How to Solve Quadratic Equations by Factoring – This is the easiest method of solving a quadratic equation as long as the binomial or trinomial is easily factorable. Otherwise, we will need other methods such as completing the square or using the quadratic formula.

A Quadratic Equation in Standard Form

(**a**, **b**, and **c** can have any value, except that **a** can’t be 0.)

“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)

Table of Contents

### 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: (x+4)(x−1) = x(x−1) + 4(x−1) = x^{2} − x + 4x − 4 = x^{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 =***1*3

And this is the graph (see how it is zero at x=0 and x=*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: Start with:2x(x+3) + (x+3) Which is:2x(x+3) + 1(x+3) And so:**(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!)

Solve the quadratic equation below using the Factoring Method.

The first thing I realize in this problem is that one side of the equation **doesn’t contain zero**. I can easily create a zero on the right side by subtracting both sides by 202020.

After doing so, the left side should have a factorable trinomial that is very similar to problem 3.

To factor out this trinomial, think of two numbers when multiplied together gives −14 – 14−14 (constant term) and when added gives +5 + 5+5 (coefficient of xxx-term). By trial and error, the numbers should be −2 – 2−2 and 777. You may verify this correct combination.

The final answers are x=2x = 2x=2 and x=− 7x = – \,7x=−7.

**Example 6**: Solve the quadratic equation below using the Factoring Method.

**Solution:**

Here we have x=− 6x = – \,6x=−6 and x=7x = 7x=7 as our final answers.

**Example**: Solve the quadratic equation below using the Factoring Method.

**Solution:**

Our final answers are x=5x = 5x=5 and x=1x = 1x=1.

DD