Factors of polynomials

1. Overview

Polynomials are a fundamental topic in IGCSE Additional Mathematics. You'll need to master factorising, finding remainders, and solving polynomial equations (especially cubics) to succeed in coordinate geometry, calculus, and other advanced topics. This section covers the Remainder and Factor Theorems, algebraic long division, and how to use these tools to solve cubic equations efficiently. Expect to see these skills tested in both Paper 1 (without a calculator) and Paper 2.

Key Definitions

• Polynomial: An expression of the form $P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$, where $n$ is a non-negative integer.
• Degree: The highest power of the variable $x$ in the polynomial.
• Remainder Theorem: A rule used to find the remainder when a polynomial is divided by a linear factor $(ax - b)$ without using long division.
• Factor Theorem: A specific case of the Remainder Theorem where the remainder is zero, indicating that $(ax - b)$ is a factor.
• Root (or Zero): The value of $x$ for which $P(x) = 0$.

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Core Content

The Remainder Theorem

When a polynomial $P(x)$ is divided by $(ax - b)$, the remainder is given by $P\left(\frac{b}{a}\right)$.

• If dividing by $(x - c)$, the remainder is $P(c)$.
• If dividing by $(x + c)$, the remainder is $P(-c)$.

The Factor Theorem

$(ax - b)$ is a factor of $P(x)$ if and only if $P\left(\frac{b}{a}\right) = 0$.
This is the most powerful tool for breaking down cubic equations into manageable parts.

Factoring Cubic Polynomials

To factorize a cubic $P(x) = ax^3 + bx^2 + cx + d$:

1. Find the first linear factor: Use the Factor Theorem by testing factors of the constant term $d$ (e.g., $\pm 1, \pm 2$).
2. Reduce to quadratic: Once a factor $(x - k)$ is found, divide $P(x)$ by $(x - k)$ using algebraic long division or the method of equating coefficients to find the quadratic quotient.
3. Factorize the quadratic: Use standard methods (inspection, completing the square, or the quadratic formula) to find the remaining factors.

Worked Example 1 — Remainder Theorem

Find the remainder when $P(x) = 2x^3 - 5x^2 + x - 7$ is divided by $(x - 3)$.

Step 1: Identify the value to substitute.
By the Remainder Theorem, $x - 3 = 0 \implies x = 3$.

Step 2: Substitute into $P(x)$.
$P(3) = 2(3)^3 - 5(3)^2 + (3) - 7$

Step 3: Evaluate.
$P(3) = 2(27) - 5(9) + 3 - 7$
$P(3) = 54 - 45 + 3 - 7$
$P(3) = 5$

Result: The remainder is $\boxed{5}$.

Worked Example 2 — Fully Factorizing a Cubic

Fully factorize $f(x) = x^3 - 6x^2 + 11x - 6$.

Step 1: Find a linear factor using the Factor Theorem.
Try $f(1)$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$.
Since $f(1) = 0$, $(x - 1)$ is a factor.

Step 2: Use algebraic long division (or inspection) to find the quadratic.
$x^3 - 6x^2 + 11x - 6 = (x - 1)(ax^2 + bx + c)$
By observation:

• $x \times ax^2 = x^3 \implies a = 1$
• $-1 \times c = -6 \implies c = 6$
  Comparing $x^2$ terms: $-1x^2 + bx^2 = -6x^2 \implies b = -5$
  So, $f(x) = (x - 1)(x^2 - 5x + 6)$.

Step 3: Factorize the quadratic.
$x^2 - 5x + 6 = (x - 2)(x - 3)$

Final Answer: $f(x) = (x - 1)(x - 2)(x - 3)$

Worked Example 3 — Solving Equations and the Discriminant

Solve $x^3 - 3x + 2 = 0$.

Step 1: Find a root.
Try $x = 1$: $1^3 - 3(1) + 2 = 0$. Thus $(x - 1)$ is a factor.

Step 2: Division.
Divide $(x^3 - 3x + 2)$ by $(x - 1)$ to get $(x^2 + x - 2)$.
Equation becomes: $(x - 1)(x^2 + x - 2) = 0$.

