Factor Theorem Revision Notes for Grade 12 NSC Matric Mathematics

Factor Theorem

Understanding the factor theorem

The factor theorem is a powerful tool that helps us determine whether one polynomial is a factor of another. To understand this concept, let's first think about how it works with regular numbers.

When we divide an integer a by an integer b, we can write this as:

$a = b \times q + r$

where q is the quotient and r is the remainder. If the remainder r = 0, then we know that b is a factor of a.

This same principle applies to polynomials. When we divide a polynomial a(x) by another polynomial b(x), we get:

$a(x) = b(x) \cdot Q(x) + R(x)$

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The beauty of the factor theorem is that it provides a direct connection between polynomial division and factorisation. Just like with integers, when the remainder is zero, we have found a factor!

If the remainder $R(x) = 0$ , then $b(x)$ is a factor of $a(x)$ .

The factor theorem statement

The factor theorem provides us with a simple way to test whether a linear expression is a factor of a polynomial.

chatImportant

The Factor Theorem

If the polynomial $p(x)$ is divided by $cx - d$ and the remainder, given by $p\left(\frac{d}{c}\right)$ , equals zero, then $cx - d$ is a factor of $p(x)$ .

Converse: If $(cx - d)$ is a factor of $p(x)$ , then $p\left(\frac{d}{c}\right) = 0$ .

This theorem tells us that we can determine if a linear expression like $(x - 3)$ or $(x + 2)$ is a factor simply by substituting the root value into the polynomial. If the result is zero, then it's a factor!

General approach to factorising cubic polynomials

When factorising a cubic polynomial, follow these systematic steps:

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Step-by-Step Approach to Cubic Factorisation

Find one factor by trial and error: Examine the coefficients of the polynomial $p(x)$ and guess a possible root value
Use the factor theorem to confirm: Show that $p(\text{root value}) = 0$ to verify it's actually a root
Divide by the linear factor: Use polynomial division or factorisation by inspection to find the remaining quadratic factor
Factorise the quadratic: Apply standard quadratic factorisation methods to complete the process

Worked examples

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Worked Example 1: Proving a factor using the factor theorem

Question: Using the factor theorem, show that $y + 4$ is a factor of $g(y) = 5y^4 + 16y^3 - 15y^2 + 8y + 16$ .

Solution:

Step 1: Determine the approach
For $y + 4$ to be a factor, we need $g(-4) = 0$ .

Step 2: Calculate $g(-4)$
$g(y) = 5y^4 + 16y^3 - 15y^2 + 8y + 16$
$g(-4) = 5(-4)^4 + 16(-4)^3 - 15(-4)^2 + 8(-4) + 16$
$= 5(256) + 16(-64) - 15(16) + 8(-4) + 16$
$= 1280 - 1024 - 240 - 32 + 16$
$= 0$

Step 3: Conclusion
Since $g(-4) = 0$ , y + 4 is a factor of g(y).

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Worked Example 2: Testing if an expression is a factor

Question: Use the factor theorem to determine if $y - 1$ is a factor of $f(y) = 2y^4 + 3y^2 - 5y + 7$ .

Solution:

Step 1: Determine the approach
For $y - 1$ to be a factor, we need $f(1) = 0$ .

Step 2: Calculate $f(1)$
$f(y) = 2y^4 + 3y^2 - 5y + 7$
$f(1) = 2(1)^4 + 3(1)^2 - 5(1) + 7$
$= 2 + 3 - 5 + 7$
$= 7$

Step 3: Conclusion
Since $f(1) \neq 0$ , y - 1 is not a factor of f(y).

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Worked Example 3: Complete factorisation of a cubic polynomial

Question: Factorise completely: $f(x) = x^3 + x^2 - 9x - 9$

Solution:

Step 1: Find a factor by trial and error
Try $f(1) = (1)^3 + (1)^2 - 9(1) - 9 = 1 + 1 - 9 - 9 = -16$
Therefore $(x - 1)$ is not a factor.

Let's try $f(-1) = (-1)^3 + (-1)^2 - 9(-1) - 9 = -1 + 1 + 9 - 9 = 0$
Therefore (x + 1) is a factor.

Step 2: Factorise by inspection
$x^3 + x^2 - 9x - 9 = (x + 1)(x^2 + ?x - 9)$

To find the middle term coefficient:

First term in second bracket must be $x^2$ to give $x^3$
Last term must be $-9$ because $(+1)(-9) = -9$
For the $x$ -term: $(+1)(x^2)$ gives no $x$ -term, so coefficient must be $0$

Therefore: $f(x) = (x + 1)(x^2 - 9)$

Step 3: Complete the factorisation
$x^2 - 9$ is a difference of two squares: $x^2 - 9 = (x - 3)(x + 3)$

Final answer: f(x) = (x + 1)(x - 3)(x + 3)

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Worked Example 4: Another cubic factorisation

Question: Use the factor theorem to factorise $f(x) = x^3 - 2x^2 - 5x + 6$ .

Solution:

Step 1: Find a factor by trial and error
Try $f(1) = (1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$
Therefore (x - 1) is a factor.

Step 2: Factorise by inspection
$x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 + ?x - 6)$

Working backwards:

First term must be $x^2$ to give $x^3$
Last term must be $-6$ because $(-1)(-6) = +6$
For the coefficient of $x$ : $(-1)(x^2)$ gives $-x^2$ , so we need $+x$ from the middle term to get overall $-x^2$

So we have: $f(x) = (x - 1)(x^2 + x - 6)$

Step 3: Factorise the quadratic
$x^2 + x - 6 = (x - 2)(x + 3)$

Final answer: f(x) = (x - 1)(x - 2)(x + 3)

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Key Points to Remember:

The factor theorem states: if $p(a) = 0$ , then $(x - a)$ is a factor of $p(x)$
To test if $(x - a)$ is a factor, substitute x = a into the polynomial - if the result is zero, it's a factor
For cubic polynomials, find one linear factor first, then divide to get a quadratic which can be factorised using standard methods
Always check your factorisation by expanding back to the original polynomial
Common exam trap: remember that if $(x + a)$ is a factor, you substitute $x = -a$ into the polynomial

Factor Theorem (Grade 12 NSC Matric Mathematics): Revision Notes

Factor Theorem

Understanding the factor theorem

The factor theorem statement

General approach to factorising cubic polynomials

Worked examples

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