Find the rational roots of x^{3} + x^{2} + 4x + 4 = 0 using the rational root theorem.
Step 1: List all the possible rational roots.
The constant term is 4 and the leading coefficient is 1. By the rational root theorem, the only possible rational roots of x^{3} + x^{2} + 4x + 4 = 0 have the following form:
(factor of a_{0} / factor of a_{n}) or (factor of 4 / factor of 1)
The positive factors of 4 are 1, 2, and 4
The negative factors of 4 are -1, -2, and -4
The factor of 1 are 1 and -1
All the possible rational roots are shown below:
Rational roots with positive factors of 4: 1/-1, 1/1, 2/-1, 2/1, 4/-1, 4/1
Rational roots with negative factors of 4: -1/1, -1/-1, -2/1, -2/-1, -4/1, -4/-1
Notice that the set of rational roots with positive factors of 4 is equal to the set of rational roots with negative factors of 4. Therefore, just use one set to find all the rational roots in order to avoid redundancy. In general, to find rational roots, you can just use the positive factors of the constant term or a_{0} and then use both positive and negative factors of a_{n}.
The possible rational roots are 1/-1, 1/1, 2/-1, 2/1, 4/-1, 4/1
After simplification, we get -1, 1, -2, 2, -4, 4
You can also write the rational roots as ±1, ±2, ±4
Step 2: Test each possible rational root for x^{3} + x^{2} + 4x + 4 = 0
x = 1
1^{3} + 1^{2} + 4(1) + 4 = 1 + 1 + 4 + 4 = 2 + 8 = 10 ≠ 0
x = -1
(-1)^{3} + (-1)^{2} + 4(-1) + 4 = -1 + 1 + -4 + 4 = 0 + 0 = 0
Therefore, -1 is a root.
x = 2
2^{3} + 2^{2} + 4(2) + 4 = 8 + 4 + 8 + 4 = 12 + 12 = 24 ≠ 0
x = -2
(-2)^{3} + (-2)^{2} + 4(-2) + 4 = -8 + 4 + -8 + 4 = -4 + -4 = -8 ≠ 0
x = 3
3^{3} + 3^{2} + 4(3) + 4 = 27 + 9 + 12 + 4 = 36 + 16 = 52 ≠ 0
x = -3
(-3)^{3} + (-3)^{2} + 4(-3) + 4 = -27 + 9 + -12 + 4 = -18 + -8 = -26 ≠ 0
The only root is x = -1
Example #2:
Find the rational roots of 2x - 4 = 0 using the rational root theorem
The constant term is -4 and the leading coefficient is 2. By the rational root theorem, the only possible rational roots of 2x - 4 = 0 have the following form:
(factor of a_{0} / factor of a_{n}) or (factor of -4 / factor of 2)
The positive factors of -4 are 1, 2, 4
The factor of 2 are 1, 2, -1, -2
All the possible rational roots are shown below:
1/1, 1/2, 1/-1, 1/-2, 2/1, 2/2, 2/-1, 2/-2, 4/1, 4/2, 4/-1, 4/-2
or 1, 1/2, -1, 1/-2, 2, 1, -2, -1, 4, 2, -4, -2
or 1, 1/2, -1, 1/-2, 2, -2, 4, -4,
or ±1, ±2, ±4, ±(1/2)
Since you are familiar with the equation 2x - 4 = 0, It is easy to see that x = 2 is the only root.
Example #3:
Find the rational roots of 2x^{4} - 5x^{3} - 17x^{2} + 41x - 21 = 0 using the rational root theorem
The constant term is -21 and the leading coefficient is 2. By the rational root theorem, the only possible rational roots of 2x^{4} - 5x^{3} - 17x^{2} + 41x - 21 = 0 have the following form:
(factor of a_{0} / factor of a_{n}) or (factor of -21 / factor of 2)
The positive factors of -21 are 1, 3, 7, and 21
The factor of 2 are 1, 2, -1, -2
All the possible rational roots are shown below:
1/1, 1/2, 1/-1, 1/-2, 3/1, 3/2, 3/-1, 3/-2, 7/1, 7/2, 7/-1, 7/-2, 21/1, 21/2, 21/-1, 21/-2
or 1, 1/2, -1, 1/-2, 3, 3/2, -3, 3/-2, 7, 7/2, -7, 7/-2, 21, 21/2, -21, 21/-2
or ±1, ±(1/2), ±3, ±(3/2), ±7, ±(7/2), ±21, ±(21/2)
Step 2: Test each possible rational root for 2x^{4} - 5x^{3} - 17x^{2} + 41x - 21 = 0
This will involve tremendous amount of work and this work is beyond the scope of what we are trying to teach in this lesson. We will just give the answers here. Feel free to do all the necessary math yourself or use the calculator on this page to verify the answers.
x = 1, 2(1)^{4} - 5(1)^{3} - 17(1)^{2} + 41(1) - 21 = 0
Therefore, 1 is a root.
x = -1, 2(-1)^{4} - 5(-1)^{3} - 17(-1)^{2} + 41(-1) - 21 = -72 ≠ 0
x = 1/2, 2(1/2)^{4} - 5(1/2)^{3} - 17(1/2)^{2} + 41(1/2) - 21 = -21/4 ≠ 0
x = -1/2, 2(-1/2)^{4} - 5(-1/2)^{3} - 17(-1/2)^{2} + 41(-1/2) - 21 = -45 ≠ 0
x = 3, 2(3)^{4} - 5(3)^{3} - 17(3)^{2} + 41(3) - 21 = -24 ≠ 0
x = -3, 2(-3)^{4} - 5(-3)^{3} - 17(-3)^{2} + 41(-3) - 21 = 0
Therefore, -3 is a root.
x = 3/2, 2(3/2)^{4} - 5(3/2)^{3} - 17(3/2)^{2} + 41(3/2) - 21 = -9/2 ≠ 0
x = -3/2, 2(-3/2)^{4} - 5(-3/2)^{3} - 17(-3/2)^{2} + 41(-3/2) - 21 = -375/4 ≠ 0
x = 7, 2(7)^{4} - 5(7)^{3} - 17(7)^{2} + 41(7) - 21 = 2520 ≠ 0
x = -7, 2(-7)^{4} - 5(-7)^{3} - 17(-7)^{2} + 41(-7) - 21 = 5376 ≠ 0
x = 7/2, 2(7/2)^{4} - 5(7/2)^{3} - 17(7/2)^{2} + 41(7/2) - 21 = 0
Therefore, 7/2 is a root.
x = -7/2, 2(-7/2)^{4} - 5(-7/2)^{3} - 17(-7/2)^{2} + 41(-7/2) - 21 = 567/4 ≠ 0
x = 21, 2(21)^{4} - 5(21)^{3} - 17(21)^{2} + 41(21) - 21 = 336000 ≠ 0
x = -21, 2(-21)^{4} - 5(-21)^{3} - 17(-21)^{2} + 41(-21) - 21 = 426888 ≠ 0
x = 21/2, 2(21/2)^{4} - 5(21/2)^{3} - 17(21/2)^{2} + 41(21/2) - 21 = 68229/4 ≠ 0
x = -21/2, 2(-21/2)^{4} - 5(-21/2)^{3} - 17(-21/2)^{2} + 41(-21/2) - 21 = 55545/2 ≠ 0
The roots are x = 1, -3, and 7/2
Dec 01, 21 04:17 AM
What is the irrational root theorem? Definition, explanation, and easy to follow examples.