Factor & Remainder Theorem | DP IB Maths: AA HL Revision Notes 2021

Factor Theorem

The factor theorem is used to find the linear factors of polynomial equations
This topic is closely tied to finding the zeros and roots of a polynomial function/equation
- As a rule of thumb a zero refers to the polynomial function and a root refers to a polynomial equation

For any polynomial function P(x)
- (x - k) is a factor of P(x) if P(k) = 0
- P(k) = 0 if (x - k) is a factor of P(x)

Consider the polynomial function P(x) = a_nxⁿ+ a_n-1xⁿ^-1+ … + a₁x + a₀ and (x - k) is a factor

Then, due to the factor theorem P(k) = a_nkⁿ+ a_n-1kⁿ^-1+ … + a₁k + a₀ = 0
$P (x) = (x - k) \times Q (x)$ , where Q(x) is a polynomial that is a factor of P(x)
Hence, $\frac{P (x)}{x - k} = Q (x)$ , where Q(x) is another factor of P(x)

If the linear factor has a coefficient of x then you must first factorise out the coefficient

If the linear factor is $(a x - b) = a (x - \frac{b}{a}) \to P (\frac{b}{a}) = 0$

A common mistake in exams is using the incorrect sign for either the root or the factor
If you are asked to find integer solutions to a polynomial then you only need to consider factors of the constant term

Determine whether $(x - 2)$ is a factor of the following polynomials:

f (x) = x^{3} - 2 x^{2} - x + 2

g (x) = 2 x^{3} + 3 x^{2} - x + 5

2-7-1-ib-aa-hl-factor-theorem-b-we-solution

It is given that $(2 x - 3)$ is a factor of $h (x) = 2 x^{3} - b x^{2} + 7 x - 6$ .

Find the value of

b

mZEjMdDm_2-7-1-ib-aa-hl-factor-theorem-c-we-solution

The remainder theorem is used to find the remainder when we divide a polynomial function by a linear function
When any polynomial P(x) is divided by any linear function (x - k) the value of the remainder R is given by P(k) = R
- Note, when P(k) = 0 then (x - k) is a factor of P(x)

Consider the polynomial function P(x) = a_nxⁿ+ a_n-1xⁿ^-1+ … + a₁x + a₀ and the linear function (x - k)

Then, due to the remainder theorem P(k) = a_nkⁿ+ a_n-1kⁿ^-1+ … + a₁k + a₀ = R
$P (x) = (x - k) \times Q (x) + R$ , where Q(x) is a polynomial
Hence, $\frac{P (x)}{x - k} = Q (x) + \frac{R}{x - k}$ , where R is the remainder

If the linear function has a coefficient of x then you must first factorise out the coefficient

If the linear function is $(a x - b) = a (x - \frac{b}{a}) \to P (\frac{b}{a}) = R$

Let $f (x) = 2 x^{4} - 2 x^{3} - x^{2} - 3 x + 1$ , find the remainder $R$ when $f (x)$ is divided by:

x - 3

2-7-1-ib-aa-hl-remainder-theorem-a-we-solution

x + 2

2-7-1-ib-aa-hl-remainder-theorem-b-we-solution The remainder when $f (x)$ is divided by $(2 x + k)$ is $\frac{893}{8}$ .

Given that

k > 0

, find the value of

k

2-7-1-ib-aa-hl-remainder-theorem-c-we-solution