Chapter 4.4, Problem 23E

(a)

To determine

To find: The zeros of the function.

Answer to Problem 23E

  $\text{Zeros: }x=-2,2\text{ and }-1\text{with multiplicity of 2}$

Explanation of Solution

Given information: Suppose $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

Calculation:

  $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

Zeros are:

  $x=-2,2\text{ and }-1\text{ with a multiplicity of 2}$

(b)

To determine

To write: $f\left(x\right)$ as a polynomial function.

Answer to Problem 23E

  $x^4+2x^3-3x^2-8x-4$

Explanation of Solution

Given information: Suppose $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

Calculation:

  $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

  $\Rightarrow x^4+2x^3-3x^2-8x-4$

(c)

To determine

To find: The number of possible positive real roots and the number of possible negative real roots using Descartes Rule of Signs.

Answer to Problem 23E

1 positive root and 1 or 3 negative roots

Explanation of Solution

Given information: Suppose $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

Calculation:

  $x^4+2x^3-3x^2-8x-4$

Descartes gives us: 1 positive root and since:

  $f\left(-x\right)=x^4-2x^3-3x^2+8x-4$

This gives us 1 or 3 negative roots.

(d)

To determine

To compare: Your answers to part a and part c. Explain.

Answer to Problem 23E

1 positive root and 1 or 3 negative roots

Explanation of Solution

Given information: Suppose $f\left(x\right)=\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

Calculation:

Given

  $\left(x-2\right)\left(x+2\right){\left(x+1\right)}^2$

This shows $x=-2,2,-1$ (with a multiplicity of 2)

So the answer for part c is correct, we have 1 positive root and 3 negative roots.