Step 3: Solve the quadratic.
$(x - 1)(x + 2)(x - 1) = 0$
$(x - 1)^2(x + 2) = 0$

Result: $x = 1$ (repeated root) or $x = -2$.

Worked Example 4 — Using the Remainder and Factor Theorems Together

The polynomial $P(x) = 2x^3 - 3x^2 - 11x + 6$.

(a) Show that $(x - 3)$ is a factor of $P(x)$.
(b) Hence factorise $P(x)$ completely.

Step 1: Apply the Factor Theorem for part (a).
$P(3) = 2(3)^3 - 3(3)^2 - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$
Since $P(3) = 0$, $(x - 3)$ is a factor of $P(x)$. ✓

Step 2: Divide $P(x)$ by $(x - 3)$ using algebraic long division (or comparing coefficients).
$2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2)$

Check by expanding: $(x-3)(2x^2 + 3x - 2) = 2x^3 + 3x^2 - 2x - 6x^2 - 9x + 6 = 2x^3 - 3x^2 - 11x + 6$ ✓

Step 3: Factorise the quadratic $2x^2 + 3x - 2$.
$2x^2 + 3x - 2 = (2x - 1)(x + 2)$

Step 4: Write the complete factorisation.
$P(x) = (x - 3)(2x - 1)(x + 2)$

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Key Equations

• The Remainder Theorem: $P(k) = R$, where $R$ is the remainder when $P(x)$ is divided by $(x - k)$.
• The Factor Theorem: If $P(k) = 0$, then $(x - k)$ is a factor.
• The Quadratic Formula (to solve the remaining factor): $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• The Discriminant: $\Delta = b^2 - 4ac$
  • $\Delta > 0$: Two distinct real roots.
  • $\Delta = 0$: One repeated real root.
  • $\Delta < 0$: No real roots.

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Common Mistakes to Avoid

• ❌ Wrong: Guessing roots of a cubic equation and moving straight to the answer without showing $P(k) = 0$.
  ✓ Right: Always write "Since $P(k) = 0$, by the factor theorem $(x - k)$ is a factor."
• ❌ Wrong: Stopping after finding the linear and quadratic factors when the question asks to "Solve."
  ✓ Right: Set the factors to zero and find the values of $x$.
• ❌ Wrong: Forgetting to check the discriminant when a quadratic cannot be factorized easily.
  ✓ Right: If a quadratic doesn't factorize, use $b^2 - 4ac$ to prove whether further real roots exist.

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Exam Tips

• Command Words:
  • "Show that": You must evaluate $P(k)$ and explicitly state it equals zero to conclude $(x - k)$ is a factor.
  • "Find all real roots": This implies you must check if the quadratic part has real solutions using the discriminant.
• Calculator Use (Paper 2): Many modern calculators can solve cubic equations. Use this to check your final answers, but you must show the algebraic steps (Long division or Equating Coefficients) to gain full marks.
• The "Constant Term" Trick: When searching for a first whole-number root, test the factors of the constant term. If the polynomial ends in $+6$, test $\pm 1, \pm 2, \pm 3, \pm 6$. If the $x^3$ coefficient is not 1, a root can also be a fraction, such as $x = \frac{1}{2}$ for the factor $(2x - 1)$.
• Missing Terms: If a polynomial is $x^3 - 7x + 6$, remember the $x^2$ coefficient is $0$ when performing long division ($x^3 + 0x^2 - 7x + 6$).

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The polynomial $f(x) = 2x^3 + ax^2 + bx - 6$ has a factor of $(x-1)$ and leaves a remainder of 4 when divided by $(x+1)$.

(a) Find the values of $a$ and $b$. [6 marks]

(b) Find the remaining quadratic factor of $f(x)$. [2 marks]

Worked Solution:

(a)

1. Apply the factor theorem: Since $(x-1)$ is a factor, $f(1) = 0$.
   $2(1)^3 + a(1)^2 + b(1) - 6 = 0$
   $2 + a + b - 6 = 0$
   $a + b = 4$

2. Apply the remainder theorem: When divided by $(x+1)$, the remainder is 4, so $f(-1) = 4$.
   $2(-1)^3 + a(-1)^2 + b(-1) - 6 = 4$
   $-2 + a - b - 6 = 4$
   $a - b = 12$

3. Solve the simultaneous equations $a + b = 4$ and $a - b = 12$.
   Adding the two equations: $2a = 16$, so $a = 8$.
   Substituting $a = 8$ into $a + b = 4$: $8 + b = 4$, so $b = -4$.

