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I'm not sure how to title this but here's the problem;

Define $a,b,c$ so that $P(x) + Q(x) = 0$ with every value of $x$.

$P(x) = 5x^2 - ax + 4 - (bx^2 - 7x + 3)$ and $Q(x) = 8x^2 + x + c$.

So you need to find values for $a, b$ and $c$ which will make the sum of $P(x)$ and $Q(x)$ equal to zero, no matter what you will input as the value of $x$.

I've tried simplifying the sum of the polynomials but that's how far I'll get. I can deduce the correct answer by trying different values but i don't think it's how you are supposed to solve this problem.

How you should mathematically approach this problem? Can you create an equation out of this?

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    $\begingroup$ What have you tried? E.g. what is $P(x) + Q(x)$? What does it mean for this to "equal zero for every value of $x$"? $\endgroup$
    – Calvin Lin
    Commented Nov 25, 2020 at 18:48
  • $\begingroup$ @CalvinLin You need to find a value for (a), (b) and (c) that will make P(x) + Q(x) = 0, no matter what you will input as the value of (x). $\endgroup$
    – Heineken
    Commented Nov 25, 2020 at 19:01
  • $\begingroup$ Great, so what is $P(x) + Q(x)$ equal to currently? $\endgroup$
    – Calvin Lin
    Commented Nov 25, 2020 at 19:02
  • $\begingroup$ @CalvinLin You mean the simplified form? Then it's; 13x^2-bx^2+8x-ax+c+1 $\endgroup$
    – Heineken
    Commented Nov 25, 2020 at 19:06
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    $\begingroup$ Tip: don't think about this as a 4-variable problem, because as a matter of fact, you have only one variable: $x$. $a, b, c$ are parameters, and you are looking for suitable parameter values so that they give a solution for any value of $x$, so that it remains a variable. $\endgroup$
    – Dávid Laczkó
    Commented Nov 25, 2020 at 19:51

1 Answer 1

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In general what you want to do is group the like terms. What does $P(x)+Q(x)$ look like? $P(x)+Q(x)=5x^2-ax+4-bx^2+7x-3+8x^2+x+c$ now we can group the like terms to have that this is equal to: $(5-b+8)x^2+(-a+7+1)x+(4-3+c)$ we want this to equal zero, so in particular we need $(5-b+8)=0$, $(-a+7+1)=0$ and $(4-3+c)=0$.

The same strategy is how you would do this in general.

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  • $\begingroup$ Yes. This is what i didn't understand to do. Thank you. $\endgroup$
    – Heineken
    Commented Nov 25, 2020 at 19:48

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