Free Maths Tutorial: Polynomials [Explained with Examples]

This Mathematics Tutorial will focus on POLYNOMIALS. We will explain it with examples and give you exercise. At the end of the tutorial, you’ll be able to download it for FREE. Please share this page with your friends who may need it.

Polynomials are mathematical expressions consisting of variables raised to non-negative integer powers, combined using addition, subtraction, and multiplication operations.

To understand polynomials better, imagine you have some blocks with letters on them, and you can arrange these blocks in a line. Each block has a special letter and a number.

A polynomial is like a line of these blocks, where each block represents a special kind of math.

Let’s say the blocks look like this:

3x² + 2x + 1

Now, let’s break it down:

The Blocks: Each part of the line is a block. In our line, we have three blocks: 3x², 2x, and 11.
The Special Letter (x): This is like a mystery number. It can be any number you want to put in. You can pick 1, 2, 3, or any other number.
The Numbers (Coefficients): Each block has a number in front.
For 3x², the number is 3.
For 2x, the number is 2.
For 1, the number is 1.
The Powers: The little number up high (like the 2 in 3x²) says how many times you multiply the special letter by itself. It’s like saying “take x and multiply it by x again.”

So, if you want to know what the line is when you put a number into the special letter, you follow the rules. For example, if you put 2 in for x, you do this:

3×(2²) + 2×2 + 1 = 17

So when x = 2, the answer will be 17.

That’s basically how a polynomial works – it’s a way to write down a math story using blocks with special letters, numbers, and powers!

According to textbooks, a polynomial can be written in the form:
P(x) = a_nxⁿ + a_n−1xⁿ⁻¹ + … + a₂x² + a₁x + a₀

Here:

P(x) is the polynomial function.
x is the variable.
a_n,a_n−1,…,a₂,a₁,a₀ are coefficients, which can be any real numbers.
n is a non-negative integer and represents the degree of the polynomial. The highest power of x in the polynomial determines the degree.

For example, the polynomial P(x) = 3x² − 2x + 1 is a quadratic polynomial (degree 2) with coefficients 3, -2, and 1.

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Change of Subject of Formula

Changing the subject of a formula involves rearranging an equation to isolate a particular variable. For example:

Exercise 1: Given the formula A = πr² for the area of a circle, express r in terms of A.

Solution 1:

The first step is to isolate r. We do this by dividing both sides of the equation by π.

ie. A/ π= πr² / π

The result will be r² = A/ π

But what we need is r, not r²

To get just r, we take the square root of both sides, to get:

√r² = √A/ π

Therefore, r² = √A / π

HOT TIP: Square-root will always cancel out root.

Factor and Remainder Theorems

Theorem 1:

The Factor Theorem states that if f(c) = 0, then (x−c) is a factor of f(x).

Exercise 2: If f(x) = 2x³−3x²−5x+6 and f(2)=0, find the factor.

Solution 2:

If f(x) = 2x³−3x²−5x+6 and f(2)=0, it means that x=2 is a root of the polynomial.

In other words, x−2 is a factor of the polynomial.

Theorem 2:

The Remainder Theorem states that if f(c) is divided by (x−c), the remainder is f(c).

Exercise 3: Find the remainder when (x) = 2x³−3x²−5x+6 is divided by (x−2).

Solution 3:

To find the remainder when the polynomial 2x³−3x²−5x+6 is divided by (x−2), you can use the Remainder Theorem. According to the theorem, if you substitute the root (in this case, x=2) into the polynomial, the result will be the remainder.

Let’s substitute x=2 into the polynomial:

R = 2(2)³ − 3(2)² − 5(2) + 6

R = 16−12−10+6

R = 0

So, the remainder R is 0. Therefore, when the polynomial 2x³−3x²−5x+6 is divided by x−2, the remainder is 0.

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Factorization of Polynomials of Degree

Factorization of polynomials involves expressing a given polynomial as the product of its factors.

Here’s a general approach to factorizing polynomials of various degrees:

Factorization of Quadratic Polynomials (Degree 2):

Quadratic trinomials are polynomials of degree 2, meaning they have the form ax² + bx + c, where a, b, and c are constants and a is not equal to 0. The goal is to factorize them into the product of two binomials.

Quadratic Trinomial: ax² + bx + c

Factorization: (px+q)(rx+s)

Steps for Factorization:

Find p and r:
- p and r are the coefficients of x².
- Multiply them to get a: p × r = a.
Find q and s:
- q and s are factors of the constant term c.
- Multiply them to get c: q×s=c.
Write the Factorization:
- Express bx using the values of p, q, r, and s.

Example: Factorize x² − 5x + 6

Solution:

Recall that ax² + bx + c = (px+q)(rx+s)

Hence, to factorize x² − 5x + 6, we do the following steps:

p × r = 1 (the coefficient of x²).

Since 1 x 1 = 1, therefore, p = 1, r = 1

q × s = 6 (the constant term).

q = -2, s = -3 (because -2 × -3 = 6)

NOTE: The reason for choosing q=-2 and s=-3 is because it certifies both the constant and middle terms.

Let me explain:

The quadratic trinomial is x² −5x + 6. When we factorize it into (x+q)(x+s), we need to find q and s such that:

q × s = 6 (the constant term).
q + s= −5 (the coefficient of the middle term).

In this case, if we choose q=-2 and s=-3, it satisfies both conditions:

q×s = (-2) × (-3) = 6 (constant term condition is met).
q+s = (-2) + (−3)= −5 (this gives us the middle term −5x).

Therefore, the correct factorization is (x−2)(x−3)

Verification: (x-2)(x−3) = x²−3x – 2x + 6 = x² − 5x + 6

So, the factorization is correct.

Practice Exercise: Factorize 2x²−7x+3.

Your answer should be (2x−1)(x−3)

Try it out!

You can verify the solution by multiplying the factors back together to ensure you get the original quadratic trinomial.

That’s all we will cover here.

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