Solve Trinomial Problem: 2(2x^2+3x-14)

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The discussion focuses on factoring the trinomial expression 2(2x^2 + 3x - 14). The initial confusion arises from the use of the FOIL method, which is intended for multiplication rather than factoring. Participants clarify that the first step is to factor the trinomial 2x^2 + 3x - 14, suggesting various integer combinations for its factors. They emphasize the importance of checking the cross terms when verifying potential factors. Ultimately, the correct factorization, including the initial "2," is noted to be (2x + 7)(x - 2).
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I was working on this one trinomial problem-2(2x^2+3x-14) but I tried doing foil but I'm not sure what I am doing wrong. Could someone help me out? Thanks.
 
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Are you solving for x?
 
nope. Just trying to factor it.
 
Well, you can't be "solving for x" since there is no equation! Since you titled this "Factoring Problem", guess you want to factor it further. That "2" in front has already been factored out and you can ignore it (until you are ready to write the answer).

The first step, then, is to factor 2x2+ 3x- 14. When you say you tried "FOIL", I was puzzled for a moment- that's really a mnemonic for multiplying, not factoring. But I suspect you really mean is:
The only way to factor 2 (into integer factors) is 2 and 1 so you must have (2x )(x ). There are two ways to factor 14: 1*14 and 2*7 and since you have -14, that gives -1*14, 1*(-14), -2*7, and 2*(-7)
That is- if it is possible to factor 2x2+ 3x- 14 with integer coefficients it must be one of:
(2x- 1)(x+ 14)
(2x+ 1)(x- 14)
(2x- 2)(x+ 7)
(2x+ 2)(x- 7)
(2x+14)(x- 1)
(2x-14)(x+ 1)
(2x- 7)(x+ 2)
(2x+ 7)(x- 2)
(2x+ 1)(x-14)
(2x- 1)(x+ 14)

Go ahead, multiply them and see if any work. You really only need to check the "cross terms"- the "O" and "I" of FOIL. Of course, there is no guarantee that this can be factored with integer coefficients!

Once you've figured out how to factor 2x2+ 3x- 14, don't forget to write that "2" factor.
 
(2x+ 7)(x- 2)

Ahh thank you!
 
u could always start with variables (ax+b)*(cx+d) multiply through
match them to each coefficent and solve =]
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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