Chapter 4: Factors of Polynomials
4.1 Polynomials – Definition and Terminology
Definition:
A polynomial in \(x\) is an expression of the form:
\[
P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
\]
where \(n\) is a non-negative integer (the degree), and \(a_n \neq 0\).
Examples:
| Polynomial |
Degree |
Leading Coefficient |
| \(3x^2 - 5x + 2\) |
2 |
3 |
| \(4x^5 - x^3 + 7\) |
5 |
4 |
| \(x - 9\) |
1 |
1 |
| \(8\) |
0 |
8 |
Key terminology:
- Degree: The highest power of \(x\) with a non-zero coefficient.
- Leading coefficient: The coefficient of the highest power term.
- Constant term: The term with no \(x\) (i.e., \(a_0\)).
- Root / zero: A value \(r\) such that \(P(r) = 0\).
- Factor: An expression \((x - r)\) that divides exactly into \(P(x)\).
Proof: A polynomial of degree \(n\) has at most \(n\) roots
This is a fundamental result (Fundamental Theorem of Algebra).
For O Level, we accept that a cubic (degree 3) has at most 3 real roots, a quadratic has at most 2, etc.
The Factor Theorem (Section 4.3) is the tool we use to find them.
4.2 The Remainder Theorem
Statement:
When a polynomial \(P(x)\) is divided by \((x - a)\), the remainder is \(P(a)\).
In other words:
\[
P(x) = (x - a)Q(x) + R
\]
where \(Q(x)\) is the quotient (also a polynomial) and \(R\) is the remainder (a constant).
Substituting \(x = a\):
\[
P(a) = (a - a)Q(a) + R = 0 + R = R
\]
Proof: The Remainder Theorem
Given: \(P(x) = (x - a)Q(x) + R\)
To prove: The remainder is \(P(a)\).
Substitute \(x = a\) into both sides:
\[
P(a) = (a - a)Q(a) + R = 0 \cdot Q(a) + R = R
\]
Therefore, the remainder is indeed \(P(a)\).
Extended version:
When \(P(x)\) is divided by \((ax - b)\), the remainder is \(P\left(\frac{b}{a}\right)\).
Proof: Let \(ax - b = 0 \Rightarrow x = \frac{b}{a}\).
By the same reasoning as above, remainder = \(P\left(\frac{b}{a}\right)\).
Worked Example 4.1 (Using the Remainder Theorem – non-calculator)
Find the remainder when \(P(x) = 2x^3 - 5x^2 + 3x - 7\) is divided by \((x - 2)\).
Solution:
By the Remainder Theorem, remainder = \(P(2)\):
\[
P(2) = 2(2)^3 - 5(2)^2 + 3(2) - 7
\]
\[
= 2(8) - 5(4) + 6 - 7 = 16 - 20 + 6 - 7 = -5
\]
Answer: Remainder = \(-5\)
Worked Example 4.2 (Remainder when divisor is \(ax - b\))
Find the remainder when \(P(x) = 3x^3 + 4x^2 - 2x + 1\) is divided by \((2x - 3)\).
Solution:
Set \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\).
Remainder = \(P\left(\frac{3}{2}\right)\):
\[
P\left(\frac{3}{2}\right) = 3\left(\frac{3}{2}\right)^3 + 4\left(\frac{3}{2}\right)^2 - 2\left(\frac{3}{2}\right) + 1
\]
\[
= 3\left(\frac{27}{8}\right) + 4\left(\frac{9}{4}\right) - 3 + 1
\]
\[
= \frac{81}{8} + 9 - 3 + 1 = \frac{81}{8} + 7 = \frac{81}{8} + \frac{56}{8} = \frac{137}{8}
\]
Answer: Remainder = \(\frac{137}{8}\)
Worked Example 4.3 (Finding an unknown coefficient using the Remainder Theorem)
When \(P(x) = x^3 + kx^2 - 4x + 6\) is divided by \((x - 3)\), the remainder is 24. Find the value of \(k\).
Solution:
By the Remainder Theorem, remainder = \(P(3) = 24\):
\[
P(3) = (3)^3 + k(3)^2 - 4(3) + 6
\]
\[
= 27 + 9k - 12 + 6 = 21 + 9k
\]
Set \(21 + 9k = 24\):
\[
9k = 3 \Rightarrow k = \frac{1}{3}
\]
Answer: \(k = \frac{1}{3}\)
4.3 The Factor Theorem
Statement:
\((x - a)\) is a factor of \(P(x)\) if and only if \(P(a) = 0\).
Derivation (from the Remainder Theorem):
When \(P(x)\) is divided by \((x - a)\), remainder = \(P(a)\).
\((x - a)\) is a factor exactly when the remainder is 0.
Hence \((x - a)\) is a factor iff \(P(a) = 0\).
Proof: The Factor Theorem
Part 1: If \((x - a)\) is a factor, then \(P(a) = 0\).
