# Factoring Quadratics

"Factoring" (or "Factorising" in the UK) a Quadratic is:

finding what to multiply to get the Quadratic

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

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

^{2}+ 3x − 4

Let us "expand"** (x+4)** and **(x−1) **to be sure:

^{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!

OK, let's try an example where we **don't know** the factors yet:

## Common Factor

First we can check for 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

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

No common factors.

Maybe we can **guess **an answer? Then check if we are right ... we may get lucky!

Let's guess (2x+3)(x+1):

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

= 2x^{2} + 5x + 3 (Close but **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!)**

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

There is a method for 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)

**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

(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?

We can also check it using a bit of arithmetic:

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

At x = \frac{2}{3}: 6(\frac{2}{3})^{2} + 5(\frac{2}{3}) − 6 = 6×(\frac{4}{9}) + \frac{10}{3} − 6 = \frac{24}{9} + \frac{10}{3} - 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

## General Solution

Quadratic equations have symmetry, the left and right are like mirror images:

The midline is at **−b/2**, and we can calculate the value **w** with these steps:

- First, "a" must be 1, if not then divide b and c by a:
- b = b/a, c = c/a
- mid = −b/2
- w = √(mid
^{2}− c) - roots are at mid−w and mid+w

### Example:** x**^{2} + 3x − 4

^{2}+ 3x − 4

a = 1, b = 3 and c = −4

- a= 1, so we can go to next step
- mid = −\frac{3}{2}
- w = √[(\frac{3}{2})
^{2}− (−4)] = √(\frac{9}{4} + 4) = √\frac{25}{4} =**\frac{5}{2}** - roots are at −\frac{3}{2}−\frac{5}{2} =
**−4**and −\frac{3}{2}+\frac{5}{2} =**1**

So we can factor **x ^{2} + 3x − 4** into

**(x + 4)(x**

**− 1)**

## Quadratic formula

We can also use the quadratic formula:

We get two answers x_{+} and x_{−} (one is for the "+" case, and the other is for the "−" case in the "±") that gets us this factoring:

a(x − x_{+})(x − x_{−})

### 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)

Done!