Polynomial division, remainders and factors
| English | Español |
|---|---|
| factor theorem/ˈfæktə ˈθɪərəm/ | teorema del factor |
How does one root unlock a cubic?
- A cubic graph crosses the axis at a known point. That root lets us reduce a cubic equation to a quadratic rather than guess the other roots.
- This lesson studies factor theorem 因式定理: The statement that x−a is a factor of polynomial P(x) exactly when P(a)=0.
Choose the mathematical structure
- Write P(x)=(x−a)Q(x)+R, with R constant. Substituting x=a gives R=P(a). Divide leading terms, multiply the whole divisor, subtract with brackets and repeat. Include zero coefficients for missing powers.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines factor theorem?
The statement that x−a is a factor of polynomial P(x) exactly when P(a)=0.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For P(x)=x³−2x²−5x+6, division by x−1 gives Q(x)=x²−x−6 and remainder 0. Indeed (x−1)(x²−x−6)=P(x). Factor Q=(x−3)(x+2), so the roots are 1,3,−2. Dividing the same P by x−2 gives quotient x²−5 and remainder −4: P=(x−2)(x²−5)−4. Hence P/(x−2)=x²−5−4/(x−2), for x≠2.
Polynomial division, remainders and factors
Write P(x)=(x−a)Q(x)+R, with R constant
Classify the algebraic steps and identify the identity or domain condition behind each decision.
For P(x)=x³−2x²−5x+6, find P(1).
P(1)=1−2−5+6=0.
Test a tempting shortcut
- The sign in x−a matters: for x+2 test P(−2). A nonzero remainder does not disappear. Subtract the entire product at each division step; omitting brackets changes signs.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
If P(a) is nonzero, x−a is still a factor of P. This claim is false. Explain which definition or assumption it violates.
Find the remainder on division of P by x−2.
R=P(2)=8−8−10+6=−4.
If P(a) is nonzero, x−a is still a factor of P.
The sign in x−a matters: for x+2 test P(−2). A nonzero remainder does not disappear. Subtract the entire product at each division step; omitting brackets changes signs.
Interpret a new situation
- Check a quotient by expanding divisor times quotient plus remainder. Factorisation can solve P(x)=0, while division of a rational expression must retain every original excluded input.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Evaluate the quotient x²−x−6 at x=4.
Q(4)=16−4−6=6.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · B. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The statement that x−a is a factor of polynomial P(x) exactly when P(a)=0. Choose the relationship, show the method, check its assumptions and interpret the result.