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MHF4U Polynomial Equations & Inequalities
Grade 12 Unit 2 — practice questions with full solutions

Factor theorem, remainder theorem, solving cubics, and using sign charts to solve inequalities properly.

Free MHF4U polynomial equations & inequalities practice for Grade 12 students in Ontario. These 5 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MHF4U unit test, catching up on a lesson, or reviewing before the final exam.
MHF4U · Grade 12 Polynomial Functions 4 core 1 challenge
1Core
Use the remainder theorem to find the remainder when f(x) = 2x³ − 5x² + x − 3 is divided by (x − 2).
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  1. The remainder theorem says the remainder equals f(a) where the divisor is (x − a).
  2. Here a = 2, so evaluate f(2).
  3. f(2) = 2(8) − 5(4) + 2 − 3 = 16 − 20 + 2 − 3.
  4. That equals −5.
Final answerremainder −5
2Core
Factor fully:   x³ − 4x² + x + 6
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  1. Test factors of the constant 6 using the factor theorem: ±1, ±2, ±3, ±6.
  2. f(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor.
  3. Divide to get the quadratic: x³ − 4x² + x + 6 = (x + 1)(x² − 5x + 6).
  4. Factor the quadratic: numbers multiplying to 6 and adding to −5 are −2 and −3.
Final answer(x + 1)(x − 2)(x − 3)
3Core
Solve:   x³ + 2x² − 5x − 6 = 0
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  1. Test small factors of 6. f(2) = 8 + 8 − 10 − 6 = 0, so (x − 2) is a factor.
  2. Divide: x³ + 2x² − 5x − 6 = (x − 2)(x² + 4x + 3).
  3. Factor the quadratic: (x + 1)(x + 3).
  4. Set each factor to zero.
Final answerx = 2, −1, −3
4Core
-125+−+−Sign of f(x) on each intervalOpen circles mark where f(x) = 0
A sign chart for a function with zeros at −1, 2 and 5 is shown. Use it to state the solution to f(x) < 0, and explain why a sign chart is needed rather than just solving f(x) = 0.
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  1. Read the intervals where the sign is negative.
  2. The chart shows minus signs between −1 and 2, and again beyond 5.
  3. So f(x) < 0 on −1 < x < 2 or x > 5.
  4. Solving f(x) = 0 only gives the boundary points; the inequality asks for the intervals between them.
  5. Because the sign can only change at a zero, testing one point per interval determines the whole interval.
Final answer−1 < x < 2 or x > 5
5Challenge
Solve the inequality:   x³ − x² − 6x > 0
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  1. Factor first: x(x² − x − 6) = x(x − 3)(x + 2).
  2. Zeros are at x = −2, 0 and 3, splitting the number line into four intervals.
  3. Test x = −3: (−3)(−6)(−1) = −18, negative.
  4. Test x = −1: (−1)(−4)(1) = 4, positive. Test x = 1: (1)(−2)(3) = −6, negative. Test x = 4: (4)(1)(6) = 24, positive.
  5. Keep the positive intervals.
Final answer−2 < x < 0 or x > 3

MHF4U Polynomial Equations & Inequalities — common questions

Short answers to the things students ask most about this unit.

What is the factor theorem?

If f(a) = 0, then (x − a) is a factor of f(x). Testing small factors of the constant term is the usual way to find the first factor of a cubic.

How do you solve a polynomial inequality?

Factor fully, find the zeros, mark them on a number line, then test one value in each interval. The sign can only change at a zero.

More MHF4U practice — grade 12 advanced functions

MHF4U Polynomial FunctionsMHF4U Rational FunctionsMHF4U Exponential & Logarithmic FunctionsMHF4U Trigonometric FunctionsAll MHF4U units
← Unit 1: Polynomial FunctionsUnit 3: Rational Functions →
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