Mathematics 9709 · Pure Mathematics 3

The remainder theorem

Same idea, one step further: the remainder when you divide $P(x)$ by $x-c$ is just $P(c)$ , again with no division.
Put $x=c$ into $P(x)=(x-c)Q(x)+R$ . The divisor term becomes 0, leaving $R=P(c)$ .

Dividing by $x-c$ ? Plug in, don't divide: $P(c)$ is the remainder. And $P(c)=0$ is the factor theorem: the same fact with the remainder set to 0.

Remainder = P(c)

$P(x)\div(x-c)\ \Rightarrow\ \text{remainder}=P(c)$

For $ax-b$ the remainder is $P\!\left(\tfrac{b}{a}\right)$ .
The factor theorem is just this case with the remainder equal to 0.

Worked example

Remainder when $4x^3-2x^2+5x-8$ is divided by $x+1$ (Worked example 1.12, textbook)

The root of $x+1$ is $x=-1$ , so the remainder is just $f(-1)$ :

$f(-1)=-4-2-5-8=-19$

Answer

\text{remainder}=-19

Two clues, two equations (the most common exam setup)

A common exam setup gives you two remainder or factor facts about a polynomial with unknowns in it.
Turn each one into $P(\text{root})=\text{value}$ (one equation each), then solve the pair together.

Worked example

Real question: $ax^3+5x^2-4x+b$ has factor $x+2$ and remainder 2 at $x+1$

9709/32 Feb/March 2021 Q2 — one factor clue, one remainder clue.

Factor $x+2$ means $p(-2)=0$ : $-8a+20+8+b=0\Rightarrow -8a+b=-28$ .
Remainder 2 at $x+1$ means $p(-1)=2$ : $-a+5+4+b=2\Rightarrow -a+b=-7$ .
Subtract the equations: $-7a=-21\Rightarrow a=3$ , then $b=-7+a=-4$ .

Answer

a=3,\quad b=-4

Worked example

Real question: $4x^3+ax^2+5x+b$ : $(2x+1)$ a factor, and $p(4)=3\,p(2)$

9709/31 Oct/Nov 2024 Q1 — the second clue compares two remainders, same method: write each as $p(\text{root})$ .

Factor $2x+1$ means $p\!\left(-\tfrac12\right)=0$ : $4(-\tfrac18)+a(\tfrac14)+5(-\tfrac12)+b=0\Rightarrow \tfrac{a}{4}+b=3$ , i.e. $a+4b=12$ .
The remainder at $x-4$ is $p(4)$ , at $x-2$ is $p(2)$ ; the clue says $p(4)=3\,p(2)$ . Expanding gives $-2a+b=75$ .
Solve the pair $a+4b=12$ and $-2a+b=75$ :

Answer

a=-32,\quad b=11

A derivative clue is just another equation

Some questions give you a fact about $p'(x)$ instead. The method is the same: differentiate, then read the clue as an equation. For $p(x)=6x^3+ax^2+bx+9$ with $(x-3)$ a factor and $p'(x)$ leaving remainder 72 at $(x-3)$ (9709/32 Feb/March 2025 Q9): the factor gives $p(3)=0$ , and $p'(x)=18x^2+2ax+b$ with $p'(3)=72$ gives the second equation. Solving the pair gives $a=-11,\ b=-24$ .

Now you try

The polynomial $p(x)=2x^3+ax^2-11x+b$ is divisible by $2x-1$ , and leaves remainder $12$ when divided by $x+1$ . Find $a$ and $b$ . (9709/32 Oct/Nov 2023 Q3)

Your turn— tap to reveal the worked answer (9709/32 Oct/Nov 2023 Q3)

Divisible by $2x-1$ means remainder $p\!\left(\tfrac12\right)=0$ ; remainder 12 at $x+1$ means $p(-1)=12$ . Two clues, two equations:

$p\!\left(\tfrac12\right)=\tfrac14+\tfrac{a}{4}-\tfrac{11}{2}+b=0\Rightarrow a+4b=21$

$p(-1)=-2+a+11+b=12\Rightarrow a+b=3$

Subtracting: $3b=18$ , then back-substitute:

Answer

a=-3,\quad b=6

Now a quadratic-divisor remainder: $2x^4+ax^3+bx-1$ leaves remainder $3x+2$ when divided by $x^2-x+1$ . Find $a$ and $b$ . (9709/32 Feb/March 2023 Q3)

Your turn— tap to reveal the worked answer (9709/32 Feb/March 2023 Q3)

The remainder theorem shortcut won't work here (the divisor is a quadratic), so divide and match the remainder to $3x+2$ . Comparing the leftover's coefficients gives two equations:

Answer

a=-3,\quad b=5

PreviousThe factor theorem NextLogs and the laws: the toolkit you apply everywhere

The remainder theorem

Same idea, one step further: the remainder when you divide $P(x)$ by $x-c$ is just $P(c)$ , again with no division.
Put $x=c$ into $P(x)=(x-c)Q(x)+R$ . The divisor term becomes 0, leaving $R=P(c)$ .

