Factor and Remainder Theorems | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

One substitution finds a linear remainder

After division by $x-a$ , the remainder $R$ is constant. Substitute $x=a$ into the division identity:

\begin{aligned}P(x)&=(x-a)Q(x)+R,\\P(a)&=0\times Q(a)+R=R.\end{aligned}

This is the remainder theorem. If $P(a)=0$ , division is exact, so $x-a$ is a factor: a polynomial that multiplies by another polynomial to give $P$ . Conversely, a factor leaves zero remainder. This is the factor theorem.

Worked example

Is x − 2 a factor of x³ + x² − 5x − 2?

P(2)=8+4-10-2=0

Yes. The substitution and zero result prove the claim; writing the final factorisation alone would not demonstrate the requested theorem.

Theorems give a remainder or test a factor; they do not by themselves find the quotient. Use division or coefficient matching if the quotient is also requested.

Your turnPractice

For $P(x)=x^3-4x+1$ , find the remainders for divisors $x-2$ and $x+1$ . Is either a factor?

Show worked answer

\begin{gathered}P(2)=8-8+1=1\\P(-1)=-1+4+1=4\end{gathered}

The remainders are 1 and 4, not zero, so neither divisor is a factor.

Set the whole divisor equal to zero

For a linear divisor $mx+b$ with $m\ne0$ , choose $x=-b/m$ so that the product $(mx+b)Q(x)$ disappears. The remainder is $P(-b/m)$ ; do not multiply or divide this value by $m$ .

Worked example

Divide 4x³ + 2x² − 3x + 1 by 2x + 1: find only the remainder.

2x+1=0\Rightarrow x=-\tfrac12

\begin{aligned}R&=4(-\tfrac12)^3+2(-\tfrac12)^2\\&\quad-3(-\tfrac12)+1\\&=-\tfrac12+\tfrac12+\tfrac32+1\\&=\tfrac52\end{aligned}

Your turnPractice

Use the factor theorem to show that $2x-1$ is a factor of $2x^3+x^2-5x+2$ .

Show worked answer

P(\tfrac12)=\tfrac14+\tfrac14-\tfrac52+2=0

Since the remainder is zero, $2x-1$ is a factor.

One substitution finds a linear remainder

After division by $x-a$ , the remainder $R$ is constant. Substitute $x=a$ into the division identity:

\begin{aligned}P(x)&=(x-a)Q(x)+R,\\P(a)&=0\times Q(a)+R=R.\end{aligned}

Worked example

Is x − 2 a factor of x³ + x² − 5x − 2?

P(2)=8+4-10-2=0

Yes. The substitution and zero result prove the claim; writing the final factorisation alone would not demonstrate the requested theorem.

Theorems give a remainder or test a factor; they do not by themselves find the quotient. Use division or coefficient matching if the quotient is also requested.

Your turnPractice

For $P(x)=x^3-4x+1$ , find the remainders for divisors $x-2$ and $x+1$ . Is either a factor?

Show worked answer

\begin{gathered}P(2)=8-8+1=1\\P(-1)=-1+4+1=4\end{gathered}

The remainders are 1 and 4, not zero, so neither divisor is a factor.

Set the whole divisor equal to zero

For a linear divisor $mx+b$ with $m\ne0$ , choose $x=-b/m$ so that the product $(mx+b)Q(x)$ disappears. The remainder is $P(-b/m)$ ; do not multiply or divide this value by $m$ .

Worked example

Divide 4x³ + 2x² − 3x + 1 by 2x + 1: find only the remainder.

2x+1=0\Rightarrow x=-\tfrac12

\begin{aligned}R&=4(-\tfrac12)^3+2(-\tfrac12)^2\\&\quad-3(-\tfrac12)+1\\&=-\tfrac12+\tfrac12+\tfrac32+1\\&=\tfrac52\end{aligned}

Your turnPractice

Use the factor theorem to show that $2x-1$ is a factor of $2x^3+x^2-5x+2$ .

Show worked answer

P(\tfrac12)=\tfrac14+\tfrac14-\tfrac52+2=0

Since the remainder is zero, $2x-1$ is a factor.