__In this note, you will learn:__

· __What is Remainder Theorem?__

· __Factor theorem: A special case of Remainder Theorem__

__What is Remainder Theorem?__

__What is Remainder Theorem?__

In the previous mathematics article on ** Polynomials**, we learnt to divide a polynomial using the long division method.

However, that is only necessary if the divisor (the polynomial which you want to divide the dividend by) is non-linear, meaning that the degree of the polynomial is greater than 1, or 2 and above.

If the divisor given is linear, there is another easier way of find the remainder of the dividend.

As you can see from the diagram above, the examples have shown a particular relationship between the **remainder** and **P(-b/a)**, and hence we can conclude the relationship to be as such:

**If a polynomial P(x) is divided by a linear divisor ax + b, the remainder of P(x) will be P(-b/a)**

This is also known as the **Remainder Theorem**, and is a **special case** of the Division Algorithm for polynomials when the divisor is linear.

Let’s take a look at an example of how we can apply the Remainder Theorem in the following question:

Find the remainder when 5x^4 + 3x^3 – 8x^2 + 7x + 4 is divided by

a) x + 1

b) x – 5

Solution:

Let P(x) = 5x^4 + 3x^3 – 8x^2 + 7x + 4,

a) Remainder = P (-1)

= 5(-1)^4 + 3(-1^)3 – 8(-1)^2 + 7(-1) + 4

= -9

b) Remainder = P [-(-5)/1]

= P (5)

= 5(5)^4 + 3(5)^3 – 8(5)^2 + 7(5) + 4

= 3339

__Factor theorem: A special case of Remainder Theorem__

__Factor theorem: A special case of Remainder Theorem__

In the previous point, we have learnt that when a polynomial **P(x)**, is divided by a linear divisor **ax + b**, then the remainder will be given by: **P(-b/a)**.

This derived from the substitution of **x = -b/a** into the Division Algorithm for polynomials:

P(x) = (ax + b) Q(x) + R

If **x = -b/a** is substituted into the equation, then:

P(-b/a) = [a(-b/a) + b] Q(x) + R

P(-b/a) = (-b + b) Q(x) + R

P(-b/a) = (0) Q(x) + R

Therefore, **P(-b/a) = R**

However, if the remainder is equals to **zero**, i.e. R = P(-b/a) = 0, then P(x) = (ax + b) Q(x),

i.e. **ax + b is a factor of P(x)**, and hence we can come to a conclusion that:

**ax + b is a factor of the polynomial P(x) if and only if P(-b/a) = 0**

This is also known as the Factor Theorem, and the Factor Theorem is a **special case** of the Remainder Theorem when the remainder is **zero**.

Let’s take a look at how we can apply the Factor Theorem in the questions shown below:

Question 1:

i) Find the value of k for which x + 1 is a factor of f(x) = 3x^3 + 11x^2 + 14x + k.

ii) Hence, find the remainder when f(x) is divided by 3x + 5.

Solution:

i) f(x) = 3x^3 + 11x^2 + 14x + k

By Factor Theorem, since x + 1 is a factor of f(x), then f (-1) = 0.

3(-1)^3 + 11(-1)^2 + 14(-1) + k = 0

(-3) + (11) – 14 = -k

-6 = -k

Therefore, k = 6

ii) f(x) = 3x^3 + 11x^2 + 14x + 6

When f(x) is divided by 3x + 5, remainder R = f (-5/3).

Therefore, R = 3(-5/3)^3 + 11(-5/3)^2 + 14(-5/3) + 6

= -2/3

Question 2:

i) Find the value of a and b for which x^2 – x – 2 is a factor of x^3 + ax^2 – 5x + b.

ii) Using the values of a and b found in part (i), factorize completely the cubic polynomial 3x^3 + ax^2 – 5x + b.

Solution:

i) Let f(x) = x^3 + ax^2 – 5x + b.

By the Factor Theorem, since **x^2 – x – 2 = (x + 1) (x – 2)** is a factor of f(x), then **x + 1** and **x – 2 **are also factors of f(x).

Therefore, f (-1) = 0 and f (2) = 0.

f (-1) = (-1)^3 + a (-1^)2 – 5(-1) + b = 0

= (-1) + a + 5 + b

= a + b + 4 -> Equation 1

f (2) = (2)^3 + a (2)^2 – 5(2) + b = 0

= 8 + 4a – 10 + b

= 4a + b – 2 -> Equation 2

From Equation 2,

4a + b – 2 = 0

Therefore, b = 2 – 4a -> Equation 3

Substituting Equation 3 into Equation 1,

a + (2 – 4a) + 4 = 0

-3a = -6

a = 2

Substituting a = 2 into Equation 2,

4(2) + b – 2 = 0

b = 2 – 8

= -6

Therefore, a = 2 and b = -6

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on Remainder Theorem and Factor Theorem, and is able to apply it to solve for questions involving polynomials more quickly!

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