Victoria Yarbrough
Factoring Quadratics
Factoring is one of three ways to solve for a quadratic equation. Here are easy steps to solve along with examples.
General Form: ax^2+bx+c
Steps:
1. Identify what ‘a’, ‘b’, and ‘c’ equal.
2. Calculate the product of ac.
3. List the factors of the product of ‘ac’.
4. Find the two factors that when added or subtracted give ‘b’.
5. Create a 2x2 box.
6. ‘A’ must go in the bottom left corner and ‘b’ in the top right. It does not matter where the two numbers that add or subtract to give ‘b’ are placed.
7. Now from right to left in both rows, factor out a common number that the two numbers have.
8. From bottom to top in both rows, do the same.
9. Factoring always consists of: f (x)= (___ +/-___)(__+/-__)
10. The numbers that are factored out on the side and top of the 2x2 box is what goes in the equation in step nine.
11. Foil out the equation in step nine to double check that the math is correct.
Example one:
f(x)= X^2+ 7x + 6
‘a’= 1
‘b’= 7
‘c’= 6
‘ac’= 6
Factors of 6: 1 and 6, and 2 and 3.
6+1=7=b
1x 6
x^2 | 6x |
x | 6 |
1x
1
f(x)= (1x+1)(1x+6)
Foiling to check:
f(x)= (1x+1)(1x+6)= f (x)= x^2+ 7x + 6
x^2+ 1x+ 6x+6
x^2+ 7x+ 6 ✓
Graphing:
When creating parabolas, it is also important to identify whether or not ‘a’ is positive, for that will tell the direction of the parabola. If a > o, the parabola will curve upward. If
a < 0, the parabola will curve downward. Below is the graphed equation from the first example problem.
Example two:
F (x)=2x^2+7x+3
a= 2
b= 7
c= 3
ac= 6
Factors of 6: 1 and 6, and 2 and 3.
6+1=7=b
x 3
2x^2 | 6x |
x | 3 |
2x
1
f(x)= (2x^2+1)(x+3)
Foiling to check:
f(x)= (2x^2+1)(x+3)= F (x)=2x^2+7x+3
2x^2+6x+1x+3
2x^2+7x+3Ö
Situations where a factor is prime:
Let’s do the problem: x^2+6x+4
Do all the same steps.
A problem should be encountered where there is no factors from the product of ‘ac’ that give ‘b’.
In a case like this, the quadratic must be solved with another method.