Dividing by $x-c$ ? Plug in, don't divide: $P(c)$ is the remainder. And $P(c)=0$ is the factor theorem: the same fact with the remainder set to 0.

Remainder = P(c)

$P(x)\div(x-c)\ \Rightarrow\ \text{remainder}=P(c)$

For $ax-b$ the remainder is $P\!\left(\tfrac{b}{a}\right)$ .
The factor theorem is just this case with the remainder equal to 0.

Worked example

Remainder when $4x^3-2x^2+5x-8$ is divided by $x+1$ (Worked example 1.12, textbook)

The root of $x+1$ is $x=-1$ , so the remainder is just $f(-1)$ :

$f(-1)=-4-2-5-8=-19$

Answer

\text{remainder}=-19

Two clues, two equations (the most common exam setup)

A common exam setup gives you two remainder or factor facts about a polynomial with unknowns in it.
Turn each one into $P(\text{root})=\text{value}$ (one equation each), then solve the pair together.

Worked example

Real question: $ax^3+5x^2-4x+b$ has factor $x+2$ and remainder 2 at $x+1$

9709/32 Feb/March 2021 Q2 — one factor clue, one remainder clue.

Factor $x+2$ means $p(-2)=0$ : $-8a+20+8+b=0\Rightarrow -8a+b=-28$ .
Remainder 2 at $x+1$ means $p(-1)=2$ : $-a+5+4+b=2\Rightarrow -a+b=-7$ .
Subtract the equations: $-7a=-21\Rightarrow a=3$ , then $b=-7+a=-4$ .

Answer

a=3,\quad b=-4

Worked example

Real question: $4x^3+ax^2+5x+b$ : $(2x+1)$ a factor, and $p(4)=3\,p(2)$

9709/31 Oct/Nov 2024 Q1 — the second clue compares two remainders, same method: write each as $p(\text{root})$ .

Factor $2x+1$ means $p\!\left(-\tfrac12\right)=0$ : $4(-\tfrac18)+a(\tfrac14)+5(-\tfrac12)+b=0\Rightarrow \tfrac{a}{4}+b=3$ , i.e. $a+4b=12$ .
The remainder at $x-4$ is $p(4)$ , at $x-2$ is $p(2)$ ; the clue says $p(4)=3\,p(2)$ . Expanding gives $-2a+b=75$ .
Solve the pair $a+4b=12$ and $-2a+b=75$ :

Answer

a=-32,\quad b=11

A derivative clue is just another equation

Now you try

The polynomial $p(x)=2x^3+ax^2-11x+b$ is divisible by $2x-1$ , and leaves remainder $12$ when divided by $x+1$ . Find $a$ and $b$ . (9709/32 Oct/Nov 2023 Q3)

Your turn— tap to reveal the worked answer (9709/32 Oct/Nov 2023 Q3)

Divisible by $2x-1$ means remainder $p\!\left(\tfrac12\right)=0$ ; remainder 12 at $x+1$ means $p(-1)=12$ . Two clues, two equations:

$p\!\left(\tfrac12\right)=\tfrac14+\tfrac{a}{4}-\tfrac{11}{2}+b=0\Rightarrow a+4b=21$

$p(-1)=-2+a+11+b=12\Rightarrow a+b=3$

Subtracting: $3b=18$ , then back-substitute:

Answer

a=-3,\quad b=6

Now a quadratic-divisor remainder: $2x^4+ax^3+bx-1$ leaves remainder $3x+2$ when divided by $x^2-x+1$ . Find $a$ and $b$ . (9709/32 Feb/March 2023 Q3)

Your turn— tap to reveal the worked answer (9709/32 Feb/March 2023 Q3)

The remainder theorem shortcut won't work here (the divisor is a quadratic), so divide and match the remainder to $3x+2$ . Comparing the leftover's coefficients gives two equations:

Answer

a=-3,\quad b=5