Chapter 3.3, Problem 57E

To determine

To show that value of $c$ is zero of $P\left(x\right)$ and find all other zeros of $P\left(x\right)$ .

Answer to Problem 57E

All other zeros of $P\left(x\right)$ is, $-1\pm \sqrt{6}$ .

Explanation of Solution

Given information:

The polynomial is given as,

  $P\left(x\right)=x^3-x^2-11x+15$

The value of $c$ is given as,

  $c=3$

Concept used:

By using factor theorem, $x-c$ is a factor of $P\left(x\right)$ ; whenc is a zero of P .

Calculation:

To show that value of $c$ is zero of $P\left(x\right)$ ; $\left(x-c\right)$ must a factor of $P\left(x\right)$ .

Since, The value of $c$ is given as, $c=3$

The value of can be $P\left(c\right)$ calculated as,

  $\begin{array}{l}P\left(x\right)=x^3-x^2-11x+15\\ P\left(3\right)={\left(3\right)}^3-{\left(3\right)}^2-11\left(3\right)+15\\ P\left(3\right)=27-9-33+15\\ P\left(3\right)=0\end{array}$

Hence, the value of $c$ is zero of $P\left(x\right)$ .

Since, $\left(x-3\right)$ is a factor of $P\left(x\right)$ ; the value $P\left(x\right)$ can be derived by division,

  $\begin{array}{l}3\overline{)1-1-1115}\\ \underset{\_}{36-15}\\ 12-50\end{array}$

The value of $P\left(x\right)=\left(x^2+2x-5\right)$

Here, $a=1\text{ , }b=2\text{ , }c=-5$

By using the quadratic formula,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(-5\right)}}{2\left(1\right)}\\ x=\frac{-2\pm \sqrt{24}}{2}\\ x=\frac{-2\pm 2\sqrt{6}}{2}\\ x=-1\pm \sqrt{6}\end{array}$

Therefore,

All other zeros of $P\left(x\right)$ is, $-1\pm \sqrt{6}$ .