Chapter 2, Problem 97CR

(a)

To determine

To find : all the zeros of function $f\left(x\right)=x^3-5x^2-7x+51$

Answer to Problem 97CR

  $x=-3,x=4+i,x=4-i$

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-5x^2-7x+51$

Concept used:

Zeros of polynomial: -

It is defined by as any real value of $x$ , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial $P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_2{x}^2+a_{1}x+a{0}_{}$ has any rational roots, then they must be of the form $\pm \left\{\frac{\text{factor of }{a}_0}{\text{factor of }{a}_n}\right\}$

Calculation:

  $f\left(x\right)=x^3-5x^2-7x+51$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 3,\pm 17,\pm 51$ . These are the possible values for $f\left(x\right)$

If $a$ is a root of the polynomial $f\left(x\right)$ , then the remainder from the division of $f\left(x\right)$ by $x-a$ should be equal $0$

Check $-1$ : divide $f\left(x\right)$ by $x+1$

  $x+1\begin{array}{c}\hfill x^2-6x-1\\ \hfill \overline{)\begin{array}{l}{x}^3-5x^2-7x+51\\ \underset{\_}{\begin{array}{l}{x}^3+x^2\\ --\end{array}}\\ -6x^2-7x+51\\ \underset{\_}{\begin{array}{l}-6x^2-6x\\ ++\end{array}}\\ -x+51\\ \underset{\_}{\begin{array}{l}-x-1\\ ++\end{array}}\\ 52\\ \end{array}}\end{array}$

The quotient is $x^2-6x-1$ , and the remainder is $52$

Check $1$ : divide $f\left(x\right)$ by $x-1$

  $x-1\begin{array}{c}\hfill x^2-4x-11\\ \hfill \overline{)\begin{array}{l}{x}^3-5x^2-7x+51\\ \underset{\_}{\begin{array}{l}{x}^3-x^2\\ -+\end{array}}\\ -4x^2-7x+51\\ \underset{\_}{\begin{array}{l}-4x^2+4x\\ +-\end{array}}\\ -11x+51\\ \underset{\_}{\begin{array}{l}-11x+11\\ +-\end{array}}\\ 40\\ \end{array}}\end{array}$

The quotient is $x^2-4x-11$ , and the remainder is $40$

Check $-3$ : divide $f\left(x\right)$ by $x+3$

  $x+3\begin{array}{c}\hfill x^2-8x+17\\ \hfill \overline{)\begin{array}{l}{x}^3-5x^2-7x+51\\ \underset{\_}{\begin{array}{l}{x}^3+3x^2\\ --\end{array}}\\ -8x^2-7x+51\\ \underset{\_}{\begin{array}{l}-8x^2-24x\\ ++\end{array}}\\ 17x+51\\ \underset{\_}{\begin{array}{l}17x+51\\ --\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $x^2-4x-11$ , and the remainder is $40$

Since the remainder is $0$ , then $-3$ is the root, and $x+3$ is the factor

  $x^3-5x^2-7x+51=\left(x+3\right)\left(x^2-8x+17\right)$

  $\begin{array}{l}\text{For}\text{a}\text{quadratic}\text{equation}\text{of}\text{the}\text{form}a{x}^2+bx+c=0\text{the}\text{solutions}\text{are}\\ x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{For}a=1,b=-8,c=17:x_{1,2}=\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\cdot 1\cdot 17}}{2\cdot 1}\\ x_{1,2}=\frac{8\pm \sqrt{64-68}}{2}\\ x_{1,2}=\frac{8+\sqrt{4}i}{2},\frac{8-\sqrt{4}i}{2}\\ x=4+i,x=4-i\end{array}$

So, the zeros of function $f\left(x\right)$ is

  $x=-3,x=4+i,x=4-i$

(b)

To determine

To find : linear factor of $f\left(x\right)=x^3-5x^2-7x+51$

Answer to Problem 97CR

  $\left(x+3\right)\left(x^2-8x+17\right)$

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-5x^2-7x+51$

Concept used:

Zeros of polynomial: -

It is defined by as any real value of $x$ , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial $P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_2{x}^2+a_{1}x+a{0}_{}$ has any rational roots, then they must be of the form $\pm \left\{\frac{\text{factor of }{a}_0}{\text{factor of }{a}_n}\right\}$

Calculation:

  $f\left(x\right)=x^3-5x^2-7x+51$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 3,\pm 17,\pm 51$ . These are the possible values for $f\left(x\right)$

If $a$ is a root of the polynomial $f\left(x\right)$ , then the remainder from the division of $f\left(x\right)$ by $x-a$ should be equal $0$

Check $-3$ : divide $f\left(x\right)$ by $x+3$

  $x+3\begin{array}{c}\hfill x^2-8x+17\\ \hfill \overline{)\begin{array}{l}{x}^3-5x^2-7x+51\\ \underset{\_}{\begin{array}{l}{x}^3+3x^2\\ --\end{array}}\\ -8x^2-7x+51\\ \underset{\_}{\begin{array}{l}-8x^2-24x\\ ++\end{array}}\\ 17x+51\\ \underset{\_}{\begin{array}{l}17x+51\\ --\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $x^2-4x-11$ , and the remainder is $40$

Since the remainder is $0$ , then $-3$ is the root, and $x+3$ is the factor

  $x^3-5x^2-7x+51=\left(x+3\right)\left(x^2-8x+17\right)$

(c)

To determine

To find : x-intercepts using the factorization of $f\left(x\right)=x^3-5x^2-7x+51$ and use graphing utility to verify that the real zeros are the only x-intercepts

Answer to Problem 97CR

  $x=\left(-3,0\right)$

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-5x^2-7x+51$

Concept used:

Zeros of polynomial: -

It is defined by as any real value of $x$ , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial $P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...}+a_2{x}^2+a_{1}x+a{0}_{}$ has any rational roots, then they must be of the form $\pm \left\{\frac{\text{factor of }{a}_0}{\text{factor of }{a}_n}\right\}$

Calculation:

  $f\left(x\right)=x^3-5x^2-7x+51$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 3,\pm 17,\pm 51$ . These are the possible values for $f\left(x\right)$

If $a$ is a root of the polynomial $f\left(x\right)$ , then the remainder from the division of $f\left(x\right)$ by $x-a$ should be equal $0$

Check $-3$ : divide $f\left(x\right)$ by $x+3$

  $x+3\begin{array}{c}\hfill x^2-8x+17\\ \hfill \overline{)\begin{array}{l}{x}^3-5x^2-7x+51\\ \underset{\_}{\begin{array}{l}{x}^3+3x^2\\ --\end{array}}\\ -8x^2-7x+51\\ \underset{\_}{\begin{array}{l}-8x^2-24x\\ ++\end{array}}\\ 17x+51\\ \underset{\_}{\begin{array}{l}17x+51\\ --\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $x^2-4x-11$ , and the remainder is $40$

Since the remainder is $0$ , then $-3$ is the root, and $x+3$ is the factor

  $x^3-5x^2-7x+51=\left(x+3\right)\left(x^2-8x+17\right)$

Graph: