Chapter 4, Problem 55RE

To determine

To calculate:

All the real zeroes of the following functions of polynomial with the help of Rational zeroes theorem and the factor $f$ over the real zeroes:

  $f\left(x\right)=4x^3+4x^2-7x+2$

Answer to Problem 55RE

The real zeroes of the given polynomial function are $f\left(x\right)=4x^3+4x^2-7x+2$ are $-2,\frac{1}{2}$

Explanation of Solution

Given information:

  $f\left(x\right)=4x^3+4x^2-7x+2$

Formula Used:

The potential rational zeroes is given by the following expression:

  $x=\pm \frac{p}{q}$

Where,

Quotients of the factors of the last term: $p$

The factors of the leading coefficient: $q$

Calculation:

The given function is:

  $f\left(x\right)=4x^3+4x^2-7x+2$

The degree of the given polynomial function is $3$ which shows that $f$ must have at most $3$ real zeroes.

Now, the values of $p,q$ are defined as:

  $\begin{array}{l}p=2\\ q=4\end{array}$

Factors of $p$ are $\pm 1,\pm 2$ and factors of $q$ is $\pm 1,\pm 2,\pm 4$ .

So the potential rational zeroes are defined as:

  $\frac{p}{q}=\pm 1,\pm 2,\pm \frac{1}{2},\pm \frac{1}{4}$

Now test the potential rational zeroes with the help of the synthetic division:

If the value of remainder is zero $f\left(1\right)=0$ then $1$ is zero of $f$ and $x-1$ is a factor.

  $\begin{array}{ccccc}1& 4& 4& -7& 2\\ & & 4& 8& 1\\ & 4& 8& 1& 3\end{array}$

Here, the remainder is $f\left(1\right)=3$ . That’s why $1$ is not zero of $q_1$ .

Now test for $-1$ by using the synthetic division:

  $\begin{array}{ccccc}1& 4& 4& -7& 2\\ & & -4& 0& 7\\ & 4& 0& -7& 9\end{array}$

Here, the remainder is $f\left(-1\right)=9$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $2$ by using the synthetic division:

  $\begin{array}{ccccc}2& 4& 4& -7& 2\\ & & 8& 24& 34\\ & 4& 12& 17& 36\end{array}$

Here, the remainder is $f\left(2\right)=36$ . That’s why $-2$ is not zero of $q_1$ .

Now test for $-2$ by using the synthetic division:

  $\begin{array}{ccccc}2& 4& 4& -7& 2\\ & & -8& 8& -2\\ & 4& -4& 1& 0\end{array}$

Here, the remainder is $f\left(-2\right)=0$ . That’s why $-2$ is zero of $q_1$ and $x-\left(-2\right)=x+2$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=4x^3+4x^2-7x+2\\ f\left(x\right)=\left(x+2\right)\left(4x^2-4x+1\right)\end{array}$

The solution of the given polynomial equation $f\left(x\right)=4x^3+4x^2-7x+2$ are the zeroes of the $q_1\left(x\right)=4x^2-4x+1$ .

The potential rational zeroes of $q_1\left(x\right)=4x^2-4x+1$ :

  $\frac{p}{q}=\pm 1,\pm \frac{1}{2},\pm \frac{1}{4}$

Now test the potential rational zeroes with the help of the synthetic division:

Now test for $1$ : :

  $\begin{array}{cccc}1& 4& -4& 1\\ & & 4& 0\\ & 4& 0& 1\end{array}$

Here, the remainder is $f\left(1\right)=1$ . That’s why $1$ is not zero of $q_1$ .

Now test for $-1$ by using the synthetic division:

  $\begin{array}{cccc}1& 4& -4& 1\\ & & -4& 8\\ & 4& -8& 9\end{array}$

Here, the remainder is $f\left(-1\right)=9$ . That’s why $-1$ is not zero of $q_1$ .

Now test for $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ by using the synthetic division:

  $\begin{array}{cccc}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& 4& -4& 1\\ & & 2& -1\\ & 4& -2& 0\end{array}$

Here, the remainder is $f\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=0$ . That’s why $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ is zero of $q_1$ and $x-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ is a factor. Use the bottom row of synthetic division to factor $f$ :

  $\begin{array}{l}f\left(x\right)=4x^3+4x^2-7x+2\\ f\left(x\right)=\left(x+2\right)\left(4x^2-4x+1\right)\\ f\left(x\right)=\left(x+2\right)\left(x-\frac{1}{2}\right)\left(4x-2\right)\\ f\left(x\right)=4\left(x+2\right){\left(x-\frac{1}{2}\right)}^2\end{array}$

The real zeroes of $f$ are $-2,\frac{1}{2}$ . When factored, $f$ is

  $\begin{array}{l}f\left(x\right)=4x^3+4x^2-7x+2\\ f\left(x\right)=4\left(x+2\right){\left(x-\frac{1}{2}\right)}^2\end{array}$

Hence, real zeroes are $-2,\frac{1}{2}$ .