# Finding Product of Scattered Variables

Hi I came across the following question where I need to find $$mk$$ from $$(x-2) (x+k) = x^2 + mx - 10$$ The answer is 15. Any suggestions on how I could do that ?

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Use the relation between the roots of a polynomial and its coefficients. – Host-website-on-iPage Jul 10 '12 at 15:53

The sum of the roots $(2, -k)$ equals $-m$. The product of the roots $-2k=-10$. Therefore $k=5$ and $m=3$.

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Thanks for the great solution . I realize that 2,-k are the two roots. However I did not know that sum of the roots (2,−k) equals −m. and the product of the roots -10 can be obtained from the equation. Could you show me how you got that or link to any references please.. – Rajeshwar Jul 10 '12 at 16:12
If you expand the left hand side, you get it. $(x-a)(x-b)=x^2+px+q$ $x^2-(a+b)x+ab=x^2+px+q$ It's true for polynomials over any field. – Host-website-on-iPage Jul 10 '12 at 16:13
Aneesh gave the full explanation! – PAD Jul 10 '12 at 16:19

You can find the product $(x-2)(x+k)$, getting $x^2+(k-2)x-2k$. This is supposed to be the same polynomial as $x^2+mx-10$.

So the constant terms must match, and the coefficients of $x$ must match. That gives us $-2k=-10$ amd $k-2=m$. From $-2k=10$, we conclude that $k=5$. Then from $k-2=m$ we conclude that $m=3$. It follows that $mk=15$.

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Expression$$(x-2) (x+k) = x^2 + mx - 10$$ can be rewriten as $$x^2+(k-2)x-2k = x^2 + mx - 10$$ equating the coefficients next to same power of $x$ we get that $k-2=m$ and $-2k=-10$ or$k=5$ and $m=5-2=3$ that means $$mk=5\times 3=15$$

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Your second equation (k-2) is missing an x – Rajeshwar Jul 10 '12 at 16:22
there is $x^0=1$ – Milingona Ana Jul 10 '12 at 16:25