# Splitting the Middle Term

We now learn how to split the middle term.
Splitting the middle term is a method for factoring quadratic equations.

By the end of this section we'll know how to write quadratics in factored form.

For example, we'll know how to show that: $2x^2+7x+3 = \begin{pmatrix} 2x + 1 \end{pmatrix} \begin{pmatrix} x + 3 \end{pmatrix}$ We start by watching a tutorial to learn a five-step method for factornig quadratics.

## Tutorial

In the following tutorial we learn how to factor quadratics by splitting the middle term. Watch it now.

## Exercise 1

Write each of the following quadratics in factored form:

1. $$2x^2 + 7x + 3$$
2. $$6x^2+x-2$$
3. $$3x^2 - 13x + 4$$
4. $$3x^2-19x + 20$$
5. $$4x^2+9x+5$$

Note: this exercise can be downloaded as a worksheet to practice with:

1. $$2x^2+7x + 3 = \begin{pmatrix} 2x + 1 \end{pmatrix} \begin{pmatrix} x + 3 \end{pmatrix}$$
2. $$6x^2 + x - 2 = \begin{pmatrix} 2x - 1 \end{pmatrix} \begin{pmatrix} 3x + 2 \end{pmatrix}$$
3. $$3x^2 - 13x + 4 = \begin{pmatrix}x - 4 \end{pmatrix} \begin{pmatrix} 3x - 1 \end{pmatrix}$$
4. $$3x^2 - 19 x + 20 = \begin{pmatrix} x - 5 \end{pmatrix} \begin{pmatrix} 3x - 4 \end{pmatrix}$$
5. $$4x^2 + 9x + 5 = \begin{pmatrix} 4x + 5 \end{pmatrix} \begin{pmatrix} x + 1 \end{pmatrix}$$

Now that we know how to factor quadratics, by splitting the middle term, we learn how to solve quadratic equations by factoring.

The method/technique rests on the null factor law, explained here.

### Null Factor Law

Given two quantities $$A$$ and $$B$$ such that: $A\times B = 0$ then one of the following must be true:

• $$A=0$$
• $$B=0$$
• $$A=B=0$$
This law can be used for solving quadratic equations.

#### Null Factor Law for Quadratics

Once a quadratic, $$ax^2+bx+c$$, has been written in factored form, we can use the null factor law to solve the equation: $ax^2+bx+c = 0$

### Example

In factored form, the quadratic $$2x^2+7x+3$$ can be written: $\begin{pmatrix}2x+1 \end{pmatrix} \begin{pmatrix}x+3 \end{pmatrix}$ Consequently we can write: $2x^2+7x+3 = 0$

has same solutions as: $\begin{pmatrix}2x+1 \end{pmatrix} \begin{pmatrix}x+3 \end{pmatrix} = 0$
Now, using the null factor law we can state that one of the following must be true:
• $$2x+1 = 0$$, or
• $$x+3 = 0$$, or
• $$2x+1 = 0$$ and $$x+3 = 0$$
Consequently, the solutions to the quadratic equation are the solutions to $$2x+1 = 0$$ and $$x+3 = 0$$.

Starting with $$2x+1 = 0$$: \begin{aligned} & 2x+1 = 0 \\ & 2x = -1 \\ & x = -\frac{1}{2} \end{aligned} We now solve $$x+3 = 0$$: \begin{aligned} & x+3 = 0 \\ & x = -3 \end{aligned} The equation $$2x^2+7x+3 = 0$$ has two solutions:

$$x = -\frac{1}{2}$$ and $$x = -3$$

## Tutorial

In the following tutorial we learn how to solve quadratic equations that have been written in their factored form, by splitting the middle term, using the null factor law.

## Exercise 2

Solve each of the following quadratic equations by factoring (splitting the middle term):

1. $$4x^2 + 7x + 3 = 0$$
2. $$x^2+3x - 4 = 0$$
3. $$6x^2 - 7x +2 = 0$$
4. $$6x^2 + 7x - 3 = 0$$
5. $$-2x^2 + x + 3 = 0$$

## Solution Without Working

1. In factored form, we write: $$4x^2 + 7x + 3 = (4x+3)(x+1)$$ and the solutions to $$4x^2 + 7x + 3 = 0$$ are:

$$x = -1$$ and $$x = -\frac{3}{4}$$

2. In factored form, we write: $$x^2+3x - 4 = (x+4)(x-1)$$ and the solutions to $$x^2+3x - 4 = 0$$ are:

$$x = -4$$ and $$x = 1$$

3. In factored form, we write: $$6x^2 - 7x +2 = (3x-2)(2x-1)$$ and the solutions to $$6x^2 - 7x +2 = 0$$ are:

$$x = \frac{2}{3}$$ and $$x = \frac{1}{2}$$

4. In factored form, we write: $$6x^2 + 7x - 3 = (2x+3)(3x-1)$$ and the solutions to $$6x^2 + 7x - 3 = 0$$ are:

$$x = -\frac{3}{2}$$ and $$x = \frac{1}{3}$$

5. In factored form, we write: $$-2x^2 + x + 3 = (3-2x)(x+1)$$ and the solutions to $$-2x^2 + x + 3 = 0$$ are:

$$x = \frac{3}{2}$$ and $$x = -1$$

## Solution With Working

### Subscribe to Our Channel

Subscribe Now and view all of our playlists & tutorials.