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My problem is here, the question is:

$\mathbf{16h^2+4kh-30k^2}$

How do you factor this question and please explain the step where the trinomial is factorized. I searched a lot on web and came out nothing helping me to understand how to do this question. I knew the answer was $\mathbf{2(4h-5k)(2h+3k)}$, but don't know the steps after 2 has been factorized.

Please explain thoroughly and as well as how to do two variables questions with first coefficient not being 1?

Also, I used the Quadratic equation method and the Finding All Possible Factor (don't know it's name) method apon a question: $\mathbf{-5a^2+18-27a}$ and found me two answers: $\mathbf{(a-\frac3 5)(a+6)}$ and $\mathbf{-(a+6)(5a-3)}$. Are they both accepted for an answer?

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    $\begingroup$ In what you call the Quadratic Equation method, what you probably got is $(-5)(a-3/5)(a+6)$. This is certainly technically correct, but probably you would be expected to "absorb" the $-5$ into the $a-3/5$. And being technically correct is no great consolation if you are marked as being wrong. The $(a-3/5)(a+6)$ version is actually wrong. $\endgroup$ – André Nicolas Jul 27 '13 at 14:49
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In order to solve this equation:

$${16h^2+4kh-30k^2}$$

Take the following steps:

  1. Factor out a common factor which is $2$ so then you will have $2(8h^2+2kh-15k^2)$
  2. Next, find two numbers that multiply to $-120$ and add to $2$. ($12$ and $-10$)
  3. So then you will have the following $2(8h^2+12kh-10kh-15k^2)$ Here, I have replaced $2$ with $12$ and $-10$ because their sum is equivalent to $2$, so the equation itself has not changed.
  4. Factor by grouping.

If you need all the steps then let me know and I'll expand.

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  • $\begingroup$ I am very sorry to bother, but I don't understand step 2 and how step 2 changed into step 3, please explain more. $\endgroup$ – Daniel Cheung Jul 27 '13 at 14:57
  • $\begingroup$ No problem! Don't hesitate to ask :). Okay. When you apply this method, you have to find two numbers that multiply to $-120$ because $8 * -15 = -120$ and those same numbers have to add to $2$ because that is the middle term. So in this case it's $12$ and $-10$ because $12+(-10)=2$. Now once you have these factors, replace the $2$ with them. So in case, $2$ is replaced by $12+(-10)$. $\endgroup$ – Jeel Shah Jul 27 '13 at 15:01
  • $\begingroup$ just one more, if any coefficient in the question is like in hundreds, it will be hard to find matching factors for the middle term, do you know any way to do these question without guess pairs? $\endgroup$ – Daniel Cheung Jul 28 '13 at 4:28
  • $\begingroup$ If you have questions with coefficients in the hundreds then try to find a common factor to reduce those coefficients, if that's not possible then just use the quadratic equation which much easier to do and faster than the alternative. $\endgroup$ – Jeel Shah Jul 28 '13 at 14:39
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Hint: Make a change $h=kx$ and reduce your problem to an equation with one variable.

You get: $k^2(16x^2+4x-30)$. Can you factor the polynomial $16x^2+4x-30$ ?

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  • $\begingroup$ Sorry, but I really don't understand, but how does that is going to help in this calculation? $\endgroup$ – Daniel Cheung Jul 27 '13 at 14:51
  • $\begingroup$ I complete the answer. $\endgroup$ – Boris Novikov Jul 27 '13 at 14:55

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