Division by Quadratic Polynomials | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

A quadratic divisor can leave a linear remainder

A quadratic has degree 2. Stop when what remains has degree below 2: it may be $ax+b$ , a constant, or zero. Do not force every remainder to be a number.

Worked example

Divide x⁴ + 2x³ − x² + 3x + 5 by x² + 1.

Divide the leading terms, $x^4\div x^2=x^2$ , and subtract its product with the whole divisor:

x^2(x^2+1)=x^4+x^2

The fourth-power terms cancel. The square terms give $-x^2-x^2=-2x^2$ ; all other terms stay unchanged. Calling the original dividend $P$ , we have:

\begin{gathered}P-x^2(x^2+1)\\=2x^3-2x^2+3x+5\end{gathered}

Now use $2x^3\div x^2=2x$ , then $-2x^2\div x^2=-2$ :

\begin{aligned}&(2x^3-2x^2+3x+5)\\&\quad-(2x^3+2x)\\&=-2x^2+x+5\\[6pt]&(-2x^2+x+5)\\&\quad-(-2x^2-2)\\&=x+7\end{aligned}

Answer

\begin{gathered}Q=x^2+2x-2\\R=x+7\end{gathered}

We cannot cancel the remaining $x$ using a polynomial multiple of $x^2+1$ : it would require a negative power. That is why we stop.

Your turnPractice

Divide $2x^4-x^3+3x+1$ by $x^2-x+1$ . Include the missing power before starting.

Show worked answer

Write $2x^4-x^3+0x^2+3x+1$ . Successive quotient terms are $2x^2$ , $x$ , −1.

\begin{gathered}P-2x^2D\\=x^3-2x^2+3x+1\\[4pt](x^3-2x^2+3x+1)\\{}-xD=-x^2+2x+1\\[4pt](-x^2+2x+1)-(-D)\\=x+2\end{gathered}

Answer

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

Matching coefficients is an alternative

Two polynomials that agree for every $x$ have the same coefficient of each power. To divide a cubic by a quadratic, write a linear quotient and a remainder $cx+d$ , expand, then match like powers.

Worked example

Write x³ + x² + 2x + 3 as (x² + 1)Q + R.

\begin{aligned}&x^3+x^2+2x+3\\&=(x^2+1)(ax+b)\\&\quad+cx+d\\&=ax^3+bx^2\\&\quad+(a+c)x+(b+d)\end{aligned}

Match $x^3$ : $a=1$ . Match $x^2$ : $b=1$ . Then $a+c=2$ gives $c=1$ , and $b+d=3$ gives $d=2$ .

Answer

\begin{gathered}Q=x+1\\R=x+2\end{gathered}

Your turnPractice

Use coefficient matching to divide $2x^3-x^2+5x$ by $x^2+2$ .

Show worked answer

\begin{aligned}&(x^2+2)(ax+b)\\&\quad+cx+d\\&=ax^3+bx^2\\&\quad+(2a+c)x+2b+d\end{aligned}

$a=2$ , $b=-1$ , $c=1$ , $d=2$ .

Answer

\begin{gathered}Q=2x-1\\R=x+2\end{gathered}

A quadratic divisor can leave a linear remainder

A quadratic has degree 2. Stop when what remains has degree below 2: it may be $ax+b$ , a constant, or zero. Do not force every remainder to be a number.

Worked example

Divide x⁴ + 2x³ − x² + 3x + 5 by x² + 1.

Divide the leading terms, $x^4\div x^2=x^2$ , and subtract its product with the whole divisor:

x^2(x^2+1)=x^4+x^2

The fourth-power terms cancel. The square terms give $-x^2-x^2=-2x^2$ ; all other terms stay unchanged. Calling the original dividend $P$ , we have:

\begin{gathered}P-x^2(x^2+1)\\=2x^3-2x^2+3x+5\end{gathered}

Now use $2x^3\div x^2=2x$ , then $-2x^2\div x^2=-2$ :

\begin{aligned}&(2x^3-2x^2+3x+5)\\&\quad-(2x^3+2x)\\&=-2x^2+x+5\\[6pt]&(-2x^2+x+5)\\&\quad-(-2x^2-2)\\&=x+7\end{aligned}

Answer

\begin{gathered}Q=x^2+2x-2\\R=x+7\end{gathered}

We cannot cancel the remaining $x$ using a polynomial multiple of $x^2+1$ : it would require a negative power. That is why we stop.

Your turnPractice

Divide $2x^4-x^3+3x+1$ by $x^2-x+1$ . Include the missing power before starting.

Show worked answer

Write $2x^4-x^3+0x^2+3x+1$ . Successive quotient terms are $2x^2$ , $x$ , −1.

\begin{gathered}P-2x^2D\\=x^3-2x^2+3x+1\\[4pt](x^3-2x^2+3x+1)\\{}-xD=-x^2+2x+1\\[4pt](-x^2+2x+1)-(-D)\\=x+2\end{gathered}

Answer

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

Matching coefficients is an alternative

Two polynomials that agree for every $x$ have the same coefficient of each power. To divide a cubic by a quadratic, write a linear quotient and a remainder $cx+d$ , expand, then match like powers.

Worked example

Write x³ + x² + 2x + 3 as (x² + 1)Q + R.

\begin{aligned}&x^3+x^2+2x+3\\&=(x^2+1)(ax+b)\\&\quad+cx+d\\&=ax^3+bx^2\\&\quad+(a+c)x+(b+d)\end{aligned}

Match $x^3$ : $a=1$ . Match $x^2$ : $b=1$ . Then $a+c=2$ gives $c=1$ , and $b+d=3$ gives $d=2$ .

Answer

\begin{gathered}Q=x+1\\R=x+2\end{gathered}

Your turnPractice

Use coefficient matching to divide $2x^3-x^2+5x$ by $x^2+2$ .

Show worked answer

\begin{aligned}&(x^2+2)(ax+b)\\&\quad+cx+d\\&=ax^3+bx^2\\&\quad+(2a+c)x+2b+d\end{aligned}

$a=2$ , $b=-1$ , $c=1$ , $d=2$ .

Answer

\begin{gathered}Q=2x-1\\R=x+2\end{gathered}