Solving a quadratic is routine. A cubic or higher-degree polynomial is harder, because there may be no simple formula to lean on. The Remainder Theorem gives a way in: a quick test that tells you whether a candidate number is a zero, and a path from there to a full factorization. It also connects three ideas that Algebra II treats as one: zeros, factors and remainders.
The Remainder Theorem
For a polynomial p(x) and a number a, the remainder when p(x) is divided by x − a is exactly p(a).
That means you can find a remainder without dividing: just substitute. It also gives the Factor Theorem as a direct consequence: p(a) = 0 if and only if x − a is a factor of p(x). A zero of the polynomial and a linear factor are two views of the same fact.
Finding a remainder by substitution
Worked example 1. Find the remainder when p(x) = 2x³ − 3x + 4 is divided by x − 2. Here a = 2, so the remainder is p(2) = 2(8) − 6 + 4 = 16 − 6 + 4 = 14.
Now divide by x + 1. Write x + 1 as x − (−1), so a = −1. The remainder is p(−1) = 2(−1)³ − 3(−1) + 4 = −2 + 3 + 4 = 5. Watch the sign: dividing by x + 1 uses a = −1, not a = 1.
Testing for a factor
Worked example 2. Is x − 2 a factor of x³ − x² − 4x + 4? Compute p(2) = 8 − 4 − 8 + 4 = 0. The remainder is zero, so yes.
Worked example 3. For what value of k is x − 3 a factor of x³ + kx² − x − 6? A factor means p(3) = 0: 27 + 9k − 3 − 6 = 0, so 18 + 9k = 0 and k = −2. The polynomial is x³ − 2x² − x − 6. Check: 27 − 18 − 3 − 6 = 0.
Finding all the zeros
Here is the plan for a cubic whose zeros are not obvious.
- Try small candidates in p(a), starting with a = 1, −1, 2, −2 and other factors of the constant term (and, when the leading coefficient is not 1, simple fractions of them).
- When p(a) = 0, you have a factor x − a. Divide it out to leave a quadratic.
- Factor the quadratic, or use the quadratic formula, to find the remaining zeros.
Worked example 4. Find all zeros of p(x) = x³ − 2x² − 5x + 6.
- p(1) = 1 − 2 − 5 + 6 = 0, so x − 1 is a factor.
- Divide: x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6). Check by multiplying: x³ − x² − 6x − x² + x + 6 = x³ − 2x² − 5x + 6.
- Factor the quadratic: x² − x − 6 = (x − 3)(x + 2).
- So p(x) = (x − 1)(x − 3)(x + 2), with zeros 1, 3 and −2.
Check the other two: p(3) = 27 − 18 − 15 + 6 = 0 and p(−2) = −8 − 8 + 10 + 6 = 0.
Worked example 5. Find all zeros of p(x) = 2x³ − 3x² − 11x + 6.
- Test small values: p(1) = 2 − 3 − 11 + 6 = −6 and p(2) = 16 − 12 − 22 + 6 = −12, neither zero. Try p(3) = 54 − 27 − 33 + 6 = 0.
- So x − 3 is a factor, and the quotient is 2x² + 3x − 2. Check: (x − 3)(2x² + 3x − 2) = 2x³ + 3x² − 2x − 6x² − 9x + 6 = 2x³ − 3x² − 11x + 6.
- Factor: 2x² + 3x − 2 = (2x − 1)(x + 2).
- The zeros are 3, 1/2 and −2.
Using zeros to sketch the graph
Once you know the zeros, you know where the graph crosses the x-axis. For example 4, the zeros are −2, 1 and 3, and the leading coefficient is positive with odd degree, so the graph starts low on the left and ends high on the right. Between the zeros it alternates above and below the axis: below the axis left of −2, above between −2 and 1, below between 1 and 3, above beyond 3. The y-intercept is p(0) = 6. That is enough for a rough graph that shows the zeros and the end behavior.
Common mistakes
- Sign error with x + a. The test number for x + 2 is −2, not 2.
- Skipping the zero placeholders in division. When dividing, write 0 for a missing power, such as x³ + 0x² − 4x + 1.
- Stopping at one zero. A cubic can have up to three real zeros. Divide out the first factor and keep going.
- Assuming every polynomial has nice zeros. The theorem helps only when a suitable factorization exists, so when no small candidate works, the zeros may not be rational.
- Confusing p(a) with the quotient. The substitution gives the remainder only. It does not give the quotient.
Practice Algebra II questions with an explanation for every answer, or read the guide to California Algebra II practice.
Source: the California Common Core State Standards for Mathematics, A-APR.2, A-APR.3.