Chapter 5.8, Problem 9CYU

To determine

To calculate: The simplified form of the equation $f\left(x\right)=4x^4-12x^3+25x^2-14x-15$ ,

Find all the possible zeros of rational function.

Answer to Problem 9CYU

The simplified form of the equation $f\left(x\right)=4x^4-12x^3+25x^2-14x-15$ has no real roots.

Explanation of Solution

Given information:

The given equation:

  $f\left(x\right)=4x^4-12x^3+25x^2-14x-15$

Calculation:

Consider the expression as:

  $f\left(x\right)=4x^4+13x^3-8x^2+13x-12$

Now take $f\left(x\right)=0$ ,

Here,

  $\begin{array}{l}f\left(x\right)=0\\ 4x^4-12x^3+25x^2-14x-15=0\end{array}$

This function is not a perfect cube.

So in grouping solve the equation.

As,

  $f\left(x\right)=4x^4-12x^3+25x^2-14x-15$

Now find the rational possible roots:

  $4x^4-12x^3+25x^2-14x-15=0$

The leading coefficient is $4$ and the trilling coefficient is $-15$ .

Now factor of leading coefficient is: $1,2,4$

Now to find the list of positive rational solutions for the expression look at the factors of the constants.

That are:

  $1,3,5,15$

Here the last two terms subtracting values from the cube of the first and want to get zero,

So as have to take the positive number, and mostly take the less than mid-way.

So take a startup with $1$ and check the result, then again tried to get zero.

  $\begin{array}{c}4x^4-12x^3+25x^2-14x-15=0\\ 4{\left(1\right)}^4-12{\left(1\right)}^3+25{\left(1\right)}^2-14\left(1\right)-15=0\\ 4-12+25-14-15=0\\ -12=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $3$ .

  $\begin{array}{c}4x^4-12x^3+25x^2-14x-15=0\\ 4{\left(3\right)}^4-12{\left(3\right)}^3+25{\left(3\right)}^2-14\left(3\right)-15=0\\ 324-12\left(27\right)+25\left(9\right)+42-15=0\\ 325-324+225+42-15=0\end{array}$

To get zero:

  $\begin{array}{c}325-324+225+42-15=0\\ 253=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $5$ .

  $\begin{array}{c}4x^4-12x^3+25x^2-14x-15=0\\ 4{\left(5\right)}^4-12{\left(5\right)}^3+25{\left(5\right)}^2-14\left(5\right)-15=0\\ 4\left(625\right)-12\left(125\right)+25\left(25\right)-70-15=0\\ 2500-1500+625-70-15=0\end{array}$

To get zero:

  $\begin{array}{c}2500-1500+625-70-15=0\\ 1540=0\end{array}$

So check with the value $3,2$ and $p/q=1.50$ .

  $\begin{array}{c}4x^4-12x^3+25x^2-14x-15=0\\ 4{\left(1.50\right)}^4-12{\left(1.50\right)}^3+25{\left(1.50\right)}^2-14\left(1.50\right)-15=0\\ 20.25-40.5+56.25-21-15=0\\ 0=0\end{array}$

So the value is zero its rational root.

So check with the value $-1,2$ and $p/q=-0.50$ .

  $\begin{array}{c}4x^4-12x^3+25x^2-14x-15=0\\ 4{\left(-0.50\right)}^4-12{\left(-0.50\right)}^3+25{\left(-0.50\right)}^2-14\left(-0.50\right)-15=0\\ 0.25+1.5+6.25+7-15=0\\ 0=0\end{array}$

Here got two real root, it means $\left(2x+1\right),\left(2x-3\right)$ is a factor.

Now divide the expression $4x^4-12x^3+25x^2-14x-15=0$ to the factor $\left(2x-3\right)$ , to check for the other possible solutions, because its not possible to divide with two different polynomial.

  $\begin{array}{l}2x-3\begin{array}{c}\hfill 2x^3-3x^2+8x+5\\ \hfill \overline{)\begin{array}{ccccccccc}4x^4& -& 12x^3& +& 25x^2& -& 14x& -& 15\end{array}}\end{array}\\ \begin{array}{ccc}4x^4& -& 6x^3\end{array}\\ {\begin{array}{ccc}-6x^3& +& 25x\end{array}}^2\\ \begin{array}{ccc}-6x^3& +& 9x^2\end{array}\\ \begin{array}{ccc}16x^2& -& 14x\end{array}\\ \begin{array}{ccc}16x^2& -& 24\end{array}x\\ \begin{array}{ccc}10x& -& 15\end{array}\\ \begin{array}{ccc}10x& -& 15\end{array}\\ \begin{array}{c}0\end{array}\end{array}$

So the other factor is:

  $2x^3-3x^2+8x+5=0$

Now find the factors of the expression $2x^3-3x^2+8x+5=0$ :

  $2x^3-3x^2+8x+5=0$

The leading coefficient is $2$ and the trilling coefficient is $5$ .

Now factor of leading coefficient is: $1,2$

Now to find the list of positive rational solutions for the expression look at the factors of the constants.

That are:

  $1,5$

Here the last two terms subtracting values from the cube of the first and want to get zero,

So as have to take the positive number, and mostly take the less than mid-way.

So take a startup with $1,1$ and check the result, then again tried to get zero.

  $\begin{array}{c}2x^3-3x^2+8x+5=0\\ 2{\left(1\right)}^3-3{\left(1\right)}^2+8\left(1\right)+5=0\\ 2-3+8+5=0\\ 12=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $-1,1$ .

  $\begin{array}{c}2x^3-3x^2+8x+5=0\\ 2{\left(-1\right)}^3-3{\left(-1\right)}^2+8\left(-1\right)+5=0\\ -2-3-8+5=0\\ -8=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $-1,2$ and $p/q=-0.50$ .

  $\begin{array}{c}2x^3-3x^2+8x+5=0\\ 2{\left(-0.50\right)}^3-3{\left(-0.50\right)}^2+8\left(-0.50\right)+5=0\\ -0.25-0.75-4+5=0\\ 0=0\end{array}$

The rational root is zero.

So the value is zero its rational root.

Here got two real root, it means $\left(2x+1\right)$ is a factor.

Now divide the expression $2x^3-3x^2+8x+5=0$ to the factor $\left(2x+1\right)$ , to check for the other possible solutions, because its not possible to divide with two different polynomial.

  $\begin{array}{l}2x+1\begin{array}{c}\hfill x^2-2x+5\\ \hfill \overline{)\begin{array}{ccccccc}2x^3& -& 3x^2& +& 8x& +& 5\end{array}}\end{array}\\ \begin{array}{ccc}2x^3& +& x^2\end{array}\\ \begin{array}{ccc}-4x^2& +& 8x\end{array}\\ \begin{array}{ccc}-4x^2& -& 2x\end{array}\\ \begin{array}{ccc}10x& +& 5\end{array}\\ \begin{array}{ccc}10x& +& 5\end{array}\\ \begin{array}{c}0\end{array}\end{array}$

So the other factor is:

  $x^2-2x+5$

This term has no factors.

Here these two factors are not same so this equation cannot be factored any more and had no rational roots.

Thus, the simplified form of the equation $f\left(x\right)=4x^4-12x^3+25x^2-14x-15$ has $\left(x^2-2x+5\right)\cdot \left(2x+1\right)\cdot \left(2x-3\right)$ rational roots.