This set covers the whole module: testing roots, finding remainders, dividing cubics, finding unknown coefficients and checking factorisations. Questions run from easier to harder. Work on paper, then open each answer and compare every step.
Lessons to revisit are at the end. For a wider view, see polynomial factors and remainders, and use the non-calculator working trainer and the quadratic structure explorer for checking. The mistake log and retest queue helps you track error types.
Questions
Q1. Show that (x − 2) is a factor of f(x) = x³ + 2x² − 5x − 6.
Show answer
f(2) = 8 + 8 − 10 − 6 = 0. Since f(2) = 0, (x − 2) is a factor.
Q2. Find the remainder when x³ − 4x + 1 is divided by (x − 3).
Show answer
f(3) = 27 − 12 + 1 = 16.
Q3. Find the remainder when 2x³ + 5x² − 3x + 1 is divided by (x + 2).
Show answer
x + 2 = 0 gives x = −2. f(−2) = 2(−8) + 5(4) − 3(−2) + 1 = −16 + 20 + 6 + 1 = 11.
Q4. Find the value of k so that (x + 1) is a factor of x³ + kx² − 2x − 5.
Show answer
x = −1. f(−1) = −1 + k + 2 − 5 = k − 4 = 0, so k = 4. Check: −1 + 4 + 2 − 5 = 0.
Q5. Find the remainder when 4x³ − 2x² + x − 3 is divided by (2x − 1).
Show answer
2x − 1 = 0 gives x = ½. f(½) = 4(⅛) − 2(¼) + ½ − 3 = 0.5 − 0.5 + 0.5 − 3 = −5/2 (that is, −2.5).
Q6. Divide x³ − 2x² − 9x + 18 by (x − 2), then factorise f(x) = x³ − 2x² − 9x + 18 fully.
Show answer
f(2) = 8 − 8 − 18 + 18 = 0, so (x − 2) is a factor. Coefficients 1, −2, −9, 18 with a = 2: 1; −2 + 2 = 0; −9 + 0 = −9; 18 − 18 = 0. The quotient is x² − 9 = (x − 3)(x + 3). So f(x) = (x − 2)(x − 3)(x + 3). Expand check: (x − 2)(x² − 9) = x³ − 9x − 2x² + 18. ✓
Q7. Given that (x − 2) is a factor of 2x³ + x² − 13x + 6, solve 2x³ + x² − 13x + 6 = 0.
Show answer
Check f(2) = 16 + 4 − 26 + 6 = 0. Coefficients 2, 1, −13, 6 with a = 2: 2; 1 + 4 = 5; −13 + 10 = −3; 6 − 6 = 0. The quotient is 2x² + 5x − 3 = (2x − 1)(x + 3). So f(x) = (x − 2)(2x − 1)(x + 3), and the solutions are x = 2, x = ½, x = −3. Check x = ½: 2(⅛) + ¼ − 6.5 + 6 = 0. ✓
Q8. The polynomial f(x) = x³ + ax² + bx − 4 has (x − 2) as a factor. The remainder is −12 when f(x) is divided by (x + 1). Find a and b.
Show answer
f(2) = 8 + 4a + 2b − 4 = 0, so 4a + 2b = −4, which is 2a + b = −2. f(−1) = −1 + a − b − 4 = −12, so a − b = −7. Add the two equations: 3a = −9, so a = −3. Then b = −2 − 2(−3) = 4. So a = −3, b = 4. Check: f(x) = x³ − 3x² + 4x − 4, f(2) = 8 − 12 + 8 − 4 = 0 and f(−1) = −1 − 3 − 4 − 4 = −12. ✓
Q9. The polynomial 3x³ + 4x² − x + c has (x + 2) as a factor. Find c, then write the polynomial as a product of (x + 2) and a quadratic. Explain whether the quadratic factorises over the real numbers.
Show answer
x = −2. f(−2) = −24 + 16 + 2 + c = c − 6 = 0, so c = 6. Coefficients 3, 4, −1, 6 with a = −2: 3; 4 − 6 = −2; −1 + 4 = 3; 6 − 6 = 0. The quotient is 3x² − 2x + 3, so f(x) = (x + 2)(3x² − 2x + 3). The discriminant is (−2)² − 4(3)(3) = 4 − 36 = −32, which is negative. The quadratic does not factorise over the real numbers.
Q10. A student claims 4x³ − 8x² − x + 2 = (2x − 1)(2x + 1)(x − 2). Use expansion to decide whether this is correct.
Show answer
(2x − 1)(2x + 1) = 4x² − 1. Then (4x² − 1)(x − 2) = 4x³ − 8x² − x + 2. This matches the original polynomial term by term, so the claim is correct. A quick check: the highest-power coefficient 2 × 2 × 1 = 4 and the constant (−1)(1)(−2) = 2. ✓
If you got these wrong
| What went wrong | Lesson to revisit |
|---|---|
| Wrong sign when turning a factor into a root (Q1, Q3, Q4, Q9) | Using a known root to obtain a factor |
| Substituted the wrong value, or slipped on a fraction (Q2, Q5) | Finding a remainder without full division |
| Division row wrong, or a missing power skipped (Q6, Q7, Q9) | Dividing a cubic by a linear factor |
| Could not turn the words into equations (Q8) | Reconstructing a polynomial from conditions |
| Expansion error, or no final check (Q10) | Checking a factorisation by expansion |
If two or more rows apply to you, work through the lessons in that order. Extra structured support with your written method is something our teachers provide in online one-to-one Additional Mathematics tuition.