Remainder and Factor Theorem in CAT: 3 Polynomial Patterns

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Long division of polynomials is slow, and CAT does not reward slow. The remainder and factor theorem let you skip the division and read the remainder straight from a substitution. One idea, one number, one line of working.

This guide shows the theorem, then three patterns that cover most exam questions: a direct remainder, a missing coefficient, and a remainder from a quadratic divisor. Every example below is small enough to check by hand.

The Theorem in Plain Words

If you divide a polynomial p(x) by (x minus a), the remainder is a plain number, and that number is p(a). The reason is quick to see. Write p(x) as (x minus a) times a quotient plus a remainder r. Put x equal to a, and the first term vanishes, so p(a) equals r.

The factor theorem is the special case where the remainder is zero. If p(a) equals 0, then (x minus a) divides p(x) exactly, so it is a factor. Conversely, if (x minus a) is a factor, p(a) must be 0. Both directions are used, one to test a candidate factor and the other to build an equation.

The divisor does not have to look like (x minus a). For a divisor like (2x minus 1), find the value that makes it zero, which is x equal to one half, and substitute that. The remainder is p of one half. This works for any linear divisor.

Pattern One: A Direct Remainder

The simplest pattern gives a polynomial and a linear divisor and asks for the remainder. Take p(x) equal to x cubed minus 4x squared plus x plus 6, and divide by (x minus 1). Substitute x equal to 1: 1 minus 4 plus 1 plus 6 is 4. The remainder is 4.

Divide the same polynomial by (x minus 2). Substitute x equal to 2: 8 minus 16 plus 2 plus 6 is 0. The remainder is zero, so (x minus 2) is a factor. You can confirm by factoring: the polynomial equals (x minus 2)(x squared minus 2x minus 3), and the quadratic factors again as (x minus 3)(x plus 1).

Now a divisor with a coefficient. Take 2x cubed minus 3x squared plus x minus 5, divided by (2x minus 1). The divisor is zero at x equal to one half. Substituting gives 2 times one eighth, minus 3 times one quarter, plus one half, minus 5. That is 0.25 minus 0.75 plus 0.5 minus 5, which equals minus 5. The remainder is minus 5.

The whole pattern is one substitution. The risk is arithmetic, especially with fractions and negatives, so write each term on its own and add at the end.

Pattern Two: Find the Missing Coefficient

The second pattern tells you that a given expression is a factor and asks for an unknown constant. Suppose x cubed plus k x squared minus 2x plus 8 has (x minus 2) as a factor. By the factor theorem, p(2) equals 0. That gives 8 plus 4k minus 4 plus 8 equals 0, so 4k equals minus 12 and k equals minus 3.

Check by substituting back: with k equal to minus 3, p(2) is 8 minus 12 minus 4 plus 8, which is 0. Good. A related version gives the remainder instead of zero. If the remainder on dividing by (x minus 1) is 7, set p(1) equal to 7 and solve for the unknown.

When two unknowns appear, expect two conditions. For example, if p(x) is divisible by (x minus 1) and by (x plus 2), then p(1) equals 0 and p(minus 2) equals 0. Two linear equations in two unknowns follow, and they solve quickly. The number of conditions you need equals the number of unknowns.

Pattern Three: A Quadratic Divisor

The third pattern divides by a quadratic, such as (x minus 1)(x minus 2), and asks for the remainder. The remainder now has degree at most 1, so write it as ax plus b. You need two equations to find a and b, and the roots of the divisor give them.

Suppose a polynomial leaves remainder 5 when divided by (x minus 1) and remainder 7 when divided by (x minus 2). Then p(1) equals 5 and p(2) equals 7. When divided by (x minus 1)(x minus 2), the remainder ax plus b must agree at those two points, so a plus b equals 5 and 2a plus b equals 7. Subtract to get a equal to 2, then b equal to 3. The remainder is 2x plus 3.

Another example uses x to the fifth plus 1 divided by (x squared minus 1), which is (x minus 1)(x plus 1). Here p(1) equals 2 and p(minus 1) equals 0. So a plus b equals 2 and minus a plus b equals 0. Adding gives b equal to 1, and then a equal to 1. The remainder is x plus 1.

You can check it by noting that x to the fifth minus x equals x times (x to the fourth minus 1), which has the factor (x squared minus 1). So x to the fifth is x plus a multiple of the divisor, and adding 1 gives remainder x plus 1. The check agrees.

Divisor	Remainder has the form	What you need
Linear, like (x minus a)	A number	p(a)
Quadratic with two real roots	ax plus b	p at both roots
Cubic with three real roots	ax squared plus bx plus c	p at all three roots

Two Helpful Substitutions

Two substitutions are worth memorising because they appear in many questions. Putting x equal to 1 gives the sum of all the coefficients. Putting x equal to minus 1 gives the alternating sum, which is the sum of even position coefficients minus the sum of odd ones.

These give shortcuts. For p(x) equal to (x squared plus x minus 1) cubed, the sum of the coefficients is p(1), which is 1 cubed, or 1. If a question asks for the sum of the coefficients in an expanded power, substitute 1 instead of expanding. The same idea gives the sum of the even-power coefficients, as the average of p(1) and p(minus 1).

These tricks connect the topic to other chapters, since they underlie binomial coefficient questions and some sequence questions. If you have time, see how our guide to quadratic equations formulas and PYQs uses related substitutions.

A Combined Example

Here is one question that uses two patterns together. A polynomial p(x) leaves remainder 3 when divided by (x minus 1) and is exactly divisible by (x plus 2). Find the remainder when p(x) is divided by (x minus 1)(x plus 2).

The remainder has the form ax plus b. From the first condition, p(1) equals 3, so a plus b equals 3. From the second, p(minus 2) equals 0, so minus 2a plus b equals 0. Subtract the second equation from the first: 3a equals 3, so a is 1 and b is 2. The remainder is x plus 2.

Check it with the given facts: at x equal to 1, x plus 2 is 3, and at x equal to minus 2, it is 0. Both match. The whole solution took four short lines, and no division was performed.

Questions like this are why the theorem is worth learning well. The same two ideas, substitute the root and match the remainder, handle conditions that would take pages with long division.

Connection to Number Remainders

The polynomial theorem has a twin in number theory. The remainder when 7 to the power n is divided by 6 is found by substituting, since 7 leaves remainder 1, and 1 to any power is 1. In effect, you are evaluating a polynomial at a residue.

That is why the same mental habit works in both places: replace the awkward object by a small number that behaves the same way. If you want practice on the number side, practise number system questions on remainders, and compare the logic.

Where the two meet, you may find a question like: what is the remainder when x to the 100 plus 1 is divided by (x plus 1)? The divisor is zero at x equal to minus 1, so the remainder is (minus 1) to the 100 plus 1, which is 2. One substitution and the question is done.

How to Practise

Start with ten short questions on pattern one, doing each by substitution only. Time yourself. Then do ten on pattern two, where the missing coefficient is the target. Finally, do a handful of pattern three questions, where two substitutions give a small system.

Keep an error log. If most of your mistakes are arithmetic, slow down and write each term on its own line. If they are in choosing the substitution value, redo the step of setting the divisor equal to zero. A mixed set at the end tests whether you can choose the pattern quickly.

A CAT study planner can schedule these short sessions, and the free CAT cheatsheets page has formula notes to check against. This chapter is a small share of the paper, so sequence it by return on effort and park the rare harder variants if time is short.