There's kind of a cool way to solve it with less factoring, which is nice for people like me who aren't the quickest at finding factorizations.

Note that `f(x)-g(x)=(b-q)x+(c-r).` Then since `g(alpha)=g(beta)=0,` we have

`f(alpha)=f(alpha)-g(alpha)=(b-q)alpha+(c-r)` and

`f(beta)=f(beta)-g(beta)=(b-q)beta+(c-r).`

So you can already see where the `b-q`and `c-r` terms come from in the answer (actually those terms were what led me to try subtracting the functions in the first place). Now multiply these (you have to do a little bit of factoring at times too) and use the fact that `alpha*beta=r` and `alpha+beta=-q,` and the answer just pops out. Since the question was already solved though, I'll leave the straightforward details of this to you.

`g(x) = x^2+qx+r = 0` with roots `alpha` and `beta`

So we can say;

`alpha+beta = -q`

`alphaxxbeta = r`

`f(alpha)*f(beta)`

`= (alpha^2+balpha+c)(beta^2+b(beta)+c)`

`=(alphabeta)^2+b(alpha^2beta+beta^2alpha)+c(alpha^2+beta^2)+c^2+b^2alphabeta+bc(alpha+beta)`

`=(alphabeta)^2+balphabeta(alpha+beta)+c[(alpha+beta)^2-2alphabeta]+b^2alphabeta+c^2+bc(alpha+beta)`

`= r^2+br(-q)+c(q^2-2r)+b^2r+c^2-bcq`

`= c^2-2cr+r^2-bqr+cq^2+b^2r-bcq`

`= (c-r)^2-bqr+b^2r+cq^2-bcq`

`= (c-r)^2-br(q-b)+cq(q-b)`

`= (c-r)^2-(q-b)(cq-br)`

`= (c-r)^2-(b-q)(cq-br)`

**So the required answer is obtained.**

**Further Reading**

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