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# Finding Zeros of Polynomials

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Finding Rational and Real Zeros
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A length of a piece of farmland is $2x^2 + 10$ and the width is $x + 1$ . The area of the farmland is 353 square yards. What are the possible rational solutions to the polynomial equation represented by this situation?

### Guidance

Recall from Quadratic Functions that every quadratic equation has two solutions. The degree of a quadratic equation is 2, thus leading us towards the notion that it has 2 solutions. The degree will always tell us the maximum number of solutions a polynomial has. Quadratic equations also have a few different possibilities for solutions; two real-number solutions (parabola passes through the $x-$ axis twice), one real-number solution (where the solution is the vertex, called a repeated root), or two imaginary solutions (where the graph does not touch the $x-$ axis at all).

When it comes to solutions for polynomials, all these options are possibilities. There can be rational, irrational and imaginary solutions. Irrational and imaginary solutions will always come in pairs. This is due to the fact that to find these types of solutions, you must use the Quadratic Formula and the $\pm$ sign will give two solutions. In this concept we will only address real-number solutions.

Now, you might be wondering, how do we find all these solutions? One way is to use the Rational Root Theorem.

Rational Root Theorem: For a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_1 x+a_0$ , where $a_n, a_{n - 1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$ . More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$ , then all the rational factors will have the form $\pm \frac{p}{q}$ .

In other words, the factors of the constant divided by the factors of the leading coefficient will yield all the possible rational solutions to $f(x)$ .

#### Example A

Find all the possible rational solutions to $f(x)=6x^4-43x^3+66x^2-3x-10$ .

Solution: All the possible factors of 10 are 1, 2, 5, and 10. All the possible factors of 6 are 1, 2, 3, and 6. The possible combinations are $\frac{\pm 1. \pm 2, \pm 5, \pm 10}{\pm 1, \pm 2, \pm 3, \pm 6}=\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm 10, \pm \frac{10}{3}$ . Therefore, there are 24 possibilities.

#### Example B

Find the rational solutions to $f(x)=6x^4-43x^3+66x^2-3x-10$ .

Solution: Before the days of graphing calculators, you would have to test all 24 possible solutions to find the correct solutions. Now, we can graph the function and eliminate any possibilities that seem unreasonable. Because the degree of the function is 4, there will be 4 solutions. Here is the graph:

Looking back at Example A, the reasonable solutions appear to be: $5, 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3}$ , or $\pm \frac{5}{6}$ . By just looking at the graph, the solutions between -1 and 1 are difficult to see. This is why we have listed all the solutions between -1 and 1 to test. Let’s test 5 and 2 using synthetic division.

The remainder is zero, like we thought.

Now, rather than starting over with the division by 2, continue with the leftover polynomial.

Again, the remainder is zero. Both 5 and 2 are zeros.

To find the last two zeros, we can test all the fractions above using synthetic division. OR, we can factor this leftover polynomial. Because we started with a polynomial of degree 4, this leftover polynomial is a quadratic. It is $6x^2-x-1. \ ac = -6$ and the factors of -1 that add up to -6 are -3 and 2. Expand the $x-$ term and factor by grouping.

$& \underbrace{6x^2-3x}+\underbrace{2x-1}\\& 3x(2x-1)+1(2x-1)\\& (2x-1)(3x+1)$

Setting these two factors equal to zero, we have $x=\frac{1}{2}$ and $-\frac{1}{3}$ . Therefore, the solutions to this polynomial are 5, 2, $\frac{1}{2}$ and $-\frac{1}{3}$ .

Check your answer: To check your work, you can multiply the factors together to see if you get the original polynomial.

$& \underbrace{(2x-1)(3x+1)} \underbrace{(x-5)(x-2)}\\& (6x^2-x-1)(x^2-7x+10)\\& 6x^4-43x^3+66x^2-3x-10$

#### Example C

Find all the real solutions to $f(x)=x^4+6x^3-2x^2-48x-32$ .

Solution: First, sketch a graph.

Now, use the Rational Root Theorem to determine all possible rational roots.

$\frac{\text{factors of -32}}{\text{factors of 1}}=\frac{\pm 32, \pm 16, \pm 8, \pm 4, \pm 2, \pm 1}{\pm 1}$

Using the graph, it looks like -4 is the only possible rational solution. Also, notice that the graph touches at -4 and does not pass through the $x-$ axis. That means that this solution is a repeated root. Let’s do synthetic division.

Because the root is repeated, we did synthetic division twice. At the end of the synthetic division, the leftover polynomial is $x^2-2x-2$ which is not factorable. Therefore, to find the last two real solutions, we must do the Quadratic Formula.

