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Polynomials, factor and remainder theorems

Expanding, factorising, algebraic division by linear and quadratic divisors, the Factor and Remainder Theorems.

Spec MM1.6Section 1 Part 1: AS-level pure mathematics17 questions12 real past questions
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1Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

When \(x^3 - 4x^2 + kx + 6\) is divided by \((x - 3)\) the remainder is \(-6\). Find \(k\).

  1. A\(1\)
  2. B\(5\)
  3. C\(-3\)
  4. D\(-5\)
  5. E\(3\)
  6. F\(-1\)
Worked solution

Remainder Theorem: \(27 - 36 + 3k + 6 = -6 \Rightarrow 3k - 3 = -6 \Rightarrow k = -1\).

Full method — how and why
A remainder is given, so use the Remainder Theorem: substitute the root of the divisor and set the result equal to the remainder.
  1. The divisor \((x - 3)\) is zero at \(x = 3\).
    The Remainder Theorem says the remainder on dividing by \((x - a)\) is \(f(a)\); you need the value of \(x\) that makes the divisor vanish.
  2. Substitute: \(f(3) = 27 - 36 + 3k + 6 = 3k - 3\).
    Work the numerical terms out first (\(27 - 36 + 6 = -3\)) so only the \(k\) term is left to carry.
  3. Set equal to the given remainder: \(3k - 3 = -6\).
    This is what the Remainder Theorem gives you - an equation, not an identity.
  4. Solve: \(3k = -3\), so \(k = -1\).
    Simple linear equation.
Where it goes wrong
  • Substituting \(x = -3\) (the sign of the root of \(x - 3\) is \(+3\)), which gives a different value of \(k\).
  • Setting \(f(3) = 0\) out of habit - that is the Factor Theorem, and only applies when the remainder is zero.
Takeaway Divisor \((x - a)\) \(\Rightarrow\) substitute \(x = a\); the result is the remainder, which is zero only when \((x-a)\) is a factor.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-01
2Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

\((x + 2)\) is a factor of \(2x^3 + ax^2 - 5x - 6\). Find the value of \(a\).

  1. A\(5\)
  2. B\(-2\)
  3. C\(1\)
  4. D\(-3\)
  5. E\(2\)
  6. F\(3\)
Worked solution

Factor Theorem: \(f(-2) = -16 + 4a + 10 - 6 = 4a - 12 = 0\), so \(a = 3\).

Full method — how and why
'Is a factor' means remainder zero, so the Factor Theorem turns the statement into a single equation for \(a\).
  1. \((x + 2)\) vanishes at \(x = -2\).
    Write \(x + 2 = x - (-2)\) if the sign is confusing; the root is the value that makes the bracket zero.
  2. Substitute: \(f(-2) = 2(-8) + a(4) - 5(-2) - 6 = -16 + 4a + 10 - 6\).
    Careful with the odd power: \((-2)^3 = -8\), while \((-2)^2 = +4\).
  3. Simplify: \(4a - 12\).
    \(-16 + 10 - 6 = -12\).
  4. Set to zero (factor \(\Rightarrow\) remainder zero): \(4a = 12\), so \(a = 3\).
    That is precisely the Factor Theorem.
Where it goes wrong
  • Using \(x = 2\) instead of \(x = -2\), which gives \(a = -3\) - an offered option.
  • Sign slips on \((-2)^3\) and \(-5(-2)\); write each term out separately.
Takeaway Factor \((x + k)\) means substitute \(x = -k\) and set the result to zero.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-02
3Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Fully factorise \(x^3 - 7x + 6\).

  1. A\((x-1)(x-2)(x-3)\)
  2. B\((x-1)(x-2)(x+3)\)
  3. C\((x-1)(x+2)(x-3)\)
  4. D\((x-1)(x^2 + x + 6)\)
  5. E\((x+1)(x+2)(x-3)\)
  6. F\((x+1)(x-2)(x+3)\)
Worked solution

\(f(1) = 0\), so \((x-1)\) is a factor: \(x^3 - 7x + 6 = (x-1)(x^2 + x - 6) = (x-1)(x+3)(x-2)\).

Full method — how and why
Factorising a cubic: find one root by inspection, divide it out, then factorise the quadratic that remains.
  1. Test small factors of the constant term \(6\): \(f(1) = 1 - 7 + 6 = 0\), so \((x - 1)\) is a factor.
    Any integer root must divide the constant term, so only \(\pm1, \pm2, \pm3, \pm6\) are worth testing - and \(1\) is always the quickest to try.
  2. Divide: \(x^3 - 7x + 6 = (x - 1)(x^2 + x - 6)\).
    Note the missing \(x^2\) term in the original - treat it as \(0x^2\) when dividing or comparing coefficients.
  3. Factorise the quadratic: \(x^2 + x - 6 = (x + 3)(x - 2)\).
    Two numbers multiplying to \(-6\) and adding to \(+1\).
  4. Full factorisation: \((x-1)(x-2)(x+3)\).
    Check by multiplying the constants: \((-1)(-2)(3) = 6\). ✓
Where it goes wrong
  • Forgetting the missing \(x^2\) term and producing the wrong quadratic factor.
  • Sign errors giving \((x-1)(x+2)(x-3)\), whose constant term is \(+6\) too - always check the \(x\) coefficient as well.
Takeaway Root-hunt among the divisors of the constant term, divide out, then factorise what remains; check by multiplying the constants back.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-03
4Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the remainder when \(x^{4} + 2x^{3} - 3x + 1\) is divided by \((2x - 1)\).

  1. A\(\dfrac{13}{16}\)
  2. B\(\dfrac{3}{16}\)
  3. C\(1\)
  4. D\(-\dfrac{1}{4}\)
  5. E\(-\dfrac{7}{16}\)
  6. F\(-\dfrac{3}{16}\)
Worked solution

\(f\!\left(\frac12\right) = \frac1{16} + \frac{2}{8} - \frac32 + 1 = \frac{1}{16} + \frac{4}{16} - \frac{24}{16} + \frac{16}{16} = -\frac{3}{16}\).

Full method — how and why
A divisor of the form \(ax - b\). The Remainder Theorem still applies, but the substitution value is \(\frac{b}{a}\), not \(b\).
  1. \(2x - 1 = 0\) at \(x = \frac12\).
    The remainder on dividing by \((ax - b)\) is \(f(b/a)\) - the root of the divisor, whatever the leading coefficient.
  2. Substitute: \(\left(\frac12\right)^4 + 2\left(\frac12\right)^3 - 3\left(\frac12\right) + 1 = \frac{1}{16} + \frac{2}{8} - \frac32 + 1\).
    Work each power out separately: \(\frac1{16}\), \(\frac18\), then the linear and constant terms.
  3. Put everything over \(16\): \(\frac{1}{16} + \frac{4}{16} - \frac{24}{16} + \frac{16}{16} = -\frac{3}{16}\).
    A common denominator prevents fraction slips; \(\frac{2}{8} = \frac{4}{16}\) and \(\frac32 = \frac{24}{16}\).
Where it goes wrong
  • Substituting \(x = 1\) or \(x = 2\) instead of \(\frac12\).
  • Arithmetic sign slips giving \(+\frac{3}{16}\), which is offered.
Takeaway For divisor \(ax - b\), evaluate at \(x = \frac ba\); convert everything to a common denominator before adding.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-04
5Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

The polynomial \(p(x) = x^3 + ax^2 + bx + 4\) has \((x-1)\) and \((x+2)\) as factors. Find the third linear factor.

  1. A\((x + 1)\)
  2. B\((x - 2)\)
  3. C\((x - 4)\)
  4. D\((x - 1)\)
  5. E\((x + 4)\)
  6. F\((x + 2)\)
Worked solution

Write \(p(x) = (x-1)(x+2)(x-r)\). The constant term is \((-1)(2)(-r) = 2r = 4\), so \(r = 2\): the third factor is \((x - 2)\).

