Finding a Missing Number
Part of the Polynomials & Rational Expressions section of Coddy's Algebra 2 journey. Lesson 9 of 54.
The factor theorem also runs backwards. Suppose p(x) = x^3 + 2x^2 + kx - 6 has x - 1 as a factor, but k is unknown. Then p(1) must be zero, and that is an equation for k.
Put 1 in for x:
1^3 + 2 × 1^2 + k × 1 - 6 = 0
Work out the numbers: 1 + 2 - 6 is -3, so k - 3 = 0 and k is 3. Check it: with k = 3, p(1) is 1 + 2 + 3 - 6, which is zero. The unknown can sit anywhere in the rule, and the method stays the same: the factor x - a forces p(a) = 0, and you solve that for the missing number.
Challenge
p(x) = x^3 - 3x^2 + kx + 6 has x - 2 as a factor, so its value at 2 is zero. The board holds that equation with 2 put in for x. Solve it for k.
Try it yourself
Solve for k
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