# Factoring Quadratics

To "Factor" (or "Factorise" in the UK) a Quadratic is to:

find what to multiply to get the Quadratic

It is called "Factoring" because we find the factors (a factor is something we multiply by)

### Example:

Multiplying **(x+4)** and **(x−1)** together (called Expanding) gets **x ^{2} + 3x − 4** :

So **(x+4)** and **(x−1)** are factors of **x ^{2} + 3x − 4**

Just to be sure, let us check:

^{2}− x + 4x − 4

^{2}+ 3x − 4

Yes, **(x+4)** and **(x−1)** are definitely factors of **x ^{2} + 3x − 4**

Did you see that Expanding and Factoring are opposites?

Expanding is usually easy, but Factoring can often be **tricky**.

It is like trying to find which ingredients

went into a cake to make it so delicious.

It can be hard to figure out!

So let us try an example where we **don't know** the factors yet:

## Common Factor

First check if there any common factors.

### Example: what are the factors of 6x^{2} − 2x = 0 ?

**6** and **2** have a common factor of **2**:

2(3x^{2} − x) = 0

And **x ^{2}** and

**x**have a common factor of

**x**:

2x(3x − 1) = 0

And we have done it! The factors are **2x** and **3x − 1**,

We can now also find the **roots** (where it equals zero):

- 2x is 0 when
**x = 0** - 3x − 1 is zero when
**x = \frac{1}{3}**

And this is the graph (see how it is zero at x=0 and x=\frac{1}{3}):

But it is not always that easy ...

## Guess and Check

Maybe we can guess an answer?

### Example: what are the factors of 2x^{2} + 7x + 3 ?

No common factors.

Let us try to **guess** an answer, and then check if we are right ... we might get lucky!

We could guess (2x+3)(x+1):

(2x+3)(x+1) = 2x^{2} + 2x + 3x + 3

= 2x^{2} + 5x + 3 **(WRONG)**

How about (2x+7)(x−1):

(2x+7)(x−1) = 2x^{2} − 2x + 7x − 7

= 2x^{2} + 5x − 7 **(WRONG AGAIN)**

OK, how about (2x+9)(x−1):

(2x+9)(x−1) = 2x^{2} − 2x + 9x − 9

= 2x^{2} + 7x − 9 **(WRONG AGAIN)**

Oh No! We could be guessing for a long time before we get lucky.

That is not a very good method. So let us try something else.

## A Method For Simple Cases

Luckily there is a method that works in simple cases.

With the quadratic equation in this form:

**Step 1**: Find two numbers that multiply to give ac (in other words a times c), and add to give b.

Example: 2x^{2} + 7x + 3

ac is 2×3 = **6** and b is **7**

So we want two numbers that multiply together to make 6, and add up to 7

In fact **6** and **1** do that (6×1=6, and 6+1=7)

How do we find 6 and 1?

It helps to list the factors of ac=**6**, and then try adding some to get b=**7**.

Factors of 6 include 1, 2, 3 and 6.

Aha! 1 and 6 add to 7, and 6×1=6.

**Step 2**: Rewrite the middle with those numbers:

Rewrite 7x with **6**x and **1**x:

2x^{2} + **6x + x** + 3

**Step 3**: Factor the first two and last two terms separately:

The first two terms 2x^{2} + 6x factor into 2x(x+3)

The last two terms x+3 don't actually change in this case

So we get:

2x(x+3) + (x+3)

**Step 4**: If we've done this correctly, our two new terms should have a clearly visible common factor.

In this case we can see that (x+3) is common to both terms, so we can go:

**(2x+1)(x+3)**

Done!

Check: (2x+1)(x+3) = 2x^{2} + 6x + x + 3 = **2x ^{2} + 7x + 3** (Yes)

Much better than guessing!