Full method — how and why
A cubic with two known factors has a third. The constant term alone pins it down - no division needed.
  1. Write \(p(x) = (x - 1)(x + 2)(x - r)\), where \((x-r)\) is the unknown factor.
    The leading coefficient is \(1\), so the product of the three monic linear factors is the whole cubic.
  2. Compare constant terms: the constant of the product is \((-1)(2)(-r) = 2r\), and \(p(x)\) has constant term \(4\).
    The constant term of a product of brackets is the product of their constant terms - much faster than expanding everything.
  3. Solve: \(2r = 4\), so \(r = 2\) and the third factor is \((x - 2)\).
    Check: \((-1)(2)(-2) = 4\) ✓.
Where it goes wrong
  • Reading the third factor as \((x + 2)\) by forgetting the sign flip between the root \(r\) and the factor \((x - r)\).
  • Trying to find \(a\) and \(b\) first - unnecessary work.
Takeaway Compare constant terms to find a missing factor: the product of the constants in the brackets equals the constant term.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-05
6Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the quotient when \(2x^3 - 3x^2 + 4x - 5\) is divided by \((x - 1)\).

  1. A\(2x^2 + x - 3\)
  2. B\(2x^2 - 5x + 9\)
  3. C\(2x^2 - x + 3\)
  4. D\(2x^2 - x - 3\)
  5. E\(2x^2 + x + 3\)
  6. F\(2x^2 - x + 5\)
Worked solution

\(2x^3 - 3x^2 + 4x - 5 = (x-1)(2x^2 - x + 3) - 2\): check the expansion \(2x^3 - x^2 + 3x - 2x^2 + x - 3 = 2x^3 - 3x^2 + 4x - 3\), then subtract \(2\) for the remainder.

Full method — how and why
A division question where only the quotient is wanted. Comparing coefficients is quicker and safer than long division.
  1. Write \(2x^3 - 3x^2 + 4x - 5 = (x - 1)(Ax^2 + Bx + C) + R\).
    Dividing a cubic by a linear factor gives a quadratic quotient and a constant remainder.
  2. Compare \(x^3\) terms: \(A = 2\).
    Only one product contributes to the highest power, so the leading coefficient comes free.
  3. Compare \(x^2\) terms: \(B - A = -3\), so \(B = -1\).
    The \(x^2\) term comes from \(x \cdot Bx\) and \(-1 \cdot Ax^2\).
  4. Compare \(x\) terms: \(C - B = 4\), so \(C = 3\). The quotient is \(2x^2 - x + 3\).
    The \(x\) term comes from \(x \cdot C\) and \(-1 \cdot Bx\). (The constant term then gives \(R = -2\), but the question does not need it.)
Where it goes wrong
  • Getting the sign of \(B\) wrong, giving \(2x^2 + x + 3\) - which is offered.
  • Reporting the remainder instead of the quotient.
Takeaway Comparing coefficients from the top down finds a quotient one coefficient at a time, with no long-division layout to get wrong.
Answer C · Polynomials, factor and remainder theorems (MM1.6) · question POL-07
7Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the coefficient of \(x^2\) in the expansion of \((x + 1)(x - 2)(2x + 3)\).

  1. A\(1\)
  2. B\(-3\)
  3. C\(-2\)
  4. D\(-1\)
  5. E\(2\)
  6. F\(3\)
Worked solution

\((x+1)(x-2) = x^2 - x - 2\); multiply by \((2x + 3)\): the \(x^2\) term is \(3x^2 - 2x^2 = x^2\), coefficient \(1\).

Full method — how and why
Only one coefficient is wanted, so do not expand everything: collect just the ways of making \(x^2\).
  1. Multiply the first two brackets: \((x+1)(x-2) = x^2 - x - 2\).
    Two brackets at a time keeps the bookkeeping small.
  2. To get \(x^2\) from \((x^2 - x - 2)(2x + 3)\), pair \(x^2\) with \(3\), and \(-x\) with \(2x\).
    An \(x^2\) term arises from (degree 2)\(\times\)(degree 0) and (degree 1)\(\times\)(degree 1) - list the pairings rather than expanding.
  3. Add the contributions: \(3x^2 + (-x)(2x) = 3x^2 - 2x^2 = x^2\), so the coefficient is \(1\).
    Only these two products can contribute.
Where it goes wrong
  • Missing the \((-x)(2x)\) contribution and answering \(3\).
  • Expanding everything under time pressure and losing a sign.
Takeaway For a single coefficient, list the pairs of terms whose degrees add to the one you want.
Answer A · Polynomials, factor and remainder theorems (MM1.6) · question POL-11
8Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Which of the following is a factor of \(x^4 - 5x^2 + 4\)?

