__In this note, you will learn:__

· __Multiplication of polynomials__

· __Equality of polynomials__

· __Division of polynomials__

__Multiplication of polynomials__

__Multiplication of polynomials__

Back in lower secondary math, we have learnt that the product of algebraic expressions can be found using the **Distributive Law**.

Similarly, the same law can be applied when we are trying to find the product of polynomials.

Let’s take a look at an example below:

If P(x) = 3x^5 – 6x^4 and Q(x) = 4x^2 + 8x – 7, find

i) P(x) x Q(x)

ii) the relationship between the degrees of P(x), Q(x) and P(x) x Q(x).

Solution:

i) P(x) x Q(x) = (3x^5 – 6x^4) (4x^2 + 8x – 7)

= 12x^7 + 24x^6 – 21x^5 – 24x^6 – 48x^5 + 42x^4

= 12x^7 – 69x^5 + 42x^4

ii) Degree of P(x) = 5

Degree of Q(x) = 2

Degree of P(x) x Q(x) = 7

Therefore, degree of P(x) x Q(x) = degree of P(x) + degree of Q(x)

__Equality of polynomials__

__Equality of polynomials__

In the scenario where two polynomials are given, P(x) = Ax^4 + Bx^3 + Cx^2 + Dx + E and Q(x) = 5x^4 + 2x^3 – 3x^2 + 8x – 15, and they are said to be **equivalent**, then the equation P(x) = Q(x) will be true for all real values of x. i.e. A = 5, B = 2, C = -3, D = 8 and E = -15. We call the equation a **polynomial identity**.

Let’s take a look at an example of how this works below:

Given that 6x^2 – 8x + 9 = a (x – 1) (x + 2) + b (x – 4) + c for all values of x, find the values of a, b and c.

Solution:

There are two methods for this type of questions: **Substitution** and **Equating the coefficient**.

**Method 1: Substitution**

Let x = 1: 6(1)^2 – 8(1) + 9 = a [(1) – 1] [(1) + 2] + b [(1) – 4] + c

6 – 8 + 9 = -3b + c

c = 7 + 3b -> **Equation 1**

Let x = -2: 6(-2)^2 – 8(-2) + 9 = a [(-2) – 1] [(-2) + 2] + b [(-2) – 4] + c

24 + 16 + 9 = -6b + c

49 = -6b + c -> **Equation 2**

Substituting **Equation 1** into **Equation 2**,

49 = -6b + (7 + 3b)

42 = -3b

Therefore, **b = -14**

Substituting **b = -14** into **Equation 1**,

c = 7 + 3(-14)

c = 7 – 42

Therefore, **c = -35**

Let x = 4: 6(4)^2 – 8(4) + 9 = a [(4) – 1] [(4) + 2] + (-14) [(4) – 4] + (-35)

96 – 32 + 9 = a (3) (6) - 35 **a = 6**

Therefore, **a = 6**, **b = -14** and **c = -35**

**Note:** The values we let x to be is not completely random! It is wise to pick values of x that eliminates a certain term so that only one or two unknown coefficients are left, which we can then easily work them out using substitution. i.e. Given that we have **a (x – 3) + b (x + 4)**, we will let **x = 3** to eliminate the term with coefficient **a** first so that we can find coefficient **b**.

**Method 2: Equating coefficients**

6x^2 – 8x + 9 = a (x – 1) (x + 2) + b (x – 4) + c

= a (x^2 + x – 2) + b (x – 4) + c

= ax^2 + (a + b) x – 2a – 4b + c

Equating coefficients of x^2: 6 = a

Equating coefficients of x: -8 = a + b

-8 = (6) + b

b = -14

Equating coefficients of x^0: 9 = -2a – 4b + c

9 = -2(6) – 4(-14) + c

c = 12 – 56 + 9

c = -35

Therefore, **a = 6**, **b = -14** and **c = -35**

__Division of polynomials__

__Division of polynomials__

The division of polynomials is actually really simple. Before we begin, let’s recall on how we do long division of positive integers.

3 -> quotient

i.e. divisor <- 5⟌16 -> dividend

__-15__

__ 1 ->__ remainder

We can express the dividend in terms of the divisor, quotient and remainder as follows:

16 = 5 x 3 + 1

**dividend = divisor x quotient + remainder**

This application is actually the exact method of how we can divide polynomials, which is by **long division**.

Let’s take a look at an example of how this is done:

i) Divide **3x^3 + 4x – 9** by **x – 3** and state the remainder.

ii) Hence, express **3x^3 + 4x – 9** in terms of **x – 3** using the Division Algorithm.

Solution:

i) __3x^2 + 9x + 23__

x - 3⟌3x^3 + 0x^2 + 4x – 9

__-(3x^3 – 9x^2) __

9x^2 + 4x

__ 9x^2 – 27x__

– 23x – 9

__ 23x – 69__

__ -78__

Therefore, remainder = -78

ii) 3x^3 + 4x – 9 = (x – 3) (3x^2 + 9x + 23) – 78

(dividend = divisor x quotient + remainder)

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on the different mathematical operations regarding polynomials!

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