
How do you use the rational zeros theorem to make a list of all possible rational zeros, and use Descartes’ rule of signs to list the possible positive/negative zeros of $f(x) = - 17{x^3} + 5{x^2} + 34x - 10$ ?
Answer
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Hint: 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 functions will be the list of all the possible rational zeros of the function f(x).
For Descartes’ rule of a sign, we have to put the function f(x) in the descending order of powers and then count the number of changes in sign of the coefficient of the terms of the function. That count and the number less than that by an even number will be the number of positive roots of the function. To find the number of negative roots, we will follow the same process with f(-x).
Complete step-by-step answer:
First, we have to apply the rational zeros theorem on $f(x) = - 17{x^3} + 5{x^2} + 34x - 10$ to make a list of all possible rational zeros. For that, the function should be in the descending order of the power.
As we know that the rational 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 find all the positive and negative factors of the last term that are constant.
Here, we have -10 as the constant, all of its factors will be: $ \pm 1, \pm 2, \pm 5, \pm 10,$
We have -17 as the leading coefficient. The factors will be: $ \pm 1, \pm 17$.
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,
$ \pm \dfrac{1}{1}, \pm \dfrac{2}{1}, \pm \dfrac{5}{1}, \pm \dfrac{{10}}{1}, \pm \dfrac{1}{{17}}, \pm \dfrac{2}{{17}}, \pm \dfrac{5}{{17}}, \pm \dfrac{{10}}{{17}}$
That is equal to,
$1,2,5,10,\dfrac{1}{{17}},\dfrac{2}{{17}},\dfrac{5}{{17}},\dfrac{{10}}{{17}}, - 1, - 2, - 5, - 10, - \dfrac{1}{{17}}, - \dfrac{2}{{17}}, - \dfrac{5}{{17}}, - \dfrac{{10}}{{17}}$
Now, we have to use Descartes’ rule of signs to list the possible positive and negative zeros of the function $f(x) = - 17{x^3} + 5{x^2} + 34x - 10$.
We must have our function in descending order of powers. As we know that Descartes’ rule of a sign is used to determine the number of real zeros of a polynomial function.
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.
If we write the coefficients of all the terms of f(x) in the same sequence, it would be:
$ - 17, + 5, + 34, - 10$
As we can see that the pattern of signs is -,+,+,-. Here, it is clearly visible that the sign is changing from the first term to the second term and from the third term to the last term. There are two changes. Therefore, it has two roots.
Now, to find the number of negative roots, we have to check the pattern of a sign of f(-x).
Therefore,
$f( - x) = - 17{\left( { - x} \right)^3} + 5{\left( { - x} \right)^2} + 34\left( { - x} \right) - 10$
That is equal to,
$f( - x) = 17{x^3} + 5{x^2} - 34x - 10$
If we write the coefficients of all the terms of f(x) in the same sequence, it would be:
$ + 17, + 5, - 34, - 10$
As we can see that the pattern of signs is +,+,-,-. Here, it is clearly visible that the sign is changing from the second term to the third term. There is only one change. Therefore, it has only one root.
So, using the rational zeros theorem, we obtained a list of possible real roots of the function
$f(x) = - 17{x^3} + 5{x^2} + 34x - 10$
That is,
$1,2,5,10,\dfrac{1}{{17}},\dfrac{2}{{17}},\dfrac{5}{{17}},\dfrac{{10}}{{17}}, - 1, - 2, - 5, - 10, - \dfrac{1}{{17}}, - \dfrac{2}{{17}}, - \dfrac{5}{{17}}, - \dfrac{{10}}{{17}}$.
Note:
While using the rational zeros theorem, we have to remember that we list both the positive and negative factors of the constant and the leading coefficient of the function. We divide the factors of the constant with the factors of the leading coefficient; we can get the same ratios so remove the terms which are repeated.
For Descartes’ rule of a sign, we have to put the function f(x) in the descending order of powers and then count the number of changes in sign of the coefficient of the terms of the function. That count and the number less than that by an even number will be the number of positive roots of the function. To find the number of negative roots, we will follow the same process with f(-x).
Complete step-by-step answer:
First, we have to apply the rational zeros theorem on $f(x) = - 17{x^3} + 5{x^2} + 34x - 10$ to make a list of all possible rational zeros. For that, the function should be in the descending order of the power.
As we know that the rational 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 find all the positive and negative factors of the last term that are constant.
Here, we have -10 as the constant, all of its factors will be: $ \pm 1, \pm 2, \pm 5, \pm 10,$
We have -17 as the leading coefficient. The factors will be: $ \pm 1, \pm 17$.
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,
$ \pm \dfrac{1}{1}, \pm \dfrac{2}{1}, \pm \dfrac{5}{1}, \pm \dfrac{{10}}{1}, \pm \dfrac{1}{{17}}, \pm \dfrac{2}{{17}}, \pm \dfrac{5}{{17}}, \pm \dfrac{{10}}{{17}}$
That is equal to,
$1,2,5,10,\dfrac{1}{{17}},\dfrac{2}{{17}},\dfrac{5}{{17}},\dfrac{{10}}{{17}}, - 1, - 2, - 5, - 10, - \dfrac{1}{{17}}, - \dfrac{2}{{17}}, - \dfrac{5}{{17}}, - \dfrac{{10}}{{17}}$
Now, we have to use Descartes’ rule of signs to list the possible positive and negative zeros of the function $f(x) = - 17{x^3} + 5{x^2} + 34x - 10$.
We must have our function in descending order of powers. As we know that Descartes’ rule of a sign is used to determine the number of real zeros of a polynomial function.
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.
If we write the coefficients of all the terms of f(x) in the same sequence, it would be:
$ - 17, + 5, + 34, - 10$
As we can see that the pattern of signs is -,+,+,-. Here, it is clearly visible that the sign is changing from the first term to the second term and from the third term to the last term. There are two changes. Therefore, it has two roots.
Now, to find the number of negative roots, we have to check the pattern of a sign of f(-x).
Therefore,
$f( - x) = - 17{\left( { - x} \right)^3} + 5{\left( { - x} \right)^2} + 34\left( { - x} \right) - 10$
That is equal to,
$f( - x) = 17{x^3} + 5{x^2} - 34x - 10$
If we write the coefficients of all the terms of f(x) in the same sequence, it would be:
$ + 17, + 5, - 34, - 10$
As we can see that the pattern of signs is +,+,-,-. Here, it is clearly visible that the sign is changing from the second term to the third term. There is only one change. Therefore, it has only one root.
So, using the rational zeros theorem, we obtained a list of possible real roots of the function
$f(x) = - 17{x^3} + 5{x^2} + 34x - 10$
That is,
$1,2,5,10,\dfrac{1}{{17}},\dfrac{2}{{17}},\dfrac{5}{{17}},\dfrac{{10}}{{17}}, - 1, - 2, - 5, - 10, - \dfrac{1}{{17}}, - \dfrac{2}{{17}}, - \dfrac{5}{{17}}, - \dfrac{{10}}{{17}}$.
Note:
While using the rational zeros theorem, we have to remember that we list both the positive and negative factors of the constant and the leading coefficient of the function. We divide the factors of the constant with the factors of the leading coefficient; we can get the same ratios so remove the terms which are repeated.
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