Division by Non-monic Linear Polynomials | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Divide by the actual leading term

A polynomial is monic if its leading coefficient is 1. For a non-monic divisor such as $3x+2$ , divide by $3x$ , not by $x$ . The rest of the method is unchanged.

Worked example

Divide 9x³ + 18x² + 5x + 4 by 3x + 2.

First $9x^3\div3x=3x^2$ . Subtract its product with the whole divisor:

\begin{aligned}&(9x^3+18x^2+5x+4)\\&\quad-(9x^3+6x^2)\\&=12x^2+5x+4.\end{aligned}

Next $12x^2\div3x=4x$ , then $-3x\div3x=-1$ :

\begin{gathered}(12x^2+5x+4)\\{}-(12x^2+8x)\\=-3x+4\\[4pt](-3x+4)-(-3x-2)\\=6\end{gathered}

Answer

\begin{gathered}Q=3x^2+4x-1\\R=6\end{gathered}

This is the division in 9709/22/M/J/24 Q5(a). The subtraction shows the given remainder; the final number alone does not demonstrate it.

Your turnPractice

Find quotient and remainder when $4x^3+3x^2-2x+5$ is divided by $2x+1$ .

Show worked answer

The quotient terms are $2x^2$ , $x/2$ and $-5/4$ . Fractions are allowed.

\begin{gathered}P-2x^2(2x+1)\\=x^2-2x+5\\[4pt](x^2-2x+5)\\{}-\tfrac x2(2x+1)\\=-\tfrac52x+5\\[4pt](-\tfrac52x+5)\\{}-\bigl[-\tfrac54(2x+1)\bigr]\\=\tfrac{25}4\end{gathered}

Answer

\begin{gathered}Q=2x^2+\tfrac12x-\tfrac54\\R=\tfrac{25}4\end{gathered}

Keep the divisor’s sign

$2-x=-(x-2)$ . Changing a divisor’s sign changes the quotient’s sign, not the remainder.

\begin{aligned}P&=(x-2)(x^2+2x)+1\\&=(2-x)(-x^2-2x)+1.\end{aligned}

Your turnPractice

Given $P=(2x+1)(x^2-3)+4$ , find the quotient and remainder when $P$ is divided by $-2x-1$ .

Show worked answer

Answer

Q=-x^2+3,\qquad R=4

The two sign changes in the product cancel; the added 4 stays unchanged.

Divide by the actual leading term

A polynomial is monic if its leading coefficient is 1. For a non-monic divisor such as $3x+2$ , divide by $3x$ , not by $x$ . The rest of the method is unchanged.

Worked example

Divide 9x³ + 18x² + 5x + 4 by 3x + 2.

First $9x^3\div3x=3x^2$ . Subtract its product with the whole divisor:

\begin{aligned}&(9x^3+18x^2+5x+4)\\&\quad-(9x^3+6x^2)\\&=12x^2+5x+4.\end{aligned}

Next $12x^2\div3x=4x$ , then $-3x\div3x=-1$ :

\begin{gathered}(12x^2+5x+4)\\{}-(12x^2+8x)\\=-3x+4\\[4pt](-3x+4)-(-3x-2)\\=6\end{gathered}

Answer

\begin{gathered}Q=3x^2+4x-1\\R=6\end{gathered}

This is the division in 9709/22/M/J/24 Q5(a). The subtraction shows the given remainder; the final number alone does not demonstrate it.

Your turnPractice

Find quotient and remainder when $4x^3+3x^2-2x+5$ is divided by $2x+1$ .

Show worked answer

The quotient terms are $2x^2$ , $x/2$ and $-5/4$ . Fractions are allowed.

\begin{gathered}P-2x^2(2x+1)\\=x^2-2x+5\\[4pt](x^2-2x+5)\\{}-\tfrac x2(2x+1)\\=-\tfrac52x+5\\[4pt](-\tfrac52x+5)\\{}-\bigl[-\tfrac54(2x+1)\bigr]\\=\tfrac{25}4\end{gathered}

Answer

\begin{gathered}Q=2x^2+\tfrac12x-\tfrac54\\R=\tfrac{25}4\end{gathered}

Keep the divisor’s sign

$2-x=-(x-2)$ . Changing a divisor’s sign changes the quotient’s sign, not the remainder.

\begin{aligned}P&=(x-2)(x^2+2x)+1\\&=(2-x)(-x^2-2x)+1.\end{aligned}

Your turnPractice

Given $P=(2x+1)(x^2-3)+4$ , find the quotient and remainder when $P$ is divided by $-2x-1$ .

Show worked answer

Answer

Q=-x^2+3,\qquad R=4

The two sign changes in the product cancel; the added 4 stays unchanged.