Chapter 2, Problem 98CR

(a)

To determine

To find : All the zeros of function $f\left(x\right)=-3x^3-19x^2-4x+12$

Answer to Problem 98CR

  $x=-1,x=\frac{2}{3},x=-6$

Explanation of Solution

Given information :

  $f\left(x\right)=-3x^3-19x^2-4x+12$

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:

  $\begin{array}{l}f\left(x\right)=-3x^3-19x^2-4x+12\\ f\left(x\right)=-\left(3x^3+19x^2+4x-12\right)\end{array}$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ . 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 $x+1$

  $x+1\begin{array}{c}\hfill 3x^2+16x-12\\ \hfill \overline{)\begin{array}{l}3x^3+19x^2+4x-12\\ \underset{\_}{\begin{array}{l}3x^3+3x^2\\ --\end{array}}\\ 16x^2+4x-12\\ \underset{\_}{\begin{array}{l}16x^2+16x\\ --\end{array}}\\ -12x+12\\ \underset{\_}{\begin{array}{l}-12x+12\\ +-\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $3x^2+16x-12$ , and the remainder is $0$

Since the remainder is $0$ , then $-1$ is the root, and $x+1$ is the factor

  $-\left(3x^3+19x^2+4x-12\right)=-\left(x+1\right)\left(3x^2+16x-12\right)$

Now, find the roots of $3x^2+16x-12$

  $\begin{array}{l}\text{For}\text{a}\text{quadratic}\text{equation}\text{of}\text{the}\text{form}a{x}^2+bx+c=0\text{the}\text{solutions}\text{are}\\ x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \text{For}a=3,b=16,c=-12:x_{1,2}=\frac{-16\pm \sqrt{{16}^2-4\cdot 3\left(-12\right)}}{2\cdot 3}\\ x_{1,2}=\frac{-16\pm \sqrt{256+144}}{6}\\ x_{1,2}=\frac{-16\pm \sqrt{400}}{6}\\ x_{1,2}=\frac{-16+20}{6},\frac{-16-20}{6}\\ x=\frac{2}{3},x=-6\end{array}$

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

  $x=-1,x=\frac{2}{3},x=-6$

(b)

To determine

To find : linear factor of $f\left(x\right)=-3x^3-19x^2-4x+12$

Answer to Problem 98CR

  $-\left(x+1\right)\left(x+6\right)\left(3x-2\right)$

Explanation of Solution

Given information :

  $f\left(x\right)=-3x^3-19x^2-4x+12$

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:

  $\begin{array}{l}f\left(x\right)=-3x^3-19x^2-4x+12\\ f\left(x\right)=-\left(3x^3+19x^2+4x-12\right)\end{array}$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ . 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 $x+1$

  $x+1\begin{array}{c}\hfill 3x^2+16x-12\\ \hfill \overline{)\begin{array}{l}3x^3+19x^2+4x-12\\ \underset{\_}{\begin{array}{l}3x^3+3x^2\\ --\end{array}}\\ 16x^2+4x-12\\ \underset{\_}{\begin{array}{l}16x^2+16x\\ --\end{array}}\\ -12x+12\\ \underset{\_}{\begin{array}{l}-12x+12\\ +-\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $3x^2+16x-12$ , and the remainder is $0$

Since the remainder is $0$ , then $-1$ is the root, and $x+1$ is the factor

  $-\left(3x^3+19x^2+4x-12\right)=-\left(x+1\right)\left(3x^2+16x-12\right)$

Now, factor $3x^2+16x-12$

  $\begin{array}{l}3x^2+16x-12=3x^2+18x-2x-12\\ =3x\left(x+6\right)-2\left(x+6\right)\\ =\left(x+6\right)\left(3x-2\right)\end{array}$

So, the linear factorization of $f\left(x\right)$ is

  $-\left(x+1\right)\left(x+6\right)\left(3x-2\right)$

(c)

To determine

To find : x-intercepts using the factorization of $f\left(x\right)=-3x^3-19x^2-4x+12$ and use graphing utility to verify that the real zeros are the only x-intercepts

Answer to Problem 98CR

  $x=\left(-1,0\right),\left(-6,0\right)\text{ and }\left(\frac{2}{3},0\right)$

Explanation of Solution

Given information :

  $f\left(x\right)=-3x^3-19x^2-4x+12$

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:

  $\begin{array}{l}f\left(x\right)=-3x^3-19x^2-4x+12\\ f\left(x\right)=-\left(3x^3+19x^2+4x-12\right)\end{array}$

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ . 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 $x+1$

  $x+1\begin{array}{c}\hfill 3x^2+16x-12\\ \hfill \overline{)\begin{array}{l}3x^3+19x^2+4x-12\\ \underset{\_}{\begin{array}{l}3x^3+3x^2\\ --\end{array}}\\ 16x^2+4x-12\\ \underset{\_}{\begin{array}{l}16x^2+16x\\ --\end{array}}\\ -12x+12\\ \underset{\_}{\begin{array}{l}-12x+12\\ +-\end{array}}\\ 0\\ \end{array}}\end{array}$

The quotient is $3x^2+16x-12$ , and the remainder is $0$

Since the remainder is $0$ , then $-1$ is the root, and $x+1$ is the factor

  $-\left(3x^3+19x^2+4x-12\right)=-\left(x+1\right)\left(3x^2+16x-12\right)$

Now, factor $3x^2+16x-12$

  $\begin{array}{l}3x^2+16x-12=3x^2+18x-2x-12\\ =3x\left(x+6\right)-2\left(x+6\right)\\ =\left(x+6\right)\left(3x-2\right)\end{array}$

So, the linear factorization of $f\left(x\right)$ is

  $\begin{array}{l}-\left(x+1\right)\left(x+6\right)\left(3x-2\right)=0\\ x+1=0\Rightarrow x=-1\\ x+6=0\Rightarrow x=-6\\ 3x-2=0\Rightarrow x=\frac{2}{3}\end{array}$

Graph: