Ch 3 · 3 Factors of Polynomials · O Level Additional Mathematics 4037
Topic 3 of 6 in Algebra and Functions — O Level Additional Mathematics 4037 — 58 lessons.
Introduction
A quadratic can be factorised by spotting two numbers. A cubic such as 2x^3 - 5x^2 - 4x + 3 needs a cleverer idea: if putting x = 3 makes it zero, then (x - 3) must be a factor. That single fact, the factor theorem, unlocks cubic equations and lets you find unknown coefficients. (Syllabus topic 3.)
Key ideas — The remainder theorem
When a polynomial p(x) is divided by (x - a), the remainder is p(a). More generally, the remainder on dividing by (ax + b) is p (-b/a ).
Key ideas — The factor theorem
(x - a) is a factor of p(x) if and only if p(a) = 0. Similarly (ax + b) is a factor if and only if p (-b/a ) = 0.
🧑🏫 Teacher's explanation
Imagine you have a large number like 60. If you want to know if 5 is a factor, you divide 60 by 5. If there is no remainder, 5 is a factor. Polynomials work exactly the same way. Instead of dividing by a number, we divide by a linear expression like (x - 2). The Remainder Theorem tells us that if we plug x = 2 into the polynomial and the result is 0, then (x - 2) fits perfectly into the polynomial without any leftover pieces.
📖 Definitions
🔍 Worked example — Easy
Show that (x - 3) is a factor of p(x) = x^3 - 2x^2 - 5x + 6.
🔍 Worked example — Standard
Find all the factors of p(x) = x^3 + 2x^2 - 5x - 6 given that (x + 1) is a factor.
🔍 Worked example — Exam-style
Solve the equation 2x^3 - 5x^2 - 4x + 3 = 0.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — From one factor to all three
Once (x - a) is a factor, write p(x) = (x - a)(quadratic). Find the quadratic by:
🧑🏫 Teacher's explanation
Think of a cubic polynomial like a large, complex machine made of three smaller gears. If you know one of the gears (a linear factor), you can 'unplug' it to see what the remaining mechanism looks like. Once that large mechanism is simplified into a quadratic, it becomes much easier to take apart.
📖 Definitions
🔍 Worked example — Standard
Given that (x + 1) is a factor of 2x^3 + 5x^2 - x - 6, find the other factors.
🔍 Worked example — Exam-style
Find the values of x for which x^3 - 7x + 6 = 0, given that x = 1 is a solution.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Find the value of a. (b) Hence, factorise P(x) completely. [5 marks]
Key ideas — Unknown coefficients
Each piece of information gives one equation: "(x - 2) is a factor" gives p(2) = 0; "the remainder on division by (x + 1) is -9" gives p(-1) = -9. Two facts let you find two unknowns by solving simultaneously.
🧑🏫 Teacher's explanation
Think of a polynomial like a complex machine with hidden settings (the unknown coefficients). Usually, we don't know how the machine works until we test it with specific inputs. In algebra, testing an input is like plugging a number into the function. If the machine outputs zero when you plug in a specific number, you have discovered a 'secret key' or a factor that unlocks the whole equation.
📖 Definitions
🔍 Worked example — Easy
Given p(x) = x^3 + ax^2 + bx + 6, find the values of a and b if (x - 1) is a factor and the remainder when p(x) is divided by (x + 1) is 12.
🔍 Worked example — Standard
The polynomial f(x) = 2x^3 + ax^2 + bx - 10 has a factor (x + 2). When f(x) is divided by (x - 1), the remainder is -12. Find a and b.
🔍 Worked example — Exam-style
Find the values of k if (x - 2) is a factor of x^3 + kx^2 - 5x + 6.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Solving cubic equations
Factorise fully, then set each factor to zero. A cubic has at least one real root and at most three. If the quadratic factor has a negative discriminant, the cubic has exactly one real root.
🧑🏫 Teacher's explanation
Think of a cubic polynomial like a complex machine made of three interconnected gears. To understand how the whole machine works, you first need to find one simple, single gear (a linear factor) that fits perfectly inside it. Once you've extracted that one gear, the rest of the machine becomes a much simpler quadratic mechanism that you already know how to dismantle.
📖 Definitions
🔍 Worked example — Easy
Show that (x - 2) is a factor of P(x) = x^3 - 4x^2 + 5x - 2, and hence solve P(x) = 0.
🔍 Worked example — Standard
Find all real roots of x^3 - 6x^2 + 11x - 6 = 0.
🔍 Worked example — Exam-style
A cubic polynomial P(x) = 2x^3 + ax^2 - 5x + 6 has (x + 2) as a factor. Find the value of a and solve P(x) = 0.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: solve a cubic
Solve 2x^3 - 5x^2 - 4x + 3 = 0.
🔍 Worked example — Easy
Show that (x - 2) is a factor of x^3 - 4x^2 + x + 6 and hence solve x^3 - 4x^2 + x + 6 = 0.
🔍 Worked example — Standard
Solve 2x^3 - 5x^2 - 4x + 3 = 0.
🔍 Worked example — Exam-style
A polynomial f(x) = 2x^3 + ax^2 - 5x + b has factors (x - 1) and (x + 2). Find the values of a and b, and solve f(x) = 0.
⚠️ Common errors
Worked examples — Example 2: finding a and b
p(x) = 2x^3 + ax^2 + bx - 4. (x - 2) is a factor, and the remainder on division by (x + 1) is -9.
🔍 Worked example — Easy
Given p(x) = x^3 + ax + b, find a and b if (x - 1) is a factor and the remainder when divided by (x + 1) is 4.
🔍 Worked example — Standard
p(x) = 2x^3 + ax^2 + bx - 4. (x - 2) is a factor, and the remainder on division by (x + 1) is -9. Find a and b.
🔍 Worked example — Exam-style
The polynomial f(x) = x^3 + ax^2 + bx + 6 has (x + 2) as a factor. When f(x) is divided by (x - 1), the remainder is 12. Find the values of a and b.
⚠️ Common errors
Worked examples — Example 3: long division
Divide x^3 - 2x^2 - 5x + 6 by (x - 1): the quotient is x^2 - x - 6 and the remainder is 0, so x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2).
🔍 Worked example — Easy
Use long division to divide x^3 + 4x^2 - x - 4 by (x + 1).
🔍 Worked example — Standard
Given that (x - 2) is a factor of 2x^3 - 5x^2 - x + 6, find the other factors.
🔍 Worked example — Exam-style
Solve the equation x^3 - 2x^2 - 5x + 6 = 0 completely.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 27 - 36 + 6 + 5 = 2. (2) -8 + 12 - 4 = 0 ✓; (x + 2)(x^2 + x - 2) = (x + 2)^2(x - 1). (3) p (tfrac12 ) = tfrac14 + tfrac p4 - tfrac72 + 3 = 0 ⇒ p = 1. (4) x = -1 works: (x + 1)(x^2 - x - 6) = (x + 1)(x - 3)(x + 2); x = -1, 3, -2. (5) 1 + a + b + 6 = 0 and 8 + 4a + 2b + 6 = 12: a + b = -7, 2a + b = -1, so a = 6, b = -13.