**Let's see Steps 1 to 4 again, in one go**:

2x^{2} + 7x + 3 |

2x^{2} + 6x + x + 3 |

2x(x+3) + (x+3) |

2x(x+3) + 1(x+3) |

(2x+1)(x+3) |

### OK, let us try another example:

### Example: 6x^{2} + 5x − 6

**Step 1**: ac is 6×(−6) = **−36**, and b is **5**

List the positive factors of ac = **−36**: 1, 2, 3, 4, 6, 9, 12, 18, 36

One of the numbers has to be negative to make −36, so by playing with a few different numbers I find that −4 and 9 work nicely:

−4×9 = −36 and −4+9 = 5

**Step 2**: Rewrite **5x** with −4x and 9x:

6x^{2} − 4x + 9x − 6

**Step 3**: Factor first two and last two:

2x(3x − 2) + 3(3x − 2)

**Step 4**: Common Factor is (3x − 2):

(2x+3)(3x − 2)

Check: (2x+3)(3x − 2) = 6x^{2} − 4x + 9x − 6 = **6x ^{2} + 5x − 6** (Yes)

### Finding Those Numbers

The hardest part is finding two numbers that multiply to give ac, and add to give b.

It is partly guesswork, and it helps to **list out all the factors**.

Here is another example to help you:

### Example: ac = −120 and b = 7

What two numbers **multiply to −120** and **add to 7** ?

The factors of 120 are (plus and minus):

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, and 120

We can try pairs of factors (start near the middle!) and see if they add to 7:

- −10 x 12 = −120, and −10+12 = 2 (no)
- −8 x 15 = −120 and −8+15 = 7 (YES!)

## Get Some Practice

You can practice simple quadratic factoring.

## Why Factor?

Well, one of the big benefits of factoring is that we can find the **roots** of the quadratic equation (where the equation is zero).

All we need to do (after factoring) is find where each of the two factors becomes zero

### Example: what are the roots (zeros) of 6x^{2} + 5x − 6 ?

We already know (from above) the factors are

(2x + 3)(3x − 2)

And we can figure out that

(2x + 3) is zero when x = −3/2

and

(3x − 2) is zero when x = 2/3

So the roots of 6x^{2} + 5x − 6 are:

−3/2 and 2/3

Here is a plot of 6x^{2} + 5x − 6, can you see where it equals zero?

And we can also check it using a bit of arithmetic:

At x = -3/2: 6(-3/2)^{2} + 5(-3/2) - 6 = 6×(9/4) - 15/2 - 6 = 54/4 - 15/2 - 6 = 6-6 **= 0**

At x = 2/3: 6(2/3)^{2} + 5(2/3) - 6 = 6×(4/9) + 10/3 - 6 = 24/9 + 10/3 - 6 = 6-6 **= 0**

## Graphing

We can also try graphing the quadratic equation. Seeing where it equals zero can give us clues.

### Example: (continued)

Starting with 6x^{2} + 5x − 6 and **just this plot:**

The roots are **around** x = −1.5 and x = +0.67, so we can **guess** the roots are:

−3/2 and 2/3

Which can help us work out the factors **2x + 3** and **3x − 2**

Always check though! The graph value of +0.67 might not really be 2/3

## The General Solution

There is also a general solution (useful when the above method fails), which uses the quadratic formula:

Use that formula to get the two answers x_{+} and x_{−} (one is for the "+" case, and the other is for the "−" case in the "±"), and we get this factoring:

a(x − x_{+})(x − x_{−})

Let us use the previous example to see how that works:

### Example: what are the roots of 6x^{2} + 5x − 6 ?

Substitute a=6, b=5 and c=−6 into the formula:

x = \frac{−b ± √(b^{2} − 4ac)}{2a}

= \frac{−5 ± √(5^{2} − 4×6×(−6))}{2×6}

= \frac{−5 ± √(25 + 144)}{12}

= \frac{−5 ± √169}{12}

= \frac{−5 ± 13}{12}

So the two roots are:

x_{+} = (−5 + 13) / 12 = 8/12 = 2/3,

x_{−} = (−5 − 13) / 12 = −18/12 = −3/2

(Notice that we get the same answer as when we did the factoring earlier.)

Now put those values into a(x − x_{+})(x − x_{−}):

6(x − 2/3)(x + 3/2)

We can rearrange that a little to simplify it:

3(x − 2/3) × 2(x + 3/2) = (3x − 2)(2x + 3)

And we get the same factors as we did before.

*(Thanks to "mathsyperson" for parts of this article)*