The Factor Theorem
Part of the Polynomials & Rational Expressions section of Coddy's Algebra 2 journey. Lesson 8 of 54.
A remainder of zero is the special case. If p(a) = 0, the division by x - a leaves nothing over, so p(x) = (x - a)q(x) and x - a is a factor. It works the other way round too: if x - a is a factor, then p(a) = 0. This is the factor theorem.
Take p(x) = x^3 - 2x^2 - x + 2. Is x - 1 a factor? Work out p(1): 1 - 2 - 1 + 2 is zero, so yes. Is x - 3 a factor? p(3) is 27 - 18 - 3 + 2, which is 8, so no: dividing by x - 3 would leave 8.
One evaluation replaces a whole long division. It also ties factors to roots: x - 1 is a factor exactly when 1 is a root, the place where the graph of p meets the x-axis. For x + 4, test p(-4).
Challenge
Is x + 2 a factor of p(x) = x^3 + x^2 - 4x - 4? Since x + 2 is x - (-2), the board holds the value of p at -2. Tap Use p(x) = ..., then type the arithmetic line by line.
Try it yourself
Simplify the expression
Test yourselfThis lesson includes a short quiz. Start the lesson to answer it and track your progress.
All lessons in Polynomials & Rational Expressions
1Polynomial Long Division
How Long Division WorksOne Term at a TimeA Remainder Left OverA Missing TermDividing by a QuadraticRecap: Long Division7Adding and Subtracting
Same BottomDifferent BottomsSubtractingFactor the Bottoms FirstRecap: Adding Fractions2Remainders and Factors
The Remainder TheoremThe Factor TheoremFinding a Missing NumberUsing a Known FactorRecap: Remainders and Factors5Sign Charts
Between the RootsOutside the RootsA Negative Leading TermReading a Sign ChartRecap: Sign Charts