Factor Theorem (AQA GCSE Further Maths)

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10 October 2024

Definition of a Polynomial

What is a polynomial?

A polynomial is a sum of terms with non-negative integer powers of x
- the highest power of x is its degree
The following are polynomials:
- $x^{3} + 4 x^{2} - 3 x + 1$ (degree 3)
- $5 x^{6} - 2 x^{2}$ (degree 6)
- $10$ (degree 0)
The following are not polynomials:
- $\frac{1}{x} + x^{3}$ (the $\frac{1}{x}$ can be written $x^{- 1}$ which has a negative power)
- $x^{2} + 3 x + \sqrt{x}$ (the $\sqrt{x}$ can be written $x^{\frac{1}{2}}$ which has a non-integer power)
- $x^{5} + 2 - \sin x$ ( $\sin x$ is not a power of x)
There are words to name different types of polynomials:
- degree
  name
  0
  constant
  1
  linear
  2
  quadratic
  3
  cubic
  4
  quartic
  5
  quintic
  ...
  ...
For the polynomial $6 x^{4} - 3 x^{2} + 2 x + 8$
- ... the degree is 4 (it's a quartic)
- ... the coefficient of x² is -3
- ... the quadratic term (or term involving x²) is -3x²
- ... the coefficient of x³ is 0
- ... the constant is 8

How do I add / subtract / multiply polynomials?

Adding and subtracting two polynomials requires collecting like terms
- for example, $(x^{3} + 4 x) + (x^{2} + 8 x + 3)$ gives $x^{3} + x^{2} + 12 x + 3$
- whereas $(x^{3} + 4 x) - (x^{2} + 8 x + 3)$ gives $x^{3} - x^{2} - 4 x - 3$
  - the second bracket expands to $- x^{2} - 8 x - 3$
Multiplying two polynomials can be done by expanding brackets or using a grid method (multiplying rows by columns then combining terms to get the answer)
- for example, $(x^{3} + 4 x) (x^{2} + 8 x + 3)$ can be done as follows:
- x³
  4x
  x²
  x⁵
  4x³
  8x
  8x⁴
  32x²
  3
  3x³
  12x
- add any like terms that arise: $3 x^{3} + 4 x^{3} = 7 x^{3}$
- the final answer is therefore $x^{5} + 8 x^{4} + 7 x^{3} + 32 x^{2} + 12 x$

Factor Theorem

What is a factor of a polynomial?

You already know that some quadratic expressions can be factorised
- $x^{2} + 5 x + 6$ factorises to $(x + 2) (x + 3)$
- (x + 2) and (x + 3) are called factors
  - as the power of x in each factor is 1, they can also be called linear factors
Similarly, other polynomials can be factorised
$4 x^{3} + 8 x^{2} - 9 x - 18$ factorises to $(x + 2) (2 x + 3) (2 x - 3)$
- there are three linear factors
  - or, by expanding the last two brackets, $(x + 2) (4 x^{2} - 9)$ , you could write it as one linear factor and one quadratic factor
Rational factors refer to linear factors in the form (ax + b), with a number in front of the x, like (2x + 3)

What is the Factor Theorem for (x - a)?

Let f(x) be a polynomial
- The Factor Theorem states that if f(a) = 0 then (x - a) is a factor
- It also works in reverse, so if (x - a) is a factor then f(a) = 0
For example, try substituting x = 2 and x = 4 into $f (x) = x^{3} - 6 x^{2} + 11 x - 6$
- $f (2) = 2^{3} - 6 \times 2^{2} + 11 \times 2 - 6 = 0$ (zero)
- $f (4) = 4^{3} - 6 \times 4^{2} + 11 \times 4 - 6 = 6$ (not zero)
- The Factor Theorem says that (x - 2) is a factor of f(x), but (x - 4) is not
  - It tells you without you having to factorise f(x)
Be careful with the signs
- f(2) = 0 means (x - 2) is a factor, not (x + 2)

What is the Factor Theorem for (ax - b)?

Let f(x) be a polynomial
- The Factor Theorem above can be extended to say that if $f (\frac{b}{a}) = 0$ then (ax - b) is a factor
- It also works in reverse, so if (ax - b) is a factor then $f (\frac{b}{a}) = 0$
This is sometimes called The Factor Theorem for rational factors, (ax - b)
For example, you can show that (2x - 3) is a factor of $f (x) = 4 x^{3} + 8 x^{2} - 9 x - 18$ without doing any factorising
- If (2x - 3) really is a factor, then the Factor Theorem says $f (\frac{3}{2})$ should equal zero - check to see if that's true
  - $f (\frac{3}{2}) = 4 {(\frac{3}{2})}^{3} + 8 {(\frac{3}{2})}^{2} - 9 (\frac{3}{2}) - 18 = 0$ so yes, (2x - 3) is a factor (by the Factor Theorem)
Be careful with the signs and fraction order
- $f (\frac{3}{2}) = 0$ means (3x - 2) is a factor, not any of (3x + 2), (2x - 3) or (2x + 3)

Examiner Tips and Tricks

To help remember what to substitute into f(x) when (ax - b) is a factor, a good trick is to set (ax - b) equal to zero and solve for x
- For example, to show that (7x + 5) is a factor, first try solving 7x + 5 = 0 to get $x = - \frac{5}{7}$ , which shows you what to substitute into f(x), i.e. $f (- \frac{5}{7})$

Worked Example

If $(2 x + 1)$ is a factor of $2 x^{3} + k x^{2} - 8 x - 3$ where $k$ is an integer, find the value of $k$ .

Assign the $f (x)$ notation to the given polynomial.

$\begin{array}{rcl} f (x) & = & 2 x^{3} + k x^{2} - 8 x - 3 \end{array}$

Set the factor equal to 0 and solve for $x$ to find the value that should be substituted into $f (x)$ .

$\begin{array}{rcl} 2 x + 1 & = & 0 \\ x & = & - \frac{1}{2} \end{array}$

Therefore $f (- \frac{1}{2}) = 0$ .

Substitute $x = - \frac{1}{2}$ and set to 0.

$\begin{array}{rcl} 2 {(- \frac{1}{2})}^{3} + k {(- \frac{1}{2})}^{2} - 8 (- \frac{1}{2}) - 3 & = & 0 \end{array}$

Simplify.

$\begin{array}{rcl} 2 (- \frac{1}{8}) + k (\frac{1}{4}) - 8 (- \frac{1}{2}) - 3 & = & 0 \\ - \frac{2}{8} + \frac{k}{4} + 4 - 3 & = & 0 \\ - \frac{1}{4} + \frac{k}{4} + 1 & = & 0 \end{array}$

Subtract 1 from both sides and multiplying both sides by 4.

$\begin{array}{rcl} - \frac{1}{4} + \frac{k}{4} & = & - 1 \\ - 1 + k & = & - 4 \end{array}$

Solve by adding 1 to both sides.

$k = - 3$

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Previous:Algebraic ProofNext:Applications of Factor Theorem

degree	name
0	constant
1	linear
2	quadratic
3	cubic
4	quartic
5	quintic
...	...

	x³	4x
x²	x⁵	4x³
8x	8x⁴	32x²
3	3x³	12x