SAT Practice Question:
\(P(x)=2x^3-18x\) Given the polynomial function \(P\) defined above, what are its zeros? a) \(\{-9, -6, 2, 3\}\) b) \(\{9, 0, 2\}\) c) \(\{-3, 3\}\) d) \(\{-3, 0, 3\}\) Source: Khan Academy |
Before we actually solve and find roots, we are first going to go over some important theorems for polynomials. The first is the Rational Roots Theorem which says that for any polynomial function that has integer coefficients, any rational root of that function must be of the form \(\large{\frac{p}{q}}\) where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. Remember that a rational number is any number that can be written as the ratio of two integers.
To see why this is true, consider a polynomial function in the form \(f\left(x\right)=ax^n+bx^{\left(n-1\right)}+cx^{\left(n-2\right)}+...+k\) where \(\left(qx+p\right)\) is a factor. This would mean that \(\large{\frac{p}{q}}\) is a root of \(f\left(x\right)\). If all of the coefficients are integers (which is a requirement for the Rational Roots Theorem), then \(q\) must be a factor of \(a\) (the leading coefficient) and \(p\) must be a factor of \(k\) (the constant).
Example 1: List all of the possible rational roots of the following function: \(f\left(x\right)=5x^4+7x^3-x^2+8x-4\).
First, we will list all of the factors of the constant and of the leading coefficient. We included both the positive and negative of each number.
\(\large{\frac{p}{q}:\pm\frac{1,2,4}{1,5}}\)
Then we divide each number in the numerator by each number in the denominator to get the following list of possible rational roots for \(f\left(x\right)\).
Possible Rational Roots: \(\pm1,\pm2,\pm4,\large{\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5}}\)
This focuses our list of options for possible rational roots. Keep in mind that a polynomial function could have no rational roots (only irrational or complex roots), but if it does have a rational root, it has to be in the list above.
Example 1: List all of the possible rational roots of the following function: \(f\left(x\right)=5x^4+7x^3-x^2+8x-4\).
First, we will list all of the factors of the constant and of the leading coefficient. We included both the positive and negative of each number.
\(\large{\frac{p}{q}:\pm\frac{1,2,4}{1,5}}\)
Then we divide each number in the numerator by each number in the denominator to get the following list of possible rational roots for \(f\left(x\right)\).
Possible Rational Roots: \(\pm1,\pm2,\pm4,\large{\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5}}\)
This focuses our list of options for possible rational roots. Keep in mind that a polynomial function could have no rational roots (only irrational or complex roots), but if it does have a rational root, it has to be in the list above.
The second theorem is the Fundamental Theorem of Algebra. This theorem says that a polynomial with degree \(n\) will have exactly \(n\) roots with the multiplicity of the factor deciding the number of roots that factor gives. For example, \(\left(x+3\right)^2\) only provides one distinct root of \(x=-3\) but since its multiplicity is \(2\), then \(x=-3\) accounts for two solutions for the polynomial that \(\left(x+3\right)^2\) is a factor of.
We will use many of the operations in the last target (particularly factoring and division...which are very related) to solve polynomial equations and find roots of polynomial functions. Factoring and division are related because if we know the factor of a polynomial expression, we can divide by that factor to get another factor of the polynomial. In this target, we will work with expressions that are easily recognizable to factor and others where we must identify a factor first.
If we can recognize how to factor the expression in the equation/function immediately, that will be our most efficient strategy. If the factorization is not immediately available, then we can follow the steps below.
Steps to finding all roots of a polynomial function:
Example 2: Solve the following equation: \(x^5-6x^3-16x=0\).
This equation has a polynomial that is easily factorable. So we can first factor the GCF out and then use the factoring techniques described in the last target.
\(x\left(x^4-6x^2-16\right)=0\)
\(x\left(x^2-8\right)\left(x^2+2\right)=0\)
\(x=0\) \(x^2-8=0\) \(x^2+2=0\) (set each factor equal to \(0\))
\(x^2=8\) \(x^2=-2\)
\(x=0,\pm2\sqrt{2},\pm i\sqrt{2}\)
This was a degree \(5\) polynomial equation so the Fundamental Theorem of Algebra would tell us it has \(5\) solutions, and this equation has \(5\) distinct solutions.
Example 3: Solve the following equation: \(x^3+9x^2+21x+5=0\).
