The remainder theorem turns polynomial division into substitution. The factor theorem is its zero-remainder special case. Together they let you test factors, determine unknown coefficients and extract information without carrying out full long division every time.
1. The remainder theorem
When a polynomial P(x) is divided by x−a, the remainder is P(a).
Example: find the remainder when P(x)=2x³−5x+4 is divided by x−2. Compute P(2)=16−10+4=10.
2. Why substitution works
Polynomial division gives P(x)=(x−a)Q(x)+R, where the remainder R is constant because the divisor has degree 1. Set x=a. The product term vanishes, leaving P(a)=R.
This short derivation is useful because it explains the theorem rather than making P(a) look like an arbitrary trick.
3. The factor theorem
x−a is a factor of P(x) exactly when P(a)=0. A factor means division leaves no remainder.
For P(x)=x³−4x²+x+6, P(2)=8−16+2+6=0, so x−2 is a factor.
4. Read the sign in the divisor correctly
If the divisor is x+3, rewrite it as x−(−3). The relevant substitution is x=−3.
Forgetting this sign is one of the most common theorem errors because the substitution value is the root of the divisor, not the visible constant copied without interpretation.
5. Unknown coefficient from a factor condition
Let P(x)=x³+kx²−4x−4 and suppose x−2 is a factor. Then P(2)=0:
8+4k−8−4=0, so 4k−4=0 and k=1.
6. Unknown coefficient from a remainder condition
Let P(x)=2x³+ax+5. The remainder on division by x−1 is 10. Therefore P(1)=10:
2+a+5=10, giving a=3.
7. Two conditions can determine two unknowns
Let P(x)=x³+ax²+bx+6. Suppose x−1 and x+2 are factors.
P(1)=0 gives 1+a+b+6=0, so a+b=−7.
P(−2)=0 gives −8+4a−2b+6=0, so 2a−b=1.
Solving simultaneously gives a=−2 and b=−5.
8. After finding one factor
Once x−a is known to be a factor of a cubic, divide the cubic by x−a to obtain a quadratic factor. Then factorise or solve that quadratic to find the remaining roots.
The theorem identifies a factor efficiently; it does not replace every later algebraic step.
9. A useful distinction: test versus prove
Evaluating P(3) tests whether x−3 is a factor. If the result is nonzero, it proves that x−3 is not a factor. If the result is zero, the factor theorem establishes that it is a factor.
10. Error diagnosis
- Divisor x+4 → substitute −4, not 4.
- “Remainder 7” → set P(a)=7, not P(a)=0.
- “x−a is a factor” → zero remainder condition.
- Two unknown coefficients usually require two independent conditions.
11. Practice
- Find the remainder when x³+2x−1 is divided by x−2.
- Determine whether x+1 is a factor of x³−x²−x+1.
- Find k if x−3 is a factor of x²+kx−6.
- Find a if the remainder when 2x²+ax+4 is divided by x+2 is 6.
- P(x)=x³+ax²+bx−2 has factors x−1 and x+2. Find a and b.
12. Answers
- P(2)=8+4−1=11.
- P(−1)=−1−1+1+1=0, so yes.
- 9+3k−6=0, so k=−1.
- P(−2)=8−2a+4=6, so a=3.
- P(1)=0 gives a+b=1; P(−2)=0 gives 2a−b=5; hence a=2, b=−1.
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