Factoring Quadratic Equations when a = 1
Where a, b, and c are constants and a ≠ 0. In other words there must be a x^{2} term.
Some examples are:
x^{2} + 3x  3 = 0
4x^{2} + 9 = 0 (Where b = 0)
x^{2} + 5x = 0 (where c = 0)
One way to solve a quadratic equation is by factoring the trinomial.
This part will focus on factoring a quadratic when a, the x^{2}coefficient, is 1.
Let's solve the following equation by factoring the trinomial:
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. This equation is already in the proper form where a = 1, b = 5 and c = 14. 
1x^{2} + 5x  14 = 0 
Step 2: Determine the factor pair of c that will add to give b.
2.1: List the factor pairs of c. First ask yourself what are the factors pairs of c, ignoring the negative sign for now. 2.2: Determine the signs of the factors. If c is positive then both factors will be positive or both factors will be negative. If c is negative then one factor will be positive and the other negative. Now create factor pairs 2.3: Determine the factor pair that will add to give b. If both c and b are positive, both factors will be positive. If both c and b are negative, the larger factor will be negative and the smaller will be positive. If c is positive and b is negative, both factors will be negative. If c is negative and b is positive, the larger factor will be positive and the smaller will be negative. 
(1, 14); (2, 7)
$\left(14,1\right):14+\left(1\right)=13\ne {b}$ This pair does not work. $\left(7,2\right):7+\left(2\right)=5={b}$ This pair works!!! (7, 2) 
Step 3: Create two sets of parentheses each containing a x and one of the factors. 
(x + 7)(x  2) = 0 
Now that the equation has been factored, solve for x. 

Step 4: Set each factor to zero and solve for x. 
(x + 7) = 0, or (x  2) = 0 x = 7, or x = 2 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. Where a = 1, b = 8 and c = 15. 
x^{2} + 8x + 15 = 0 
Step 2: Determine the factor pair of c that will add to give b.
2.1: List the factor pairs of c. First ask yourself what are the factors pairs of c, ignoring the negative sign for now. 2.2: Determine the signs of the factors. If c is positive then both factors will be positive or both factors will be negative. If c is negative then one factor will be positive and the other negative. Now create factor pairs 2.3: Determine the factor pair that will add to give b. If both c and b are positive, both factors will be positive. If both c and b are negative, the larger factor will be negative and the smaller will be positive. If c is positive and b is negative, both factors will be negative. If c is negative and b is positive, the larger factor will be positive and the smaller will be negative. 
(1, 15);(3,5)
$\left(15,1\right):15+1=16\ne {b}$ This pair does not work. $\left(3,5\right):3+5=8={b}$
This pair works!!! (3, 5) 
Step 3: Create two sets of parentheses each containing a x and one of the factors. 
(x + 3)(x + 5) 
Now that the equation has been factored, solve for x. 

Step 4: Set each factor to zero and solve for x. 
(x + 3) = 0, or (x + 5) = 0 x= 3, or x = 5 
Step 1: Write the equation in the general form ax^{2} + bx + c = 0. Where a = 1, b = 10 and c = 24. 
x^{2} + 10x  24 = 0 
Step 2: Determine the factor pair of c that will add to give b.
2.1: List the factor pairs of c. First ask yourself what are the factors pairs of c, ignoring the negative sign for now. 2.2: Determine the signs of the factors. If c is positive then both factors will be positive or both factors will be negative. If c is negative then one factor will be positive and the other negative. Now create factor pairs 2.3: Determine the factor pair that will add to give b. If both c and b are positive, both factors will be positive. If both c and b are negative, the larger factor will be negative and the smaller will be positive. If c is positive and b is negative, both factors will be negative. If c is negative and b is positive, the larger factor will be positive and the smaller will be negative. 
(1, 24);(2, 12);(3, 8);(6, 4)
$\left(1,24\right):1+24=23\ne {b}$ $\left(3,8\right):3+8=5\ne {b}$ $\left(4,6\right):4+6=2\ne {b}$ These pairs do not work. $\left(2,12\right):2+12=10={b}$ This pair works!!! (2, 12) 
Step 3: Create two sets of parentheses each containing a x and one of the factors. 
(x  2)(x + 12) 
Now that the equation has been factored, solve for x. 

Step 4: Set each factor to zero and solve for x. 
(x  2) = 0, or (x + 12) = 0 x = 2, or x = 12 
Related Links: Math algebra Factoring Quadratic Equations when a ≠ 1 Quadratic Formula Algebra Topics 
To link to this Factoring Quadratic Equations when a = 1 page, copy the following code to your site: