Question

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given $f\left( x \right)={{x}^{3}}+7{{x}^{2}}+7x-15?$

Answer 1

Hint: We use Descartes’ Rule of signs. Then, we will get the number of positive real roots (zeros) and negative roots (zeros). We use the Rational Roots theorem to find the possible rational roots. Then we apply each one of them to find which of them are the zeros.

Complete step-by-step solution:
We use Descartes’ Rule to find the possible number of positive real zeros and negative real zeros.
Descartes’ Rule of signs states that if the nonzero terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign changes between consecutive coefficients, or is less than it by an even number.
To find the number of negative roots, we will multiply the equation with $-1$ and the number of sign changes or less than it by an even number is equal to the number of roots.
If the number of sign changes is zero or one, then the number of roots equals to the number of sign changes.
Consider the given polynomial function $f\left( x \right)={{x}^{3}}+7{{x}^{2}}+7x-15.$
In this equation, there is only one sign change. Therefore, this polynomial function has only one positive real root.
Now consider $f\left( -x \right)={{\left( -x \right)}^{3}}+7{{\left( -x \right)}^{2}}+7\left( -x \right)-15=-{{x}^{3}}+7{{x}^{2}}-7x-15.$
There are two sign changes in this equation. Therefore, this equation has $0$ or $2$ negative roots.
To find the rational roots, we use the Rational Roots theorem. It states that a polynomial equation in one variable with integer coefficients has a solution of the form $\dfrac{p}{q},$ then $p$ divides the constant term and $q$ divides the leading coefficient.
Here, the leading coefficient is $1.$
The factors of $1$ are $\pm 1.$
The constant term is $15.$
The factors of $15$ are $\pm 3,$ and $\pm 5.$
That is, the possible values of $p=\pm 1,\pm 3,\pm 5,\pm 15$ and $q=\pm 1.$
Therefore, the possible rational roots are $\dfrac{p}{q}=\pm \dfrac{1}{1},\pm \dfrac{3}{1},\pm \dfrac{5}{1},\pm \dfrac{15}{1}=\pm 1,\pm 3,\pm 5,\pm 15.$
When $1$ is applied, $f\left( 1 \right)={{1}^{3}}+7\cdot {{1}^{2}}+7\cdot 1-15=1+7+7-15=0.$
When $-1$ is applied, $f\left( -1 \right)={{\left( -1 \right)}^{3}}+7{{\left( -1 \right)}^{2}}+7\left( -1 \right)-15=-1+7-7-15=-16\ne 0.$
When $3$ is applied, $f\left( 3 \right)={{3}^{3}}+7\cdot {{3}^{2}}+7\cdot 3-15=27+63+21-15=96\ne 0.$
When $-3$ is applied, $f\left( -3 \right)={{\left( -3 \right)}^{3}}+7{{\left( -3 \right)}^{2}}+7\left( -3 \right)-15=-27+63-21-15=0.$
When $-5$ is applied, $f\left( -5 \right)={{\left( -5 \right)}^{3}}+7{{\left( -5 \right)}^{2}}+7\left( -5 \right)-15=-125+175-35-15=0.$
Similarly, we can find $f\left( 5 \right),f\left( -15 \right)$ and $f\left( 15 \right).$
But by Descartes’ Rule, we found that the maximum number of real roots (positive or negative) is $3.$
Here, we have found $3$ real roots.
Therefore, the real roots are $1,-3$ and $-5.$
Hence, the only positive real zero is $1$ and negative zeros are $-3$ and $-5.$

Note: We can use the sum of coefficients shortcut to find the real roots of the polynomial equation as:
The sum of coefficients of the given polynomial equation is $1+7+7-15=0.$
Therefore, $1$ is a zero and $\left( x-1 \right)$ is a factor of the given equation. We find the next factor using $x-1$ and we will get,
$\Rightarrow f\left( x \right)={{x}^{3}}+7{{x}^{2}}+7x-15=\left( x-1 \right)\left( {{x}^{2}}+8x+15 \right)$
Also, ${{x}^{2}}+8x+15=\left( x+3 \right)\left( x+5 \right)$
Therefore, $f\left( x \right)={{x}^{3}}+7{{x}^{2}}+7x-15=\left( x-1 \right)\left( x+3 \right)\left( x+5 \right).$
Therefore, the factors are $x=1,-3,-5.$