   $\boxed{a=8, b=-4}$

How to earn full marks: Use the factor theorem and the remainder theorem to write one equation each, and show both equations before you solve them together.

(b)

1. We now have $f(x) = 2x^3 + 8x^2 - 4x - 6$. Since we know $(x-1)$ is a factor, we can use polynomial long division.

2. Perform polynomial division to find the quadratic factor.
   $(2x^3 + 8x^2 - 4x - 6) \div (x - 1) = 2x^2 + 10x + 6$.

3. $\boxed{2x^2 + 10x + 6}$

How to earn full marks: Show the division (or comparing coefficients), not just the final quadratic.

Common Pitfall: When using the factor theorem, make sure you substitute the correct value into the polynomial. A common mistake is to substitute the factor directly instead of solving for x first. Also, double-check your arithmetic when solving simultaneous equations to avoid errors.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The polynomial $f(x) = x^3 - x^2 - 8x - 4$.

(a) Show that $(x+2)$ is a factor of $f(x)$. [2 marks]

(b) Hence, factorise $f(x)$ completely. [3 marks]

(c) Solve the equation $f(x) = 0$. [2 marks]

Worked Solution:

(a)

1. Apply the factor theorem: If $(x+2)$ is a factor, then $f(-2) = 0$.
   $f(-2) = (-2)^3 - (-2)^2 - 8(-2) - 4 = -8 - 4 + 16 - 4 = 0$

2. Since $f(-2) = 0$, $(x+2)$ is a factor.

How to earn full marks: Show the substitution term by term, then state that $f(-2) = 0$, so $(x+2)$ is a factor.

(b)

1. Perform polynomial division: $(x^3 - x^2 - 8x - 4) \div (x+2)$.
   This gives $x^2 - 3x - 2$.

2. Thus, $x^3 - x^2 - 8x - 4 = (x+2)(x^2 - 3x - 2)$.

3. The discriminant of $x^2-3x-2$ is $(-3)^2 - 4(1)(-2) = 9+8 = 17$, which is not a perfect square, so $x^2 - 3x - 2$ does not factorise further and the factorisation is complete.
   $\boxed{(x+2)(x^2 - 3x - 2)}$

How to earn full marks: Show the division, then check whether the quadratic factor factorises further before you stop.

(c)

1. From part (b), the equation is $(x+2)(x^2 - 3x - 2) = 0$.

2. One solution is $x = -2$.

3. Use the quadratic formula to solve $x^2 - 3x - 2 = 0$.
   $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} = \frac{3 \pm \sqrt{17}}{2}$

4. $\boxed{x = -2, x = \frac{3 + \sqrt{17}}{2}, x = \frac{3 - \sqrt{17}}{2}}$

How to earn full marks: Give all three roots, including $x = -2$, and leave the other two in exact surd form.

Common Pitfall: Remember to include all solutions when solving a cubic equation. Don't forget the solution you used to perform the division in the first place! Also, be careful with signs when using the quadratic formula.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [9 marks]

Question:

The polynomial $g(x) = x^3 + px^2 + qx - 12$ is such that $(x-2)$ is a factor and when $g(x)$ is divided by $(x+1)$ the remainder is -15.

(a) Find the values of $p$ and $q$. [5 marks]

(b) Show that $x = 2$ is the only real root of the equation $g(x) = 0$. [4 marks]

Worked Solution:

(a)

1. Apply the factor theorem: Since $(x-2)$ is a factor, $g(2) = 0$.
   $(2)^3 + p(2)^2 + q(2) - 12 = 0$
   $8 + 4p + 2q - 12 = 0$
   $4p + 2q = 4$
   $2p + q = 2$

2. Apply the remainder theorem: When divided by $(x+1)$, the remainder is -15, so $g(-1) = -15$.
   $(-1)^3 + p(-1)^2 + q(-1) - 12 = -15$
   $-1 + p - q - 12 = -15$
   $p - q = -2$

3. Solve the simultaneous equations $2p + q = 2$ and $p - q = -2$.
   Adding the two equations: $3p = 0$, so $p = 0$.
   Substituting $p = 0$ into $2p + q = 2$: $2(0) + q = 2$, so $q = 2$.
   $\boxed{p=0, q=2}$

How to earn full marks: Form one equation from the factor theorem and one from the remainder theorem, then solve them together. Check your values by substituting back.

(b)

1. We now have $g(x) = x^3 + 0x^2 + 2x - 12 = x^3 + 2x - 12$. Since we know $(x-2)$ is a factor, we can perform polynomial division.

2. $(x^3 + 2x - 12) \div (x-2) = x^2 + 2x + 6$.

3. So, $x^3 + 2x - 12 = (x-2)(x^2 + 2x + 6)$.

4. Check whether $x^2 + 2x + 6 = 0$ has real roots using the discriminant:
   Discriminant $= 2^2 - 4(1)(6) = 4 - 24 = -20$

5. The discriminant is negative, so $x^2 + 2x + 6 = 0$ has no real roots. The only real root of $g(x) = 0$ is $x = 2$.

How to earn full marks: Find the quadratic factor, work out its discriminant and state clearly that it is negative, so there are no more real roots.

Common Pitfall: When solving for the coefficients in part (a), be extremely careful with your signs. A small error there will propagate through the rest of the problem. Also, a negative discriminant does not mean you made a mistake: it means the quadratic factor gives no more real roots.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [8 marks]

Question:

The polynomial $h(x) = x^3 + ax + b$ has a factor of $(x-3)$. When $h(x)$ is divided by $(x+2)$, the remainder is -10.

(a) Find the values of $a$ and $b$. [4 marks]

(b) Show that $x = 3$ is the only real root of $h(x) = 0$. [4 marks]

Worked Solution:

(a)

1. Apply the factor theorem: Since $(x-3)$ is a factor, $h(3) = 0$.
   $(3)^3 + a(3) + b = 0$
   $27 + 3a + b = 0$
   $3a + b = -27$

2. Apply the remainder theorem: When divided by $(x+2)$, the remainder is -10, so $h(-2) = -10$.
   $(-2)^3 + a(-2) + b = -10$
   $-8 - 2a + b = -10$
   $-2a + b = -2$

3. Solve the simultaneous equations $3a + b = -27$ and $-2a + b = -2$.
   Subtracting the second equation from the first: $(3a + b) - (-2a + b) = -27 - (-2)$
   $5a = -25$, so $a = -5$.
   Substituting $a = -5$ into $-2a + b = -2$: $-2(-5) + b = -2$, so $10 + b = -2$, and $b = -12$.

   $\boxed{a=-5, b=-12}$

How to earn full marks: Write one equation from each theorem, solve them together, and check that $h(3) = 0$ with your values.

(b)

1. We now have $h(x) = x^3 - 5x - 12$. Since we know $(x-3)$ is a factor, we can perform polynomial division.

2. $(x^3 - 5x - 12) \div (x-3) = x^2 + 3x + 4$.

3. So, $x^3 - 5x - 12 = (x-3)(x^2 + 3x + 4)$.

4. Check whether $x^2 + 3x + 4 = 0$ has real roots using the discriminant:
   Discriminant $= 3^2 - 4(1)(4) = 9 - 16 = -7$

5. The discriminant is negative, so $x^2 + 3x + 4 = 0$ has no real roots. The only real root of $h(x) = 0$ is $x = 3$.

How to earn full marks: Show the division, work out the discriminant of the quadratic factor and state clearly that it is negative, so there are no more real roots.

Common Pitfall: Don't assume that all cubic equations will have three real roots. Always check the discriminant of the quadratic factor to determine the nature of the remaining roots. If the discriminant is negative, the quadratic factor gives no more real roots.