# Polynomials

## Polynomials Revision

**Polynomials**

A **polynomial** is an expression of algebraic terms. You will see and use polynomials all throughout this course

**Expanding Brackets**

There are different types of brackets that you will need to expand. For each, you need to use the **FOIL** method that you will have seen in GCSE Maths. Here are some examples:

**Single Brackets**

\textcolor{limegreen}{a}(x+y+z) = \textcolor{limegreen}{a}x + \textcolor{limegreen}{a}y + \textcolor{limegreen}{a}z

**Double brackets**

(\textcolor{limegreen}{a}+\textcolor{blue}{b})(x+y) = \textcolor{limegreen}{a}x + \textcolor{limegreen}{a}y + \textcolor{blue}{b}x + \textcolor{blue}{b}y

**Squared Brackets**

(\textcolor{limegreen}{a}+\textcolor{blue}{b})^2 = a^2 + ab + ab + b^2 = \textcolor{limegreen}{a}^2 + 2\textcolor{limegreen}{a}\textcolor{blue}{b} + \textcolor{blue}{b}^2

**Difference of Two Squares**

(\textcolor{limegreen}{a}+\textcolor{blue}{b})(\textcolor{limegreen}{a}-\textcolor{blue}{b}) = a^2 - ab + ab - b^2 = \textcolor{limegreen}{a}^2 - \textcolor{blue}{b}^2

**Longer Brackets**

(\textcolor{limegreen}{a}+\textcolor{blue}{b}+\textcolor{orange}{c})(w+x+y+z) = \textcolor{limegreen}{a}(w+x+y+z) + \textcolor{blue}{b}(w+x+y+z) + \textcolor{orange}{c}(w+x+y+z)

You then multiply out each bracket in the same way for single brackets.

**Simplifying Expressions**

A number, bracket or variable that is in each term of an expression is a **common factor**. You can take common factors outside a bracket.

**Example:** Simplify x(x+2)(x-3) + x^2 (x+2) - x(x+2) fully.

\textcolor{red}{x} is in each term of the expression, so we can take this out as a common factor

\textcolor{red}{x} \left[(x+2)(x-3) + x(x+2) - (x+2) \right]

Also, \textcolor{blue}{(x+2)} is a common factor:

\textcolor{red}{x}\textcolor{blue}{(x+2)} \left[x - 3 + x - 1 \right]

We can then simplify the terms inside the square brackets:

\textcolor{red}{x}\textcolor{blue}{(x+2)}\textcolor{black}{(2x-4)}

**Factorising Tip:**

For a quadratic in the form ax^2 + bx + c:

- When b and c are positive, both brackets will contain a +
- When b is negative and c is positive, both brackets will contain a -
- When c is negative, one bracket will contain a + and the other will contain a -

**Factorising Quadratics (a=1)**

**Factorising quadratics** of the form ax^2+bx+c when **a=1** is fairly straightforward.

**Example:** Factorise the following quadratic: x^2\textcolor{blue}{-7}x+\textcolor{red}{12}

**Step 1:** Firstly, write two brackets with an x placed at the start each bracket.

(x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,) (x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,)

**Step 2:** b is negative and c is positive, so both numbers in the brackets will need to be negative.

**Step 3: **Find two numbers which **multiply to make **\textcolor{red}{12} and **add or subtract to make** \textcolor{blue}{-7}

We know both numbers will be negative:

-3 \times -4 = \textcolor{red}{12}

-3 + -4= \textcolor{blue}{-7}

Finally, add these numbers to the brackets:

\textcolor{black}{(x-3) (x-4)}

**Factorising Quadratics (a> 1)**

It becomes a little trickier when **factorising quadratics** of the form ax^2+bx+c where **a>1**.

