Factoring Quadratics from Standard Form to Factored Form
You will need to know how to change quadratics from the standard from to the factored form for your algebra test.
Factoring quadratics is also called “unfoiling” since it is like doing the FOIL method in reverse.
Exercises on Standard Form to Factored Form
Instructions: Factor the following quadratic equations using the steps provided in the next section. Then check your work with the solutions provided after each question. First, you may want to view the methods for factoring, which are provided below the quiz.
[WpProQuiz 14]
Factoring Quadratics – Box Method
Factoring quadratics is sometimes done using the box method.
The box method involves making a grid and then multiplying the numbers inside the boxes.
Factoring quadratics with the box method can be a bit confusing.
However, if you are confident with algebra and would like to try it, you can attempt factoring quadratics with the box method.
You should use the simplified method for factoring quadratics if you are not as confident with algebra.
We illustrate the simplified method for factoring quadratics in the following sections of this page.
Factoring Quadratics when a =1
When a = 1, you will see the quadratic equation in the following form:
x2 + bx + c = 0
So, lets look at an example equation in this format and try to factor it.
Factor the following quadratic: x2 + 3x – 28 = 0
STEP 1:
Look at the third term of the equation, which is –28 in our problem.
We need to find two integers that equal –28 when they are multiplied.
–1 × 28 = –28
–2 × 14 = –28
–4 × 7 = –28
–7 × 4 = –28
–14 × 2 = –28
–28 × 1 = –28
STEP 2:
Then look at the second term of the equation, which is 3x.
We need to be sure that our two integers also equal 3 when they are added.
–1 + 28 = 27
–2 + 14 = 12
–4 + 7 = 3
–7 + 4 = –3
–14 + 2 = –12
–28 + 1 = –27
So, we need the two integers –4 and 7 to factor our quadratic.
STEP 3:
Finally, take the two integers that meet both conditions above, and set out the factored form of the quadratic equation.
(x – 4)(x + 7) = 0
Standard Form to Factored Form – Example
You will remember from our posts on quadratic forms and the quadratic equation, that the standard quadratic equation is:
ax2 + bx + c = 0
Now let’s look at a more difficult example.
Factor the following quadratic: 2x2 + 5x – 12 = 0
STEP 1:
Look at the integer in the first term of the equation, which is 2 in our problem.
We need to find two integers that equal 2 when they are multiplied.
Since 2 is a prime number, there is only one way to do the multiplication:
2 × 1 = 2
STEP 2:
Look at the third term of the equation, which is –12 in our problem.
We ignore the sign of the term since we deal with this in the last step.
So, we need to find two integers that equal 12 when they are multiplied.
Taking the order of the integers into account, we have:
1 × 12 = 12
2 × 6 = 12
3 × 4 = 12
4 × 3 = 12
6 × 2 = 12
12 × 1 = 12
STEP 3:
Take the first number (2) from the multiplication in step 1 (which was 2 × 1 = 2).
Take this number (2) and multiply by the first number from each of the multiplications in step 2.
2 × 1 = 2
2 × 2 = 4
2 × 3 = 6
2 × 4 = 8
2 × 6 = 12
2 × 12 = 24
STEP 4:
Take the second number (1) from the multiplication in step 1 (which was 2 × 1 = 2).
Take this number (1) and multiply by the second number from each of the multiplications in step 2.
1 × 12 = 12
1 × 6 = 6
1 × 4 = 4
1 × 3 = 3
1 × 2 = 2
1 × 1 = 1
STEP 5:
We need to find a result from step 3 and subtract from it a result from step 4 so that the difference will equal variable b.
Step 3 result: 2 × 4 = 8
Step 4 result: 1 × 3 = 3
Difference: 8 – 3 = 5
The difference must be equal to variable b in our original equation, which was 2x2 + 5x – 12 = 0.
STEP 6:
Set out the possible factored forms of the quadratic equation.
The signs you use in factoring quadratics will depend on the signs in the original equation.
Rules for factoring quadratics:
Look at the signs on the second and third terms of the equation.
If you have + c and + b, then factor like this: (ax + )(x + )
If you have + c and – b, then factor like this: (ax – )(x – )
If you have – c and + b or – b , you have to use trial and error to work out the correct combination of signs in factoring quadratics. Your result will be like this: (nx – )(x + ) or this: (nx + )(x – )
First
Take the two integers that we have multiplied in step 3 above (2 and 4), as well as the second number from step 4 (3).
Outside
Integers used in step 3 above: 2 × 4 = 8 (These two numbers go on the “outside” positions of the sets of parentheses.)
Inside
Second number from step 4: 1 × 3 = 3 (This number goes on the “inside” position of the first set of parentheses.)
Equation
Our original equation was: 2x2 + 5x – 12 = 0, so b = + 5.
Here we have a negative sign on c and a positive sign on b.
So the possible factored forms are: (2x – 3)(x + 4) or (2x + 3)(x – 4)
(2x – 3)(x + 4) = 2x2 + 5x – 12 = 0
(2x + 3)(x – 4) = 2x2 – 5x – 12 = 0
So our factored quadratic is: (2x – 3)(x + 4) = 0
