# Factor a Quadratic Trinomial

How to factor a quadratic trinomial: 5 examples and their solutions.

## Example 1

### Example

### Solution

Think of a pair of numbers

whose product is the last term, +2,

and whose sum is the coefficient of the middle term, +3.

Start from finding the factors of +2.

The last term is plus.

And the middle term's coefficient is also plus.

So the pair of numbers are both plus.

(+)⋅(+) = (+)

(+) + (+) = (+)

+1 and +2 seems to be good.

1⋅2 = +2

1 + 2 = +3

So +1 and +2 are the numbers you're looking for.

The first term's variable is x^{2}.

The middle term's variable is x.

And there's no variable in the last term.

So write a form like this.

Write +1 and +2 in the form.

Then (x + 1)(x + 2).

So (x + 1)(x + 2) is the answer.

## Example 2

### Example

### Solution

Think of a pair of numbers

whose product is the last term, +6,

and whose sum is the coefficient of the middle term, -5.

Start from finding the factors of +6.

The last term is plus.

And the middle term's coefficient is minus.

So the pair of numbers are both minus.

(-)⋅(-) = (+)

(-) + (-) = (-)

-1 and -6 seems to be good.

(-1)⋅(-6) = +6

-1 - 6 ≠ -5

So -1 and -6 are not the numbers you're looking for.

If you choose the wrong pair of numbers,

find another pair of numbers

that satisfy the conditions.

This is a process of trial and error.

So it's natural to choose the wrong numbers.

-2 and -3 seems to be good.

(-2)⋅(-3) = +6

-2 - 3 = -5

So -2 and -3 are the numbers you're looking for.

The first term's variable is x^{2}.

The middle term's variable is x.

And there's no variable in the last term.

So write a form like this.

Write -2 and -3 in the form.

Then (x - 2)(x - 3).

So (x - 2)(x - 3) is the answer.

## Example 3

### Example

### Solution

Think of a pair of numbers

whose product is the last term, -12,

and whose sum is the coefficient of the middle term, -1.

Start from finding the factors of -12.

The last term is minus.

So the pair of numbers are different.

One is plus.

And the other is minus.

(+)⋅(-) = (-)

+3 and -4 seems to be good.

3⋅(-4) = -12

3 - 4 = -1

So +3 and -4 are the numbers you're looking for.

The first term's variable is x^{2}.

The middle term's variable is x.

And there's no variable in the last term.

So write a form like this.

Write +3 and -4 in the form.

Then (x + 3)(x - 4).

So (x + 3)(x - 4) is the answer.

## Example 4

### Example

### Solution

Think of a pair of numbers

whose product is the coefficient of the last term, -24,

and whose sum is the coefficient of the middle term, +5.

Start from finding the factors of -24.

The last term's coefficient is minus.

So the pair of numbers are different.

One is plus.

And the other is minus.

(+)⋅(-) = (-)

-3 and +8 seems to be good.

(-3)⋅8 = -24

-3 + 8 = +5

So -3 and +8 are the numbers you're looking for.

The first term's variable is x^{2}.

The middle term's variable is xy.

And the last term's variable is y^{2}.

So write a form like this.

Write -3 and +8 in the form.

Then (x - 3y)(x + 8y).

So (x - 3y)(x + 8y) is the answer.

## Example 5

### Example

### Solution

If the first term's coefficient is not 1,

the condition is slightly different.

Think of a pair of numbers

whose product is the last term, +6,

and whose sum satisfies

[number 1] + [first term's coefficient, 2]⋅[number 2] = +7.

Start from finding the factors of +6.

The last term is plus.

And the middle term's coefficient is also plus.

So the pair of numbers are both plus.

(+)⋅(+) = (+)

(+) + 2⋅(+) = (+)

+3 and +2 seems to be good.

3⋅2 = +6

3 + 2⋅2 = +7

So +3 and +2 are the numbers you're looking for.

The first term is 2x^{2}.

The middle term's variable is x.

And there's no variable in the last term.

So write a form like this.

The first term's coefficient 2

should be in the first factor.

Write +3 and +2 in the form.

Then (2x + 3)(x + 2).

Keep the order of +3 and +2.

So (2x + 3)(x + 2) is the answer.