Factor & Remainder Theorems: Why P(a) Matters

Start with the polynomial division identity

If P(x) is divided by x−a, write P(x)=(x−a)Q(x)+R. The remainder R is constant because its degree must be lower than the divisor’s degree, which is 1.

Set x=a. The product (a−a)Q(a) vanishes, leaving P(a)=R. This explains the remainder theorem rather than presenting substitution as a lucky shortcut.

Read the divisor’s sign carefully

For divisor x−4, substitute 4. For x+4=x−(−4), substitute −4. Keep negative inputs in brackets when evaluating powers. A sign error can make a factor test look convincing but wrong.

For x+2, the required value is P(−2), not P(2). The value that matters is the root of the divisor.

Connect zero remainder to a factor

If P(a)=0, the division identity becomes P(x)=(x−a)Q(x), so x−a is a factor. Conversely, if x−a is a factor, substitution at a gives zero.

Finding one factor does not finish factorising a higher-degree polynomial. Divide out the factor or find the remaining quadratic, then check the full product.

Evaluate a polynomial systematically

For P(x)=2x³−3x²+4x−5, organise terms before substituting. At x=−1 the values are −2,−3,−4,−5, summing to −14. The even power is positive before its negative coefficient is applied.

If a term is missing, its coefficient is zero. That matters in long division or synthetic division. A direct substitution should still include every stated term.

Use a factor condition to find a coefficient

If x−2 is a factor of x³+kx+6, substitute 2 to get 8+2k+6=0. Thus k=−7. Reinsert that value and check the zero before accepting it.

Two unknown coefficients usually need two independent conditions. One factor condition gives one equation and is not enough by itself to determine two unknowns uniquely.

Extend the remainder idea to a non-monic linear divisor

For a non-zero constant m, the root of mx−n is n/m. In P(x)=(mx−n)Q(x)+R, substitution at n/m still leaves R=P(n/m).

For divisor 2x−1, use x=1/2. Do not multiply the remainder by 2 just because the divisor has a leading coefficient of 2. Polynomial coefficients here may be rational numbers.

Use the theorem within its scope

A constant remainder is appropriate for division by a linear polynomial. Division by a quadratic can leave a linear remainder, so one substitution will not generally determine it.

A zero at a also does not tell you how many times the factor repeats. Repeated factors require further work. Keep a single factor test distinct from a complete root or multiplicity analysis.

Finish with two different checks

After finding a factor, verify P(a)=0 and expand your proposed factorisation. When only a remainder is required, write the divisor root and the substituted expression so that the sign choice is visible.

This guide extends the idea of a polynomial zero. Learners who struggle with it may need substitution and bracket practice first, rather than more polynomial division.

See the reasoning

Worked examples and checks

Try, explain, check

12 questions with explained answers

Try each task before opening its explanation. These are original BananaLearn questions, not official exam or past-paper questions. Where several answers can work, the explanation gives the criteria.

1. Which input tests divisor x−5?

   Check answer and reasoning

   5
   It makes x−5 equal zero.

2. Which input tests divisor x+5?

   Check answer and reasoning

   −5
   x+5=x−(−5).

3. Find the remainder of x²+3x+1 divided by x−2.

   Check answer and reasoning

   11
   P(2)=4+6+1.

4. Find the remainder of x²+3x+1 divided by x+2.

   Check answer and reasoning

   −1
   P(−2)=4−6+1.

5. Is x−3 a factor of x²−9?

   Check answer and reasoning

   Yes
   P(3)=9−9=0.

6. Is x+3 also a factor?

   Check answer and reasoning

   Yes
   P(−3)=9−9=0.

7. Find k if x−1 is a factor of x²+kx−4.

   Check answer and reasoning

   3
   1+k−4=0.

8. Find the remainder of x²+1 on division by 2x−1.

   Check answer and reasoning

   5/4
   P(1/2)=1/4+1.

9. Can a quadratic divisor leave a non-constant remainder?

   Check answer and reasoning

   Yes
   Its remainder may have degree 1.

10. Does P(2)=0 prove that (x−2)² is a factor?

    Check answer and reasoning

    No
    It proves at least one x−2 factor, not repetition.

11. If P(x)=(x+1)(x−4), what is P(−1)?

    Check answer and reasoning

    0
    The first factor vanishes.

12. Can one factor condition usually determine two unknown coefficients?

    Check answer and reasoning

    No
    It supplies only one equation unless additional information is given.

A checking record to use in your notebook

Copy these headings and fill them for one worked task, then for a fresh attempt. Use your own evidence rather than copying the model answer.

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Choose the next useful step

• Polynomial zeros and substitution
• Rebuild a product by factorising
• Algebra 2 polynomial support

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Original BananaLearn explanations, examples, illustrations and practice. Source facts checked 7 November 2026. Suggested levels are learning guides, not a claim that every board or textbook uses the same sequence.

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