Solving by Factoring
A.REI.4b
Read the standard text
A.REI.4b Solve quadratic equations as appropriate to the initial form of the equation by inspection, e.g., for x² = 49; taking square roots; completing the square; applying the quadratic formula; or utilizing the Zero-Product Property after factoring.
Section 1 · LearnLearn the idea
In this lesson you will: Write a quadratic equation in standard form before factoring it; Factor, then apply the zero-product property to each factor; Check solutions against the original equation, or against the x-intercepts of the graph.
Zero on one side, always
Factoring was Chapter 15's business. Here it becomes a solving method, and it comes with one non-negotiable first step: everything on one side, zero on the other.
x^2+5x=24 is not ready. Subtract 24: x^2+5x-24=0. Now factor: (x+8)(x-3)=0, and the zero-product property gives x=-8 or x=3.
Which factoring do you need?
The four patterns from Chapter 15 cover everything Ohio asks here. Look for a common factor first — it makes every later step smaller.
Ohio's content limits keep general trinomials friendly: the leading coefficient is usually 1, and where it is not, it is one of -1, +/-2 or +/-3. Special forms such as a difference of squares may carry any suitable rational leading coefficient.
The solutions are the x-intercepts
If x^2-2x-8=0 has solutions 4 and -2, then the graph of y=x^2-2x-8 meets the x-axis at 4 and -2. Solving the equation and finding where the curve crosses zero are the same task.
That gives you a free check whenever a graph is available, and it explains the vocabulary: solutions, roots and x-intercepts all name the same numbers.
Common mistake — Splitting a product that is not zero
Asked to solve (x-2)(x+3)=14, a student writes x-2=14 or x+3=14, and reports x=16 or x=11.
The move is copied from a method that works. On (x-2)(x+3)=0 splitting the factors is exactly right, and nothing on the page looks different except one number.
But the property is about zero and only zero. Fourteen can be built from a great many pairs — 2 and 7, 28 and 1/2, -1 and -14 — so knowing the product is 14 says nothing about either factor.
The repair is to expand and collect: x^2+x-6=14, so x^2+x-20=0, which factors as (x+5)(x-4)=0. The solutions are x=-5 and x=4. Neither of the student's answers is close: (16-2)(16+3)=266.
How to catch it: Before you split anything, look at the right-hand side. If it is not a bare 0, you are not allowed to split yet. Expand, move everything to one side, and factor again.
Section 2 · DoPractice this standard
These published practice sets are scored on the student site and cover A.REI.4b directly.