  1. A\(x - 3\)
  2. B\(x^2 + 1\)
  3. C\(x - 4\)
  4. D\(x + 4\)
  5. E\(x + 2\)
  6. F\(x^2 + 4\)
Worked solution

\(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)\).

Full method — how and why
A quartic that is quadratic in \(x^2\). Factorise fully and read off which of the offered factors appears.
  1. Substitute \(u = x^2\): \(u^2 - 5u + 4 = (u - 1)(u - 4)\).
    Only even powers appear, so the substitution works.
  2. Undo it: \(x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4)\).
    Replace \(u\) by \(x^2\) in each bracket.
  3. Factorise each as a difference of two squares: \((x-1)(x+1)(x-2)(x+2)\).
    \(x^2 - a^2 = (x-a)(x+a)\) - do not stop at the quadratics.
  4. Compare with the options: \((x + 2)\) appears.
    Equivalently, a factor \((x - a)\) exists iff \(f(a) = 0\): \(f(-2) = 16 - 20 + 4 = 0\) ✓.
Where it goes wrong
  • Choosing \(x^2 + 1\) or \(x^2 + 4\) - these are sums, not differences, and are not factors.
  • Testing only positive values when checking \(f(a) = 0\).
Takeaway Factorise as far as possible; a quick check of \(f(a) = 0\) confirms any candidate factor \((x-a)\).
Answer E · Polynomials, factor and remainder theorems (MM1.6) · question POL-13
9Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

\(p(x)\) is a cubic polynomial with \(p(1) = p(2) = p(3) = 0\) and \(p(0) = 12\). Find \(p(4)\).

  1. A\(24\)
  2. B\(-24\)
  3. C\(12\)
  4. D\(0\)
  5. E\(-12\)
  6. F\(6\)
Worked solution

\(p(x) = a(x-1)(x-2)(x-3)\); \(p(0) = -6a = 12 \Rightarrow a = -2\). \(p(4) = -2 \cdot 3\cdot2\cdot1 = -12\).

Full method — how and why
Three roots determine a cubic up to a constant multiple. One extra value fixes the multiple.
  1. Write \(p(x) = a(x-1)(x-2)(x-3)\).
    Given roots, start from the factorised form; the leading coefficient \(a\) is the only freedom left.
  2. Use \(p(0) = 12\): \(a(-1)(-2)(-3) = -6a = 12\), so \(a = -2\).
    Three negative factors give a negative product - a very common sign slip.
  3. Evaluate \(p(4) = -2(3)(2)(1) = -12\).
    Substitute directly into the factorised form; no expansion needed.
Where it goes wrong
  • Taking \(a = 2\) from \(6a = 12\) and answering \(+12\).
  • Assuming \(a = 1\) because the question says 'a cubic' - the leading coefficient is not given.
Takeaway Roots give the shape, one extra point gives the scale; keep everything in factorised form.
Answer E · Polynomials, factor and remainder theorems (MM1.6) · question POL-14
10Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

The polynomial \(x^3 + ax^2 + bx + c\) has roots \(2\), \(-1\) and \(3\). Find the value of \(a + b + c\).

  1. A\(5\)
  2. B\(-5\)
  3. C\(-3\)
  4. D\(3\)
  5. E\(-7\)
  6. F\(7\)
Worked solution

\(p(1) = 1 + a + b + c\), and \(p(1) = (1-2)(1+1)(1-3) = (-1)(2)(-2) = 4\). So \(a + b + c = 3\). (Directly: \((x-2)(x+1)(x-3) = x^3 - 4x^2 + x + 6\).)

