Polynomial division, remainders and factors
Введено| English | Значение |
|---|---|
| factor theorem/ˈfæktə ˈθɪərəm/ | теорема о множителе |
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.
The method
- 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.
Why does P(1)=0 tell you x-1 is a factor of P(x)?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Substituting x=a into P=(x-a)Q+R kills the quotient term, leaving R=P(a). Zero remainder means exact division - the factor theorem in action.
A worked case
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).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
P(1)=1−2−5+6=0.
Where it goes wrong
- 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.
Not true: If P(a) is nonzero, x−a is still a factor of P. By the factor theorem, x−a is a factor only when P(a)=0.
Find the remainder on division of P by x−2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
R=P(2)=8−8−10+6=−4.
By the factor theorem, x−a is a factor only when P(a)=0.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. By the factor theorem, x−a is a factor only when P(a)=0.
Applying it
- 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.
Evaluate the quotient x²−x−6 at x=4.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Q(4)=16−4−6=6.
Match each division concept to its statement.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Watch the sign: x+2 corresponds to testing P(-2). A nonzero remainder survives into the final expression.
In your course
- AQA A-level Mathematics (7357), specification section B.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Factor theorem: The statement that x−a is a factor of polynomial P(x) exactly when P(a)=0.
- Write P(x)=(x−a)Q(x)+R, with R constant.
- Watch for: the sign in x−a matters: for x+2 test P(−2).