Algebra II Recipe: Solving Quadratic Equations by Factoring

1. Find the largest number common to every coefficient or number.
2. Find the GCF of each variable.
   • It will always be the variable raised to the smallest exponent.
3. Find the terms that the GCF would be multiplied by to equal the original polynomial.
   • It looks like the distributive property when in factored form...GCF(terms).

1. The factors will always be (a + b)(a - b).
2. The "a" and "b" represent terms.
   • "a" is the square root of the first term.
   • "b" is the square root of the second term.

1. Characteristics
• "ax²" term is a perfect square.
• "c" term is a perfect square.
• "c" term is positive.
• Factors into two identical binomials: (a + b)².
2. Steps to Factor
• Since it factors into two identical binomials - write it as (a ± b)².
• "a" is the square root of the "ax²" term.
• "b" is the square root of the "c" term.
• The operation in the binomial factor is the same as the operation in front of the "x" term.

1. Multiply "a" and "c".
2. Find two numbers that multiply to equal this product, but adds to equal "b".
   • When a = 1 stop here. The two numbers chosen will be the numbers in the two binomial factors
3. Use these two numbers to rewrite the "x" term when writing out the problem again.
   • When you have a choice, write the negative term first.
4. Group the first two terms and the last two terms together.
   • If the third term from the left has subtraction in front, add the opposite before grouping.
5. Factor the GCF out of each set of parentheses.
   • If you added the opposite in step 4, factor a negative GCF out of the second set of parentheses.
6. The two GCFs make up one binomial factor and the common set of parentheses is the other binomial factor.

1. Set all terms equal to zero.
2. Factor the quadratic completely.
3. Set each factor having a variable equal to zero.
4. Solve each equation.
5. If we were to graph the quadratic equation, these values would be the x-intercepts. The numbers you get are the solutions.