Full method — how and why
You could expand, but \(a + b + c\) is the sum of the non-leading coefficients, which \(p(1)\) delivers instantly.
  1. Note \(p(1) = 1 + a + b + c\), so \(a + b + c = p(1) - 1\).
    Substituting \(x = 1\) turns every power into \(1\), so \(p(1)\) is the sum of all coefficients including the leading \(1\).
  2. Write the factorised form from the roots: \(p(x) = (x-2)(x+1)(x-3)\).
    Monic, because the given polynomial starts with \(x^3\).
  3. Evaluate: \(p(1) = (-1)(2)(-2) = 4\).
    Two negatives and one positive give a positive.
  4. So \(a + b + c = 4 - 1 = 3\).
    Remember to subtract the leading coefficient.
Where it goes wrong
  • Answering \(4\) by forgetting the leading \(1\) in \(p(1) = 1 + a + b + c\).
  • Expanding and mis-collecting; the expansion is \(x^3 - 4x^2 + x + 6\), so \(-4 + 1 + 6 = 3\) ✓.
Takeaway \(p(1)\) is the sum of all coefficients - a one-second route to questions like this.
Answer D · Polynomials, factor and remainder theorems (MM1.6) · question POL-15
11Polynomials, factor and remainder theorems · spec MM1.6 · Paper 2 style · ★☆☆

\(p(x)\) is a polynomial. \(P\): '\(p(2) = 0\)'. \(Q\): '\((x - 2)^2\) is a factor of \(p(x)\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

If \((x-2)^2\) divides \(p\) then certainly \(p(2) = 0\), so \(P\) is necessary for \(Q\). But \(p(x) = x - 2\) has \(p(2) = 0\) without a repeated factor, so \(P\) is not sufficient.

Full method — how and why
The factor theorem gives one direction; a repeated factor is strictly more than a root.
  1. If \((x-2)^2\) is a factor, then \((x-2)\) is a factor, so \(p(2) = 0\): \(Q \Rightarrow P\). \(P\) is necessary.
    Factor theorem: \((x-a)\) is a factor exactly when \(p(a)=0\).
  2. Is \(P\) sufficient? Take \(p(x) = x-2\): \(p(2)=0\), but \((x-2)^2\) does not divide a degree-\(1\) polynomial.
    A single root is not a repeated root.
  3. So \(P\) is necessary but not sufficient.
    Note the direction: the weaker condition is necessary for the stronger one.
Where it goes wrong
  • Answering 'sufficient but not necessary' - the common pattern, but reversed here.
  • Thinking \(p(2)=0\) and \(p'(2)=0\) together would be needed; that is exactly the extra condition for a repeated root, and its absence is the point.
Takeaway \(p(a)=0 \iff (x-a)\) is a factor; a repeated factor also needs \(p'(a)=0\).
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question X2-08
12Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

When \(p(x)\) is divided by \((x-1)\) the remainder is \(3\), and when divided by \((x+1)\) the remainder is \(-1\). Find the remainder when \(p(x)\) is divided by \((x^2 - 1)\).

  1. A\(2\)
  2. B\(2x + 1\)
  3. C\(-x + 2\)
  4. D\(2x - 1\)
  5. E\(x + 2\)
  6. F\(3x - 1\)
Worked solution

The remainder has degree \(< 2\): \(ax + b\) with \(a + b = 3\) and \(-a + b = -1\). So \(b = 1\), \(a = 2\): remainder \(2x + 1\).

