 2 when it can't be

Answer 7

what can be mult. to make -56 but added to make 1

Answer 8

-7 and 4 multiply to get -28, okay?

So then...if one side is (2x+/-_)(x+/-_) we can plug in those two numbers to try and see which works.

Answer 9

im guessing that another way to do it Eli?

Answer 10

So...let's try (2x-7)(x+4) first.

We get \(2x^2+8x-7x-28\) which simplified, would work. So, (2x-7)(x+4) is the correct.

Answer 11

correct answer* ^^

Answer 12

so 2x-7=0 is x=7/2 or 3 1/2

Answer 13

and x=-4

Answer 14

Normally, it is left as an improper fraction, so x=7/2 is better. and x=-4 is correct as well.

Answer 15

You could have also used the quadratic formula.

Answer 16

It is a lot easier. Or, even complete the square works too

Answer 17

\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] \[ax^2+bx+c=0\]

Answer 18

Ok, i got a question... the formula is ax^2+bx+c... i was told that you mult. a*c and the number that you get has to be mult. by x number and x number, but those two number must also be able to be added together to equal B... did that make sense

Answer 19

Is there an exception to this??

Answer 20

Huh?

Answer 21

You don't have to do anything to the a b or c. just plug the coefficients into the quadratic formula and solve. Lets use your problem for example. We were given \[2x^2+x-28\] A would be 2, b would be 1, and c would be -28.

Answer 22

\[\frac{ -(1) \pm \sqrt{(1)^2-4(2)(-28)} }{ 2(2) }\]\[\frac{ -1 \pm \sqrt{1+224} }{ 4 }\]and so on

Answer 23

Did that make sense?

Answer 24

@Beleaguer did you figure out What the two numbers are?

Answer 25

In other words, can you solve
m+n =1
mn =-56