The structure of this polynomial expression does not allow us to easily factor it so we will use the steps above.
The possible rational solutions are \(\pm1,\pm5\). When we graph the function \(f\left(x\right)=x^3+9x^2+21x+5\), we obtain the following graph:
If we can recognize how to factor the expression in the equation/function immediately, that will be our most efficient strategy. If the factorization is not immediately available, then we can follow the steps below.
Steps to finding all roots of a polynomial function:
- List all possible roots using the Rational Roots Theorem.
- If you are not given a calculator or graph, use your list of possible roots and substitute those values in until you find a zero.
- Repeat the process until you have factored it down to a quadratic expression (or another factorable expression). You can then find the other roots by factoring, completing the square, or using the quadratic formula.
Example 2: Solve the following equation: \(x^5-6x^3-16x=0\).
This equation has a polynomial that is easily factorable. So we can first factor the GCF out and then use the factoring techniques described in the last target.
\(x\left(x^4-6x^2-16\right)=0\)
\(x\left(x^2-8\right)\left(x^2+2\right)=0\)
\(x=0\) \(x^2-8=0\) \(x^2+2=0\) (set each factor equal to \(0\))
\(x^2=8\) \(x^2=-2\)
\(x=0,\pm2\sqrt{2},\pm i\sqrt{2}\)
This was a degree \(5\) polynomial equation so the Fundamental Theorem of Algebra would tell us it has \(5\) solutions, and this equation has \(5\) distinct solutions.
Example 3: Solve the following equation: \(x^3+9x^2+21x+5=0\).
The structure of this polynomial expression does not allow us to easily factor it so we will use the steps above.
The possible rational solutions are \(\pm1,\pm5\). When we graph the function \(f\left(x\right)=x^3+9x^2+21x+5\), we obtain the following graph:
Although there are three real solutions, only one is a rational one and we know that \(x=-5\) is that solution. This means that \(\left(x+5\right)\) is a factor of the polynomial expression. We will use synthetic division to divide by that factor.
We knew that the remainder should be \(0\) since we knew that \(-5\) was a root. If you didn’t have a graph to work from, you could guess and check with the possible rational roots until you get a remainder of \(0\).
We can now write our equation as \(\left(x+5\right)\left(x^2+4x+1\right)=0\) and since our remaining factor is a quadratic, we can solve that using quadratic formula or completing the square (since it is not factorable). Let’s use completing the square for this one.
\(\begin{align}x^2+4x+1&=0\\\left(x^2+4x+4\right)+1-4&=0\\\left(x+2\right)^2-3&=0\\\left(x+2\right)^2&=3\\x+2&=\pm\sqrt{3}\\x&=-2\pm\sqrt{3}\end{align}\)
So the solutions to the equation are \(x=-5,-2\pm\sqrt{3}\).
Example 4: Given \(f\left(x\right)=3x^5+5x^4+9x^3+15x^2+6x+10\), one of the roots of \(f\left(x\right)\) is \(x=-\frac{5}{3}\). Find the remaining roots.
We can now write our equation as \(\left(x+5\right)\left(x^2+4x+1\right)=0\) and since our remaining factor is a quadratic, we can solve that using quadratic formula or completing the square (since it is not factorable). Let’s use completing the square for this one.
\(\begin{align}x^2+4x+1&=0\\\left(x^2+4x+4\right)+1-4&=0\\\left(x+2\right)^2-3&=0\\\left(x+2\right)^2&=3\\x+2&=\pm\sqrt{3}\\x&=-2\pm\sqrt{3}\end{align}\)
So the solutions to the equation are \(x=-5,-2\pm\sqrt{3}\).
Example 4: Given \(f\left(x\right)=3x^5+5x^4+9x^3+15x^2+6x+10\), one of the roots of \(f\left(x\right)\) is \(x=-\frac{5}{3}\). Find the remaining roots.
Example 5: Find all of the roots of \(f\left(x\right)=125x^3-8\).
Quick Check
Quick Check Solution
- Solve the equation. \(x^6-100x^2=0\).
- Find all of the roots of \(f\left(x\right)\) if \(f\left(x\right)=x^3-x^2-7x-20\).
- Given that \(x=-1\) is a root of \(f\left(x\right)=x^3-6x^2+5x+12\), find the remaining roots.
Quick Check Solution