Polynomials

In this note, you will learn:

· Multiplication of polynomials

· Equality of polynomials

· Division 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)

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

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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