If \((x - a)\) is a factor, then \(P(x) = (x - a)Q(x)\) for some polynomial \(Q(x)\).
Substitute \(x = a\): \(P(a) = (a - a)Q(a) = 0\).
Part 2: If \(P(a) = 0\), then \((x - a)\) is a factor.
By the Remainder Theorem, remainder = \(P(a) = 0\).
Since the remainder is 0, \(P(x)\) is exactly divisible by \((x - a)\).
Thus \((x - a)\) is a factor.
Extended version:
\((ax - b)\) is a factor of \(P(x)\) iff \(P\left(\frac{b}{a}\right) = 0\).
Worked Example 4.4 (Testing for a factor – non-calculator)
Show that \((x - 2)\) is a factor of \(P(x) = x^3 - 6x^2 + 11x - 6\).
Solution:
By the Factor Theorem, test \(P(2)\):
\[
P(2) = (2)^3 - 6(2)^2 + 11(2) - 6
\]
\[
= 8 - 24 + 22 - 6 = 0
\]
Since \(P(2) = 0\), \((x - 2)\) is indeed a factor.
Worked Example 4.5 (Finding a factor and factorising fully)
Given that \((x - 2)\) is a factor of \(P(x) = x^3 - 6x^2 + 11x - 6\), factorise completely.
Solution:
Divide \(P(x)\) by \((x - 2)\) using synthetic division:
|
1 |
-6 |
11 |
-6 |
| 2 |
|
2 |
-8 |
6 |
|
1 |
-4 |
3 |
0 |
Quotient: \(x^2 - 4x + 3 = (x - 1)(x - 3)\)
Therefore:
\[
P(x) = (x - 2)(x^2 - 4x + 3) = (x - 2)(x - 1)(x - 3)
\]
Answer: \(P(x) = (x - 2)(x - 1)(x - 3)\)
Worked Example 4.6 (Mauritian context – finding unknown coefficient)
A company's profit function is \(P(x) = x^3 + 3x^2 - kx + 12\), where \(x\) is the number of units sold (in hundreds). If \((x - 4)\) is a factor of \(P(x)\), find \(k\).
Solution:
By the Factor Theorem, if \((x - 4)\) is a factor, then \(P(4) = 0\):
\[
P(4) = (4)^3 + 3(4)^2 - k(4) + 12
\]
\[
= 64 + 48 - 4k + 12 = 124 - 4k
\]
Set \(124 - 4k = 0 \Rightarrow 4k = 124 \Rightarrow k = 31\).
Answer: \(k = 31\)
4.4 Finding Factors of Polynomials
Strategy for factorising a cubic:
- Guess a root \(r\) by testing factors of the constant term.
- Use the Factor Theorem to check if \(P(r) = 0\).
- If \(P(r) = 0\), then \((x - r)\) is a factor.
- Divide \(P(x)\) by \((x - r)\) (using long division or synthetic division) to get a quadratic quotient.
- Factorise the quadratic (if possible).
Rational Root Theorem (optional – for guidance):
If \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0\) has integer coefficients, then any rational root \(\frac{p}{q}\) must have:
- \(p\) dividing \(a_0\) (constant term)
- \(q\) dividing \(a_n\) (leading coefficient)
For O Level, we typically test integer factors of the constant term.
Worked Example 4.7 (Factorising a cubic)
Factorise fully: \(P(x) = x^3 + 2x^2 - 5x - 6\).
Solution:
Possible roots: \(\pm 1, \pm 2, \pm 3, \pm 6\)
Test \(x = -1\):
\[
(-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0 \quad \checkmark
\]
So \((x + 1)\) is a factor.
Divide \(P(x)\) by \((x + 1)\) using synthetic division:
|
1 |
2 |
-5 |
-6 |
| -1 |
|
-1 |
-1 |
6 |
|
1 |
1 |
-6 |
0 |
Quotient: \(x^2 + x - 6 = (x + 3)(x - 2)\)
Therefore:
\[
P(x) = (x + 1)(x + 3)(x - 2)
\]
Answer: \(P(x) = (x + 1)(x + 3)(x - 2)\)
Worked Example 4.8 (Cubic with leading coefficient \(\neq 1\))
Factorise: \(P(x) = 2x^3 + x^2 - 13x + 6\).
Solution:
Possible rational roots: \(\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}\)
Test \(x = 2\):
\[
2(8) + 4 - 26 + 6 = 16 + 4 - 26 + 6 = 0 \quad \checkmark
\]
So \((x - 2)\) is a factor.
Divide \(P(x)\) by \((x - 2)\) using synthetic division:
|
2 |
1 |
-13 |
6 |
| 2 |
|
4 |
10 |
-6 |
|
2 |
5 |
-3 |
0 |
Quotient: \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\)
Therefore:
\[
P(x) = (x - 2)(2x - 1)(x + 3)
\]
Answer: \(P(x) = (x - 2)(2x - 1)(x + 3)\)
4.5 Solving Cubic Equations
Steps:
- Factorise \(P(x)\) completely (using the Factor Theorem).
