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

## Example 1

### 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 x2.
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

### 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 x2.
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

### 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 x2.
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

### 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 x2.
The middle term's variable is xy.
And the last term's variable is y2.

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

### 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 2x2.
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.