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Find $p$ and $q$, if $x^3-2x^2+p+q$ is divided by $x^2+x-2$. I have found $x =-2$ and $x=1$ however when placed into polynomial I am unable to find a value for $p$ and $q$.

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    $\begingroup$ I suspect that you mean $x^3-2x^2+px+q$. Otherwise there is no need for $q$. $\endgroup$ – Dietrich Burde Sep 27 '18 at 18:54
  • $\begingroup$ Hint: Compute $\left(x^2+x-2\right)(x-3)$ $\endgroup$ – robjohn Sep 27 '18 at 19:51
  • $\begingroup$ By Remainder thm $ f(1)=f(2)=0 $ and solve. Or perform synthetic division. $\endgroup$ – Narasimham Sep 27 '18 at 22:05
  • $\begingroup$ @Sherma You sholud accept one of answers you get. $\endgroup$ – Aqua Sep 28 '18 at 14:57
  • $\begingroup$ I accept Greedoid's answer. questions should be x^3-2x^2+px+q $\endgroup$ – Sherma Oct 1 '18 at 15:05
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Hint: If $x-a$ divide $P(x)$ then $P(a)=0$, so

since $x^2+x-2=(x+2)(x-1)$ divide $P(x)= x^3-2x^2+p+q$ we have $P(-2)=0$ and $P(1)=0$

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  • $\begingroup$ Yeah, you are right. @bobajob $\endgroup$ – Aqua Sep 28 '18 at 9:28
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Or you can use Vieta formulas: You find $x_1=-2$ and $x_2=1$ so, since $$x_1+x_2+x_3 = 2\implies x_3= 3$$

so $$p = x_1x_2+x_2x_3+x_3x_1=...\;\;\;{\rm and}\;\;\;q =-x_1x_2x_3 = 6$$

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Hint: Multiply out $(x^2+x-2)(x-a)=x^3-2x^2+px+q$ and compare coefficients.

For the polynomial $x^3-2x^2+px+q$ the comparison gives a solution, namely $a=3$, $p=-5$ and $q=6$. For the above polynomial (with typo) there is no solution.

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  • $\begingroup$ Thanks much "no solution". Maybe teacher had a typo in posting the question. Thanks much $\endgroup$ – Sherma Sep 27 '18 at 19:05
  • $\begingroup$ @Sherma You could just correct the typo in your question, then it works. $\endgroup$ – Dietrich Burde Sep 27 '18 at 19:17
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Precalculus tag, people.

Assuming the typo fix, you can simply do long division directly. Maybe you're learning synthetic division, which is probably easier, but I haven't done it since high school so I forget the details.

              x   -    3
            _____________________
x^2 + x - 2 | x^3 - 2x^2 + px + q
              x^3 +  x^2 - 2x
            - _______________
                   -3x^2 + (2+p)x + q
                   -3x^2 - 3x     + 6
                 - __________________
                           (5+p)x + (q-6)

We want that final expression on the bottom to equal $0$ so there's no remainder.

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  • $\begingroup$ How about $p=-5$ and $q=6$? $\endgroup$ – Toby Mak Sep 27 '18 at 22:10
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One can also reduce $x^3 - 2 x^2 + p x + q$ modulo $x^2 + x - 2$ and then demand that the resulting polynomial is identical to zero. This then means that all the coefficients of that polynomial are zero, and that yields the solution. Modulo $q(x) = x^2 + x - 2$ we have:

$$x^2 \bmod q(x)= -x + 2$$

and

$$x^3 \bmod q(x)= \left(-x^2 + 2 x\right) \bmod q(x) = 3 x-2 $$

Therefore:

$\left(x^3 - 2 x^2 + p x + q\right) \bmod q(x) = (5+p) x + q-6$

which implies that $p = -5$ and $q = 6$.

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