**Example:** Factorise the following quadratic: 4x^2-\textcolor{blue}{11x}\textcolor{red}{-3}

**Step 1: **When a>1 it makes factorising more complicated. It is not immediately obvious what the **coefficient **of each x term should be. Here, there are two possible options:

(4x \kern{1 cm} ) (x \kern{1 cm} ) or (2x \kern{1 cm} ) (2x \kern{1 cm} )

**Step 2: **c is negative, so one bracket will contain a + and the other a -

There are two options to place the + and - for the brackets containing 4x and x, and one option for the brackets containing 2x, since putting the symbols the other way round would give the same result.

\begin{aligned}(4x \kern{0.4 cm} +\kern{0.4 cm} )&(x \kern{0.4 cm}-\kern{0.4 cm} )\\ (4x \kern{0.4 cm} -\kern{0.4 cm} )&(x \kern{0.4 cm}+\kern{0.4 cm} )\\(2x \kern{0.4 cm}+\kern{0.4 cm} )&(2x \kern{0.35 cm}-\kern{0.35 cm})\end{aligned}

**Step 3:** We need to find **two numbers** which **multiply to make** \textcolor{red}{-3}

-3 \times 1 = -3 or -1 \times 3

**Step 4: **We need to find a combination which when **added or subtracted** gives \textcolor{blue}{-11x}

We can try the 6 possibilities:

\begin{aligned} \textcolor{black}{(4x+1)(x-3)} &= \textcolor{black}{4x^2 - 11x -3} \\ (4x+3)(x-1) &= 4x^2 - x -3 \\ (4x-3)(x+1) &= 4x^2 +x -3 \\ (4x-1)(x+3) &= 4x^2 + 11x -3 \\ (2x+1)(2x-3) &= 4x^2 -4x -3 \\ (2x+3)(2x-1) &= 4x^2 +4x -3 \end{aligned}

\textcolor{black}{(4x+1) (x-3)} gives the correct expansion, so this is the answer.

**Example: Expanding Brackets**

Expand and simplify the following brackets:

**a)** (2x+y)^2

**b)** (3x^2 + 2x)(x^3 + 5x + 2)

**c)** (3x+4)(3x-4)

**[6 marks]**

**a)** (2x+y)^2 = (2x)^2 + 2(2x)(y) + (y)^2 = 4x^2 + 4xy + y^2

**b)** (3x^2 + 2x)(x^3 + 5x + 2) \\ = 3x^2(x^3 + 5x + 2) + 2x(x^3 + 5x + 2) \\ = 3x^5 + 15x^3 + 6x^2 + 2x^4 + 10x^2 + 4x \\ = 3x^5 + 2x^4 + 15x^3 + 16x^2 + 4x

**c)** (3x+4)(3x-4) = (3x)^2 - (4)^2 = 9x^2 - 16

## Polynomials Example Questions

**Question 1:** Simplify the following expression: (x-3)(x-4) + x(x-4) + (x-4)^2

**[2 marks]**

**Question 2:** Expand the following brackets: (2x^2 + 6x + 4)(3x^2 - x - 3)

**[2 marks]**

**Question 3:** Factorise the following quadratic: p^2 + 4p - 21

**[2 marks]**

c is negative, so one number will be positive and the other will be negative.

We need two numbers which add to make 4 and multiply to make -21

The numbers that satisfy this are 7 and -3

Therefore, the factorisation of p^2 + 4p - 21 is

(p+7)(p-3)

**Question 4:** Factorise the following quadratic: 3y^2 + 5y + 2

**[2 marks]**

b and c are both positive, therefore both numbers in the brackets will be positive.

Therefore, there is one possible option for the brackets:

(3y \quad + \quad)(y \quad + \quad)

We need to find two numbers which multiply to make 2:

1 \times 2 = 2

Now, we can try the combinations to see which will result in +5y:

\begin{aligned} (3y+1)(y+2) &= 3y^2 + 7y + 2 \\ (3y+2)(y+1) &= 3y^2 + 5y + 2 \end{aligned}

So, the correct factorisation of 3y^2 + 5y + 2 is

(3y+2)(y+1)

**Question 5:** Fully factorise the following: 3x^4 - 75x^2

**[2 marks]**

We can simplify this expression:

3x^4 - 75x^2 = x^2(3x^2 - 75) = 3x^2(x^2 - 25)

We can see that x^2-25 is the difference of two squares:

x^2 - 25 = (x+5)(x-5)

Hence,

3x^2(x^2 - 25) = 3x^2(x+5)(x-5)