Full method — how and why
Dividing by a quadratic leaves a remainder of degree at most \(1\). Write it as \(ax + b\) and use the two given remainders to find \(a\) and \(b\).
  1. Write the division statement: \(p(x) = (x^2 - 1)Q(x) + ax + b\).
    The remainder always has lower degree than the divisor, so with a quadratic divisor it is linear (or constant).
  2. Note \(x^2 - 1 = (x-1)(x+1)\), so substituting \(x = 1\) or \(x = -1\) kills the \(Q(x)\) term.
    Choosing the roots of the divisor is what makes the unknown quotient disappear.
  3. \(x = 1\): \(p(1) = a + b\). The Remainder Theorem gives \(p(1) = 3\), so \(a + b = 3\).
    The remainder on dividing by \((x-1)\) is \(p(1)\), which is given as \(3\).
  4. \(x = -1\): \(p(-1) = -a + b = -1\).
    Same reasoning with the other root.
  5. Solve: adding gives \(2b = 2\), so \(b = 1\) and \(a = 2\). Remainder \(2x + 1\).
    Two linear equations in two unknowns.
Where it goes wrong
  • Assuming the remainder is a constant and trying to force a single number.
  • Sign error at \(x = -1\): \(a(-1) + b = -a + b\), not \(a - b\), which would give \(2x - 1\) (an offered option).
Takeaway Remainder on division by a degree-\(n\) divisor has degree \(< n\); substitute the divisor's roots to find its coefficients.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-06
13Polynomials, factor and remainder theorems · spec MM1.6 · Paper 2 style · ★★☆

Which of the following statements about the polynomial \(p(x) = x^3 - 6x^2 + 11x - 6\) is/are true?

  1. I\(p(x)\) has three distinct real roots.
  2. II\((x - 4)\) is a factor of \(p(x) - 6\).
  3. III\(p(x) > 0\) for all \(x > 3\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(p(x) = (x-1)(x-2)(x-3)\), so I is true. \(p(4) - 6 = 3\cdot2\cdot1 - 6 = 0\), so II is true. For \(x > 3\) all three factors are positive, so III is true.

Full method — how and why
Three statements about one cubic. Factorising it first makes all three immediate.
  1. Factorise: \(p(1) = 1 - 6 + 11 - 6 = 0\), so \((x-1)\) is a factor and \(p(x) = (x-1)(x-2)(x-3)\).
    Test the divisors of \(6\); the resulting quadratic \(x^2 - 5x + 6\) factorises easily.
  2. I: the roots \(1\), \(2\), \(3\) are distinct. True.
    Three distinct linear factors means three distinct roots.
  3. II: \((x-4)\) is a factor of \(p(x) - 6\) iff \(p(4) - 6 = 0\). Now \(p(4) = 3 \times 2 \times 1 = 6\), so \(p(4) - 6 = 0\). True.
    Factor Theorem applied to the shifted polynomial - evaluate the whole expression \(p(x) - 6\) at \(x = 4\).
  4. III: for \(x > 3\) each factor \(x-1\), \(x-2\), \(x-3\) is positive, so the product is positive. True.
    Sign analysis from the factorised form; beyond the largest root everything is positive.
  5. All three: 'I, II and III'.
    Match verdicts to options.
Where it goes wrong
  • Expanding \(p(x) - 6\) and re-testing from scratch instead of just evaluating \(p(4) - 6\).
  • Checking III at one value like \(x = 10\) and calling it proved - the factor argument covers every \(x > 3\).
Takeaway Factorise first: roots, sign and shifted-factor questions then all read off the same form.
Answer H · Polynomials, factor and remainder theorems (MM1.6) · question POL-08
14Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

The expression \(x^4 + 4\) can be factorised into two quadratics with integer coefficients. What is the sum of the coefficients of \(x\) in these two quadratics?

  1. A\(1\)
  2. B\(0\)
  3. C\(2\)
  4. D\(4\)
  5. E\(-4\)
  6. FIt cannot be factorised.
Worked solution

\(x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)\). The \(x\)-coefficients are \(2\) and \(-2\), summing to \(0\).

