Factoring 28s^2 + 8st - 20t^2: A Simplified Explanation and Solution

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The discussion focuses on factoring the quadratic expression 28s^2 + 8st - 20t^2. The correct factorization is identified as 4(7s - 5t)(s + t). Participants suggest combining terms to simplify the expression, leading to a clearer path to the solution. The original poster expresses relief upon realizing the mistake and appreciates the help received. The interaction highlights the importance of careful term management in polynomial factorization.
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Homework Statement



28s^2 + 8st - 20t^2

The answer in the back of the book is 4(7s - 5t)(s + t)

2. The attempt at a solution

4(7s^2 + st + st - 5t^2)
4(7s^2 + st + st - 20t^2)
4(s(7s + t) + t(s - 20t))

I'm lost, can someone please help me?

Thanks in advance.
 
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In your first attempt at the solution, why not combine the second and third terms inside the (), it might make things more obvious.
 
Is there some trick to this that I am missing?

4(7s^2 + 2st - 5t^2)
 
Well, you could note that 2st=7st-5st, which makes your expression 4(7s2+7st-5st-5t2). Is this any easier to factorise?
 
OHH OHH OHHHH OHHHHHH!

Oh my god, how could I be so blind? That t^2 at the end threw me off. I feel like such an idiot. Thanks a billion cristo and robb, and I apologize for wasting your time :)
 
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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