Question

How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descartes’ rule of signs to list the possible positive/negative zeros of $f(x) = 3{x^3} + 3{x^2} - 11x - 10$?

Answer 1

To use the Rational Zeros theorem, first we have to find all the factors of the constant in the function and all the factors of the leading coefficient and then we have to find their ratio in every possible combination.

Those fractions will be the list of all the possible rational zeros of the function $f(x)$. For Descartes’ rule of sign, we have to put the function $f(x)$ in the descending order of powers and then count the number of changes in signs of the coefficient of the terms of the function.

That count and the numbers less than that by an even number will be the number of positive roots of the function. To find the number of the negative roots, we will follow the same process with $f( - x)$.

Complete step by step solution:
(i)
Firstly, we have to apply the Rational Zeros theorem on $f(x) = 3{x^3} + 3{x^2} - 11x - 10$ to make a list of all possible rational zeros. For that the function should be in the descending order of powers. Since, the given a function is already in the descending order of its power, we will move forward.

As we know that the Rational Zeros theorem says that we can find all the possible rational roots of a polynomial by dividing all the positive and negative factors of the last term by all the positive and negative factors of the first term. Therefore, we will first find all the positive and negative factors of the last term i.e., the constant.

Since, here we have $- 10$ as the constant, all of its factors will be:
$\pm 1, \pm 2, \pm 5, \pm 10$

And we have $3$ as the leading coefficient i.e., the coefficient of the highest power term. Its factors will be:
$\pm 1, \pm 3$

(ii)
Since we have all the factors of both the constant and the leading coefficient, we will find the ratio of them keeping factors of the constant in the numerator and the factors of the leading coefficient in the denominator.

Therefore, we will get the following ratios:

$\pm \dfrac{1}{1}, \pm \dfrac{2}{1}, \pm \dfrac{5}{1}, \pm \dfrac{{10}}{1}, \pm \dfrac{1}{3}, \pm \dfrac{2}{3}, \pm \dfrac{5}{3}, \pm \dfrac{{10}}{3}$

On simplifying the fractions and separating them according to their signs, we get:
$1,2,5,10,\dfrac{1}{3},\dfrac{2}{3},\dfrac{5}{3},\dfrac{{10}}{3}, - 1, - 2, - 5, - 10, - \dfrac{1}{3}, - \dfrac{2}{3}, - \dfrac{5}{3}, - \dfrac{{10}}{3}$

Therefore, this is the list of all the possible rational roots of the function $f(x) = 3{x^3} + 3{x^2} - 11x - 10$.

(iii)
Now, we have to use Descartes’ rule of signs to list the possible positive and negative zeros of the function $f(x) = 3{x^3} + 3{x^2} - 11x - 10$. For this rule too, we must have our function in a descending order of powers which here, we already have. As we know that Descartes’ rule of sign is used to determine the number of real zeros of a polynomial function.

It tells us that the number of positive real zeros in the given function $f(x)$ is the same or less than by an even number as the number of changes in the sign of the coefficients of the terms.

Similarly, The number of negative real zeros of the function $f(x)$ is the same as the number of changes in the sign of the coefficients of the terms of $f( - x)$ or less than this by an even number.

So, we are given:
$f(x) = 3{x^3} + 3{x^2} - 11x - 10$

If we write the coefficients of all the terms in the same sequence, it would be:
$+ 3, + 3, - 11, - 10$

As we can see that the pattern of signs is $+ , + , - , -$ it is clearly visible that since the sign is changing from the second term to the third term, there is only one change. Therefore, it has only one positive
root.

(iv)
To find the number of negative roots, we have to check the pattern of signs of $f( - x)$. Therefore, we have $f( - x)$ as:

$f( - x) = 3{( - x)^3} + 3{( - x)^2} - 11( - x) - 10 \\\ f( - x) = - 3{x^3} + 3{x^2} + 11x - 10 \\\$

If we write the coefficients of all the terms in the same sequence, it would be:
$- 3, + 3, + 11, - 10$

As we can see that the pattern of signs is $- , + , + , -$ it is clearly visible that first, the sign changes from first term to second term and then it changes again from the third term to fourth term. Since the change occurred two times, we can say that there are either two negative roots or zero negative roots as zero is an even number less than by two.

Hence, using the Rational Zeros theorem, we obtained a list of possible real roots of the function $f(x) =

3{x^3} + 3{x^2} - 11x - 10$i.e.,$1,2,5,10,\dfrac{1}{3},\dfrac{2}{3},\dfrac{5}{3},\dfrac{{10}}{3}, - 1, - 2, - 5, - 10, - \dfrac{1}{3}, - \dfrac{2}{3}, - \dfrac{5}{3}, - \dfrac{{10}}{3}$

And, using the Descartes’ rule of signs we concluded that there is only one positive root of the function and either two or zero negative roots of the function.

Note: While using the Rational Zeros Theorem, we have to keep in mind that we list both the positive as well as negative factors of the constant and the leading coefficient of the function. Also, when we divide the factors of the constant with the factors of the leading coefficients, we can get the same ratios so remove the terms which are repeated. In Descartes’ rule, the function needs to be in the descending order of the powers i.e., the highest power term should be at first and then the following terms should be in decreasing order of their powers otherwise the whole solution will get wrong.