Full method — how and why
\(x^4 + 4\) has no real roots, so it cannot be split into linear factors - but it can be split into two quadratics via a difference of two squares.
  1. Complete the square in \(x^2\): \(x^4 + 4 = (x^2 + 2)^2 - 4x^2\).
    \((x^2+2)^2 = x^4 + 4x^2 + 4\), which is \(4x^2\) too big - so subtract it back. This manufactured square is the standard trick (the Sophie Germain identity).
  2. Recognise a difference of two squares: \(A^2 - B^2\) with \(A = x^2 + 2\), \(B = 2x\).
    \(4x^2 = (2x)^2\), so the expression is now in the form that factorises.
  3. Factorise: \((x^2 + 2 - 2x)(x^2 + 2 + 2x) = (x^2 - 2x + 2)(x^2 + 2x + 2)\).
    \(A^2 - B^2 = (A-B)(A+B)\).
  4. Add the \(x\)-coefficients: \(-2 + 2 = 0\).
    The question asks only for this sum, which the symmetry makes obvious.
Where it goes wrong
  • Concluding it cannot be factorised because it has no real roots - a quartic with no real roots still factorises into two quadratics.
  • Adding the constant terms (\(2 + 2 = 4\)) instead of the \(x\)-coefficients.
Takeaway \(x^4 + 4a^4 = (x^2 - 2ax + 2a^2)(x^2 + 2ax + 2a^2)\): manufacture a square, then use the difference of two squares.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-09
15Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

Given that \(x^3 + px + q\) is divisible by \((x - 2)^2\), find the value of \(q\).

  1. A\(16\)
  2. B\(12\)
  3. C\(8\)
  4. D\(-8\)
  5. E\(-16\)
  6. F\(-12\)
Worked solution

\(x^3 + px + q = (x-2)^2(x - r)\). Comparing \(x^2\) coefficients: \(-4 - r = 0 \Rightarrow r = -4\). Constant term: \((4)(4) = 16 = q\). (Also \(p = 4 + 4r = -12\).)

Full method — how and why
A repeated factor \((x-2)^2\) means \(x = 2\) is a double root. Comparing coefficients in the factorised form gives \(q\) without any calculus.
  1. Write \(x^3 + px + q = (x - 2)^2(x - r)\).
    A cubic with a repeated root \(2\) and one other root \(r\) must take this form (leading coefficient \(1\)).
  2. Compare \(x^2\) coefficients. The left-hand side has none, so the coefficient is \(0\); expanding the right gives \(-(4 + r)\,\)... more precisely \((x^2 - 4x + 4)(x - r)\) has \(x^2\) coefficient \(-r - 4\).
    The absence of a term is information: its coefficient is zero.
  3. So \(-r - 4 = 0\), giving \(r = -4\).
    One comparison is enough to find the third root.
  4. Constant term: \(4 \times (-r) = 4 \times 4 = 16\), so \(q = 16\).
    The constant of \((x-2)^2(x-r)\) is \(4 \times (-r)\); with \(r = -4\) that is \(16\).
Where it goes wrong
  • Forgetting that the missing \(x^2\) term forces a condition, and trying to solve with only the constant term.
  • Sign slip on \(-r\), giving \(q = -16\).
Takeaway A missing power in a polynomial is a coefficient equal to zero - often the most useful equation available.
Answer A · Polynomials, factor and remainder theorems (MM1.6) · question POL-10
16Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

When \(x^3 - 3x^2 + 2x + 7\) is divided by \(x^2 + 1\), what is the remainder?

  1. A\(x + 4\)
  2. B\(x + 10\)
  3. C\(10\)
  4. D\(-x + 10\)
  5. E\(x + 7\)
  6. F\(3x + 4\)
Worked solution

\(x^3 - 3x^2 + 2x + 7 = (x^2 + 1)(x - 3) + (x + 10)\): expanding gives \(x^3 - 3x^2 + x - 3 + x + 10\). Remainder \(x + 10\).

