Question

How do you find all the rational zeros of a polynomial function?

Answer 1

Hint: In this question we are asked to find how to find the rational zeros of the polynomial function,
We can find the rational zeros of a polynomial function using rational root theorem, by using the definition of the rational root theorem we will find the rational zeros of any polynomial function.

Complete step by step solution:
We can find the rational roots of a polynomial function using a rational root theorem,
Now we will use the definition of rational root theorem which states that, given a polynomial in a single variable with integer coefficients:
${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........ + {a_0}$, with ${a_n} \ne 0$ and ${a_0} \ne 0$, any rational root of that polynomial are expressible in the form $\dfrac{p}{q}$ for integers $p$, $q$ where $p$ is a divisor of the constant term ${a_0}$ and $q$ is a divisor of the coefficient term ${a_n}$ of the leading term.
Let’s take example: ${x^3} + 2{x^2} + 3x + 6 = 0$,
Here, the coefficient of leading term = 1,
So, the factors of 1 are is also 1,
Coefficient of constant term = 6,
So, the factors of 6 are 1,2,3,6,
Therefore the rational roots of the equation may be $\dfrac{{ \pm 1}}{1}$,$\dfrac{{ \pm 2}}{1}$,$\dfrac{{ \pm 3}}{1}$,$\dfrac{{ \pm 6}}{1}$,
The rational roots are $ \pm 1$,$ \pm 2$,$ \pm 3$and $ \pm 6$,
So, now we will check that these roots satisfy the given polynomial, which is ${x^3} + 2{x^2} + 3x + 6 = 0$,
Let us take $f\left( x \right) = {x^3} + 2{x^2} + 3x + 6$,
First substitute $x = 1$in the polynomial,
$ \Rightarrow $$f\left( 1 \right) = {1^3} + 2{\left( 1 \right)^2} + 3\left( 1 \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( 1 \right) = 1 + 2 + 3 + 6 = 11$, which is not equal to zero so, 1 is not the rational root of the polynomial.
Now substitute $x = - 1$in the polynomial,
$ \Rightarrow $$f\left( { - 1} \right) = {\left( { - 1} \right)^3} + 2{\left( { - 1} \right)^2} + 3\left( { - 1} \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( { - 1} \right) = - 1 + 2 - 3 + 6 = 4$, which is not equal to zero so, -1 is not the rational root of the polynomial.
Now substitute $x = 2$ in the polynomial,
$ \Rightarrow $$f\left( 2 \right) = {\left( 2 \right)^3} + 2{\left( 2 \right)^2} + 3\left( 2 \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( 2 \right) = 8 + 8 + 6 + 6 = 28$, which is not equal to zero so, 2 is not the rational root of the polynomial.
Now substitute $x = - 2$ in the polynomial,
$ \Rightarrow $$f\left( { - 2} \right) = {\left( { - 2} \right)^3} + 2{\left( { - 2} \right)^2} + 3\left( { - 2} \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( { - 2} \right) = - 8 + 8 - 6 + 6 = 0$, which is equal to zero so, -2 is the rational root of the polynomial.
Now substitute $x = 3$in the polynomial,
$ \Rightarrow $$f\left( 3 \right) = {\left( 3 \right)^3} + 2{\left( 3 \right)^2} + 3\left( 3 \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( 2 \right) = 27 + 18 + 9 + 6 = 60$, which is not equal to zero so, 3 is not the rational root of the polynomial.
Now substitute ..in the polynomial,
$ \Rightarrow $$f\left( { - 3} \right) = {\left( { - 3} \right)^3} + 2{\left( { - 3} \right)^2} + 3\left( { - 3} \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( { - 3} \right) = - 27 + 18 - 9 + 6 = - 12$, which is not equal to zero so, -3 is not the rational root of the polynomial.
Now substitute $x = 6$ in the polynomial,
$ \Rightarrow $$f\left( 6 \right) = {\left( 6 \right)^3} + 2{\left( 6 \right)^2} + 3\left( 6 \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( 6 \right) = 216 + 72 + 18 + 6 = 312$, which is not equal to zero, so 6 is not the rational root of the polynomial.
Now substitute $x = - 6$ in the polynomial,
$ \Rightarrow $$f\left( { - 6} \right) = {\left( { - 6} \right)^3} + 2{\left( { - 6} \right)^2} + 3\left( { - 6} \right) + 6$,
Now simplifying we get,
$ \Rightarrow f\left( { - 6} \right) = - 216 + 72 - 18 + 6 = - 156$, which is not equal to zero so, -6 is not the rational root of the polynomial.
Out of these values $x = - 2$ makes the polynomial on the left side 0, hence $x + 2$ would be a factor of the polynomial. Now divide the given polynomial by $x + 2$,
$ \Rightarrow \left. {x + 2} \right){x^3} + 2{x^2} + 3x + 6\left( {{x^2} + 3} \right.$
                 \[\underline {{x^3} + \,2{x^2}} \]
                                     \[3x + 6\]
                                     \[\underline {3x + 6} \]
                                           \[0\]
So, the quotient is \[{x^2} + 3\], so, this is another factor of the polynomial, now factoring the quotient we get,
\[ \Rightarrow {x^2} + 3 = 0\],
Now subtract 3 from both sides the equation we get,
\[ \Rightarrow {x^2} + 3 - 3 = 0 - 3\],
Now simplifying we get,
\[ \Rightarrow {x^2} = - 3\],
Further simplifying we get,
\[ \Rightarrow x = \sqrt { - 3} = \pm 3i\],
So, the roots of the given polynomial are \[ - 2\] and \[ \pm 3i\].

\[\therefore \] We used rational root theorem to find the roots of the given polynomial i.e., ${x^3} + 2{x^2} + 3x + 6 = 0$ which are \[ - 2\] and \[ \pm 3i\].


Note: The rational roots theorem is a very useful theorem. It tells you that given a polynomial function with integer or whole number coefficients, a list of possible solutions can be found by listing the factors of the constant, or last term, over the factors of the coefficient of the leading term.