$x &= \frac{2 \pm \sqrt{(-2)^2-4(1)(-2)}}{2(1)}\\&= \frac{2 \pm \sqrt{4+8}}{2}=\frac{2 \pm \sqrt{12}}{2}=\frac{2 \pm 2\sqrt{3}}{2}=1 \pm \sqrt{3} \approx 2.73, -0.73$

The roots, or zeros, of $f(x)=x^4+6x^3-2x^2-48x-32$ are -4 (twice), 2.73, and -0.73. Looking back at the graph, we see that this is where the function crosses the $x-$ axis. The graph is always a good way to double-check your work.

Intro Problem Revisit First, we need to set up the equation.

$(x + 1)(2x^2 + 10) = 353\\= 2x^3 + 10x + 2x^2 + 10 = 353\\= 2x^3 + 2x^2 + 10x - 343 = 0$

All the possible factors of 343 are 1, 7, 49, and 343. All the possible factors of 2 are 1 and 2. The possible combinations are $\frac{\pm 1, \pm 7, \pm 49, \pm 343}{\pm 1, \pm 2}=\pm 1, \pm 7, \pm 49, \pm 343, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{49}{2}, \pm \frac{343}{2}$ . Therefore, there are 16 possibilities.

### Guided Practice

Find all the real solutions of the following functions.

1. $f(x)=x^3-2x^2-15x+30$

2. $f(x)=6x^3+19x^2+11x-6$

3. $f(x)=x^5-4x^4-18x^3+38x^2-11x-6$

1. Using the Rational Root Theorem, the possible rational roots are: $\pm 30, \pm 15, \pm 10, \pm 6, \pm 5, \pm 3, \pm 2, \pm 1$ . Now, graph the function.

By looking at the graph, the only reasonable rational root is 2. We can rule out 4 and -4 because they are not included in the list of rational roots. Therefore, these two roots will be irrational. Do the synthetic division for 2.

The leftover polynomial is $x^2-15=0$ . This polynomial can be solved by using square roots.

$x^2-15 &= 0\\x^2 &= 15\\x &= \pm \sqrt{15} \approx \pm 3.87$

$^*$ Instead of using the Rational Root Theorem and synthetic division, this problem could have also been solved using factoring by grouping.

2. Using the Rational Root Theorem, the possible rational roots are: $\pm 6, \pm 3, \pm 2, \pm \frac{3}{2}, \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$ .

By looking at the graph, the reasonable rational roots are $-2, -\frac{3}{2}, \frac{1}{3}$ or $\frac{1}{6}$ . The rational answers are difficult to see because they do not cross exactly the $x-$ axis on an integer. Therefore, we will do the synthetic division for -2 first.

The leftover polynomial is $6x^2+7x-3$ , which is factorable. You can decide if you would like to factor this polynomial, use the Quadratic Formula, or test the rational possibilities from above. Let’s factor.

$& 6x^2+7x-3\\& 6x^2+9x-2x-3\\& 3x(2x+3)-1(2x+3)\\& (3x-1)(2x+3)$

From these factors, the rational solutions are $\frac{1}{3}$ and $-\frac{3}{2}$ .

3. Using the Rational Root Theorem, the possible rational roots are: $\pm 6, \pm 3, \pm 2, \pm 1$ .

From the graph, the possible roots are 6 and 1. It looks like 1 is a double root because the function reaches the $x-$ axis at 1, but does not pass through it. Do synthetic division with 6, 1, and 1 again.

The leftover polynomial is $x^2+4x+1$ . This is not a factorable polynomial, so use the Quadratic Formula to find the last two roots.

$x=\frac{-4 \pm \sqrt{4^2-4(1)(1)}}{2(1)}=\frac{-4 \pm \sqrt{12}}{2}=\frac{-4 \pm 2\sqrt{3}}{2}=-2 \pm \sqrt{3} \approx -0.27, -3.73$

### Vocabulary

Rational Root Theorem: For a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$ . More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$ , then all the rational factors will have the form $\pm \frac{p}{q}$ .

### Practice

Find all the possible rational solutions for the following polynomials. Use the Rational Root Theorem.

1. $f(x)=x^3+6x^2-18x+20$
2. $f(x)=4x^4+x^2-15$
3. $f(x)=-2x^3+7x^2-x+8$
4. $f(x)=x^4-3x^3-4x^2+15x+9$
5. $f(x)=8x^4 -5x^3+16x^2+37x-24$

Find all the real-number solutions for each function below. Use any method you like.

1. $f(x)=6x^3-17x^2+11x-2$
2. $f(x)=x^4+7x^3+6x^2-32x-32$
3. $f(x)=16x^3+40x^2-25x-3$
4. $f(x)=2x^3-9x^2+21x-18$
5. $f(x)=4x^3-16x^2+39x-295$
6. $f(x)=18x^4+3x^3-17x^2+17x-55$
7. $f(x)=x^5+7x^4-3x^3-65x^2-8x-156$
8. $f(x)=4x^4+20x^3-23x^2-120x+144$
9. $f(x)=9x^4-226x^2+25$
10. Solve $f(x)=3x^4-x^2-14$ by factoring. How many real solutions does this function have? What type of solution(s) could the others be?