Full method — how and why
Division by a quadratic again: remainder is linear. Comparing coefficients is fastest because the divisor has no \(x\) term.
  1. Write \(x^3 - 3x^2 + 2x + 7 = (x^2 + 1)(x + c) + (ax + b)\).
    Cubic divided by quadratic gives a linear quotient and a linear remainder.
  2. Expand the right: \(x^3 + cx^2 + x + c + ax + b\).
    \((x^2+1)(x+c) = x^3 + cx^2 + x + c\).
  3. Compare \(x^2\): \(c = -3\).
    Only the quotient contributes to \(x^2\).
  4. Compare \(x\): \(1 + a = 2\), so \(a = 1\). Compare constants: \(c + b = 7\), so \(b = 7 + 3 = 10\).
    Work down the powers; each comparison introduces exactly one new unknown.
  5. Remainder: \(x + 10\).
    Degree \(1 <\) degree \(2\) ✓.
Where it goes wrong
  • Forgetting the \(+x\) that \((x^2+1)(x)\) contributes, giving \(a = 2\) and the remainder \(2x + 10\).
  • Using the Remainder Theorem with \(x^2 + 1\) - it has no real roots, so that route needs complex numbers and is off-specification.
Takeaway Comparing coefficients handles any divisor; only use root substitution when the divisor has real roots.
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question POL-12
17Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★★☆

For which value of \(k\) does \(x^2 - x + k\) divide exactly into \(x^3 - 2x^2 + 3x - 4\)? (Assume such a value exists.)

  1. A\(-2\)
  2. B\(3\)
  3. C\(4\)
  4. D\(2\)
  5. E\(1\)
  6. Fno such value exists
Worked solution

Try \(x^3 - 2x^2 + 3x - 4 = (x^2 - x + k)(x + m)\): comparing \(x^2\): \(m - 1 = -2 \Rightarrow m = -1\). Then \(x\): \(k - m = k + 1 = 3 \Rightarrow k = 2\). Constant: \(km = -2 \neq -4\). No value of \(k\) works.

Full method — how and why
An 'assume it exists' question where the honest answer is that it does not. Comparing coefficients over-determines \(k\), and the system is inconsistent.
  1. If \(x^2 - x + k\) divides exactly, the quotient is linear: \(x^3 - 2x^2 + 3x - 4 = (x^2 - x + k)(x + m)\).
    Cubic \(\div\) quadratic \(=\) linear, and 'divides exactly' means no remainder term.
  2. Compare \(x^2\): \(m - 1 = -2\), so \(m = -1\).
    The \(x^2\) terms are \(mx^2\) and \(-x^2\).
  3. Compare \(x\): \(k - m = 3\), so \(k = 2\).
    The \(x\) terms are \(kx\) and \(-mx\).
  4. Check the constant term: \(km = 2 \times (-1) = -2\), but the cubic has \(-4\). Contradiction.
    With three equations and two unknowns the system is over-determined - the last equation is a genuine test, not a formality.
  5. So no value of \(k\) works: the answer is 'no such value exists'.
    The bracketed 'assume such a value exists' is a deliberate misdirection.
Where it goes wrong
  • Stopping at \(k = 2\) and never checking the constant term.
  • Trusting the question's hint that a value exists - always verify all the coefficients.
Takeaway When comparing coefficients gives more equations than unknowns, the leftover equations must be checked; they can reveal that no solution exists.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-16

How this topic has been examined

Real TMUA questions on Polynomials, factor and remainder theorems, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 7polynomials P(x) and Q(x)question papersit online
2023 · Paper 2Question 19roots of f(x) and x f(x)question papersit online
2020 · Paper 1Question 2factors of 2x^3 + px^2 + qquestion papersit online
2020 · Paper 1Question 20(x - a)(x^2 - x + a) = 0: exactly two rootsquestion papersit online
2020 · Paper 2Question 18f - g > 0 for cubicsquestion papersit online
2019 · Paper 2Question 2factors of 2x^3 + px^2 + qquestion papersit online
October 2018 · Paper 1Question 5cubic with coefficients 1, 2, 3 in some orderquestion papersit online
November 2017 · Paper 1Question 4remainder theorem with (px - 1)question papersit online
Practice papers · Paper 1Question 2factorise a cubic with factor (x + 2)question papersit online
Specimen papers · Paper 1Question 6(x - 2) is a factor: sum of possible valuesquestion papersit online
Specimen papers · Paper 1Question 11sum of roots of a factorised quarticquestion papersit online
Specimen papers · Paper 2Question 12deductions from p(1) = 2question papersit online