FACTOR 2x^2 + x - 28 = 0 There should be 2 x's. I am stuck at 2(x^2+x-14)=0

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FACTOR 2x^2 + x - 28 = 0 There should be 2 x's. I am stuck at 2(x^2+x-14)=0

Mathematics
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@haleyelizabeth2017 Lists steps
Wait, it was originally \(2x^2+2x-28?\)
What you have is the simplest it can go.

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Originally 2x^2+x-28=0
Okay, then what you have so far is not correct.
oh yea i made a mistake, i thought the GCF was 2 when it can't be
what can be mult. to make -56 but added to make 1
-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.
im guessing that another way to do it Eli?
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.
correct answer* ^^
so 2x-7=0 is x=7/2 or 3 1/2
and x=-4
Normally, it is left as an improper fraction, so x=7/2 is better. and x=-4 is correct as well.
You could have also used the quadratic formula.
It is a lot easier. Or, even complete the square works too
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] \[ax^2+bx+c=0\]
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
Is there an exception to this??
Huh?
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.
\[\frac{ -(1) \pm \sqrt{(1)^2-4(2)(-28)} }{ 2(2) }\]\[\frac{ -1 \pm \sqrt{1+224} }{ 4 }\]and so on
Did that make sense?
@Beleaguer did you figure out What the two numbers are?
In other words, can you solve m+n =1 mn =-56

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