Chapter EP, Problem 6.7.11EP

To determine

To calculate: The zeros of the function $f\left(x\right)=10x^3+7x^2-82x+56$ .

Answer to Problem 6.7.11EP

The zeros of the function $f\left(x\right)=10x^3+7x^2-82x+56$ are $\frac{4}{5},-\frac{7}{2}\text{ and }2$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=10x^3+7x^2-82x+56$ .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(x\right)=10x^3+7x^2-82x+56$ .

Observe that degree of polynomial is 3, so the functions has 3 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(x\right)=10x^3+7x^2-82x+56$

The coefficients are $+10,+7,-82,+56$ .

There are 2 sign changes, so there are 2 or 0 positive real zeros.

Now,

  $\begin{array}{c}f\left(-x\right)=10{\left(-x\right)}^3+7{\left(-x\right)}^2-82\left(-x\right)+56\\ f\left(-x\right)=-10x^3+7x^2+82x+56\end{array}$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-x\right)=-x^3-7x^2-16x-10$ .

The coefficients are $-10,+7,+82,+56$ .

There is 1 sign change, so there is 1 negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 2& 1& 0& 2+1+0=3\\ 0& 1& 2& 0+1+2=3\end{array}$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(x\right)$ for real values of $x$ .

  $\begin{array}{ccccc}x& 10& 7& -82& 56\\ -1& 10& -3& -79& 135\\ 0& 10& 7& -82& 56\\ 1& 10& 17& -65& -9\\ 2& 10& 27& -28& 0\end{array}$

As observed one zero is resulted at $x=2$ . Since, only one real positive zero is obtained so the remaining one positive and one negative real zero is obtained with help of quadratic formula.

Now, the depressed polynomial is obtained is $f\left(x\right)=10x^2+27x-28$ .

Use the quadratic formula to obtain the remaining zeros.

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-27\pm \sqrt{729+1120}}{20}\\ x=\frac{-27\pm 43}{20}\end{array}$

Either $x=\frac{-27+43}{20}\text{ or }x=\frac{-27-43}{20}$ .

That is, $x=\frac{4}{5},-\frac{7}{2}$ .

Thus, the zeros of the function $f\left(x\right)=10x^3+7x^2-82x+56$ are $\frac{4}{5},-\frac{7}{2}\text{ and }2$ .