Appendix C - Solving Polynomial Equations - Exercise Set C - Page A33: 6

$p(1) = -24$ $p(-1) = -12$ $p(3) = 12$ $p(-3) = 0$ $p(7) = 420$ $p(-7) = -168$ $p(21) = 10416$ $p(-21) = -7812$

We are given the polynomial $p(x) = x^3 + 3x^2 -7x - 21$ and asked to find $p(x)$ for $x=\pm 1, \pm 3, \pm 7,\pm21$ using synthetic division and the Remainder Theorem. We can directly evaluate the polynomial at each point. Find $p(1)$: We divide $p(x)$ by $x-1$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline 1 & & 1 & 4 & -3 \\ \hline & 1 & 4 & -3 & -24 \\ \end{array} \] The rest is $-24$, so $p(1)=-24$. Find $p(-1)$: We divide $p(x)$ by $x+1$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline -1 & & -1 & -2 & 9 \\ \hline & 1 & 2 & -9 & -12\\ \end{array} \] The rest is $-12$, so $p(-1)=-12$. Find $p(3)$: We divide $p(x)$ by $x-3$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline 3 & & 3 & 18 & 33 \\ \hline & 1 & 6 & 11 & 12\\ \end{array} \] The rest is $12$, so $p(3)=12$. Find $p(-3)$: We divide $p(x)$ by $x+3$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline -3 & & -3 & 0 & 21 \\ \hline & 1 & 0 & -7 & 0\\ \end{array} \] The rest is $0$, so $p(-3)=0$. Find $p(7)$: We divide $p(x)$ by $x-7$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline 7 & & 7 & 70 & 441 \\ \hline & 1 & 10 & 63 & 420\\ \end{array} \] The rest is $420$, so $p(7)=420$. Find $p(-7)$: We divide $p(x)$ by $x+7$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline -7 & & -7 & 28 & -147 \\ \hline & 1 & -4 & 21 & -168\\ \end{array} \] The rest is $-168$, so $p(-7)=-168$. Find $p(21)$: We divide $p(x)$ by $x-21$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline 21 & & 21 & 504 & 10437 \\ \hline & 1 & 24 & 497 & 10416\\ \end{array} \] The rest is $10416$, so $p(21)=10416$. Find $p(-21)$: We divide $p(x)$ by $x+21$: \[ \begin{array}{r|rrrr} & 1 & 3 & -7 &-21 \\ \hline -21 & & -21 & 378& -7791 \\ \hline & 1 & -18 & 371 & -7812\\ \end{array} \] The rest is $-7812$, so $p(-21)=-7812$.