Chapter 2.2, Problem 55E

To determine

To find a polynomial that has zeroes $0,-2,-4$ .

Answer to Problem 55E

  $x^3+6x^2+8x$

Explanation of Solution

Given:

Zeroes of polynomial are $0,-2,-4$ .

Concept Used:

If $x=a$ is a zero of a polynomial, then $\left(x-a\right)$ is a factor of that polynomial.

Calculation:

To find a polynomial of degree that has zeroes $0,-2,-4$ . Use the concept that if a is a zero of a polynomial, then $\left(x-a\right)$ is a factor of that polynomial.

Since, $0,-2,-4$ are zeroes of the polynomial thus

  $\begin{array}{l}\left(x-0\right)=x,\\ \left(x-\left(-2\right)\right)=\left(x+2\right),\text{ and }\\ \left(x-\left(-4\right)\right)=\left(x+4\right)\end{array}$

are the factors corresponding to the zeros of the polynomial respectively.

Now, to find a polynomial that has factors $x,\left(x+2\right),\text{ and }\left(x+4\right)$ , multiply these factors, as

  $\begin{array}{c}x\left(x+2\right)\left(x+4\right)=\left(x^2+2x\right)\left(x+4\right)\\ =x\left(x^2+2x\right)+4\left(x^2+2x\right)\\ =x^3+2x^2+4x^2+8x\\ =x^3+6x^2+8x\end{array}$

Thus, a polynomial that has zeroes $0,-2,-4$ will be multiple of polynomial $x^3+6x^2+8x$ .

Thus, a polynomial that has zeroes $0,-2,-4$ is $x^3+6x^2+8x$ .