Solving Polynomial Equations Worksheet
Solve polynomial equations using factors, root candidates, and division. Includes a worked example, printable practice, and an answer key with explanations.
The answer key follows the practice below. In the editor, remove the final answer-key section before sharing a student copy.
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Solving Polynomial Equations
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Learning goal and worked example
Solve polynomial equations using factors, root candidates, and division.
x³−9x = x(x−3)(x+3), so its roots are 0,3,−3. A candidate from the Rational Root Theorem must still be checked by substitution.
Prerequisites: factoring and synthetic division. The final extension uses i² = −1 to include complex roots.
Read each question carefully and solve the polynomial equations using the most appropriate method. Show all your work for full credit.
1. Solve the following polynomial equation by factoring: x³ - 4x² + 4x = 0
2. Find all real solutions for the equation: x⁴ - 16 = 0
3. According to the Rational Root Theorem, the possible rational roots of the polynomial P(x) = 2x³ + 3x² - 8x + 3 are .
4. If (x - 1) is a factor of the polynomial P(x) = x³ - kx² + 3x - 1, then the value of k must be .
5. Use synthetic division to find the quotient and remainder when P(x) = x³ + 2x² - 5x - 6 is divided by (x + 1).
6. Given that x = 2 is a root of the polynomial equation x³ - 7x + 6 = 0, find the other roots.
7. Which number is a candidate allowed by the Rational Root Theorem for P(x) = 3x³ - 5x² + 4x - 2? A candidate need not be an actual root.
1/3
2/5
3/2
4
8. How many real roots does the polynomial x⁴ + 5x² + 6 = 0 have?
0
1
2
4
9. Find all roots (real and complex) of the polynomial equation: x⁴ - 5x³ + 7x² - 5x + 6 = 0. You may use any method you deem appropriate, but show all steps clearly.
Answer key and worked solutions
1. x(x−2)²=0, so x=0 or 2; 2 has multiplicity 2.
2. (x²−4)(x²+4)=0, so real solutions are x=−2,2.
3. Candidates ±1, ±3, ±1/2, ±3/2. They are possibilities, not a list of confirmed roots.
4. P(1)=1−k+3−1=3−k=0, so k=3.
5. Synthetic division at −1 gives quotient x²+x−6 and remainder 0. Check: (x+1)(x²+x−6)=x³+2x²−5x−6.
6. Divide by x−2 to get x²+2x−3=(x+3)(x−1); other roots are −3 and 1.
7. a (1/3 is an allowed candidate). 8. a (0 real roots): x⁴+5x²+6=(x²+2)(x²+3)>0 for real x.
9. Factor as (x−2)(x−3)(x²+1). Thus roots are 2,3,i,−i. Multiplying (x²−5x+6)(x²+1) reproduces the given polynomial.
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