- Set each factor equal to 0 and solve.
Key fact: A cubic equation has at most 3 real roots.
Worked Example 4.9 (Solving a cubic equation – non-calculator)
Solve \(x^3 - 6x^2 + 11x - 6 = 0\).
Solution:
From Example 4.5, factorised form: \((x - 1)(x - 2)(x - 3) = 0\)
Set each factor to zero:
\[
x - 1 = 0 \Rightarrow x = 1
\]
\[
x - 2 = 0 \Rightarrow x = 2
\]
\[
x - 3 = 0 \Rightarrow x = 3
\]
Answer: \(x = 1, 2, 3\)
Worked Example 4.10 (Cubic with a quadratic factor that has no real roots)
Solve \(x^3 - 3x^2 + 4x - 12 = 0\).
Solution:
Test \(x = 3\):
\[
27 - 27 + 12 - 12 = 0 \quad \checkmark
\]
So \((x - 3)\) is a factor.
Divide \(P(x)\) by \((x - 3)\):
|
1 |
-3 |
4 |
-12 |
| 3 |
|
3 |
0 |
12 |
|
1 |
0 |
4 |
0 |
Quotient: \(x^2 + 4 = 0 \Rightarrow x^2 = -4\)
This has no real solutions (since \(x^2 \ge 0\) cannot equal \(-4\)).
Answer: \(x = 3\) (only real root)
Worked Example 4.11 (Mauritian context – break-even points)
A factory's profit (in thousands of rupees) is given by:
\[
P(x) = x^3 - 10x^2 + 31x - 30
\]
where \(x\) is the production level (in hundreds of units). Find the break-even points (where profit = 0).
Solution:
Solve \(P(x) = 0\):
Test \(x = 2\):
\[
8 - 40 + 62 - 30 = 0 \quad \checkmark
\]
So \((x - 2)\) is a factor.
Divide \(P(x)\) by \((x - 2)\):
|
1 |
-10 |
31 |
-30 |
| 2 |
|
2 |
-16 |
30 |
|
1 |
-8 |
15 |
0 |
Quotient: \(x^2 - 8x + 15 = (x - 3)(x - 5)\)
Factorised: \(P(x) = (x - 2)(x - 3)(x - 5)\)
Set to zero: \(x = 2, 3, 5\)
Answer: Break-even at production levels of 200, 300, and 500 units.
Chapter 4 Summary
| Concept |
Key points |
| Polynomial |
\(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0\) |
| Degree |
Highest power of \(x\) |
| Remainder Theorem |
\(P(x) \div (x - a)\) \(\rightarrow\) remainder = \(P(a)\) |
| Factor Theorem |
\((x - a)\) is a factor iff \(P(a) = 0\) |
| Finding factors |
Test factors of constant term; divide to get quadratic |
| Solving cubics |
Factorise fully \(\rightarrow\) set each factor = 0 |
Exercises – Chapter 4
Easy (Drill – Non-Calculator)
- Find the remainder when:
(a) \(P(x) = x^3 - 4x^2 + 2x - 5\) is divided by \((x - 3)\)
(b) \(P(x) = 2x^3 + 3x^2 - x + 7\) is divided by \((x + 2)\)
- Show that \((x - 1)\) is a factor of \(P(x) = x^3 - 2x^2 - x + 2\).
- Factorise fully:
(a) \(x^3 - 6x^2 + 11x - 6\)
(b) \(x^3 + 2x^2 - x - 2\)
Medium (Examination Style)
- Find the remainder when \(P(x) = 2x^3 - 3x^2 + 5x - 2\) is divided by \((2x - 1)\).
- Given that \((x - 3)\) is a factor of \(P(x) = x^3 + kx^2 - 8x - 15\), find \(k\).
- Factorise fully:
(a) \(x^3 + 4x^2 + x - 6\)
(b) \(2x^3 - 3x^2 - 11x + 6\)
- Solve:
(a) \(x^3 - 7x^2 + 14x - 8 = 0\)
(b) \(x^3 - 4x^2 + x + 6 = 0\)
Hard (Challenge for A*)
- The polynomial \(P(x) = x^3 + ax^2 + bx - 20\) has factors \((x - 2)\) and \((x + 4)\). Find \(a\) and \(b\), and factorise fully.
- Prove that \((x - 2)\) is a factor of \(x^n - 2^n\) for all positive integers \(n\).
Hint: use the Factor Theorem.
- Mauritian context: The volume of a rectangular box (in cm³) is given by:
\[
V(x) = x^3 - 12x^2 + 47x - 60
\]
where \(x\) is the height in cm. The box has integer dimensions.
(a) Factorise \(V(x)\).
(b) Find the three possible heights if the volume is 0.
(c) If the volume must be positive, what range of \(x\) is possible?
(d) The box is made from a sheet of metal. If the length is 7 cm and width is 5 cm, what is the height? (Assume volume = length \(\times\) width \(\times\) height.)