Quadratic Factors and Unknown Roots | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

A whole quadratic factor gives more information

If a cubic has a known quadratic factor, its quotient is linear. You can write that factorisation as an identity and match coefficients. This also works when the quadratic has no real roots.

Worked example

x² + 1 is a factor of 2x³ + 3x² + ax + b. Find a and b.

The leading term forces the quotient’s first term to be $2x$ . Write its constant as $c$ .

\begin{gathered}(x^2+1)(2x+c)\\=2x^3+cx^2+2x+c\end{gathered}

Matching $x^2$ gives $c=3$ . Matching the remaining terms gives $a=2$ , $b=3$ .

If the quadratic factors into distinct linear factors, their roots give two separate zero-value conditions. For example, a factor $(x-1)(x+2)$ requires both $P(1)=0$ and $P(-2)=0$ .

Your turnPractice

$x^2+x-2$ is a factor of $x^3+2x^2+ax+b$ . Find $a,b$ and factorise the cubic completely.

Show worked answer

$x^2+x-2=(x-1)(x+2)$ . Substitution gives $a+b=-3$ and $-2a+b=0$ ; hence $a=-1$ , $b=-2$ .

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

Repeating one root does not give two equations

A repeated factor such as $(x-1)^2$ contains $x-1$ twice. Writing $P(1)=0$ twice still gives only one condition. Instead use the entire quadratic factor.

Worked example

(x − 1)² is a factor of x³ + ax + b. Find a and b.

\begin{aligned}&(x-1)^2(x+c)\\&=(x^2-2x+1)(x+c)\\&=x^3+(c-2)x^2\\&\quad+(1-2c)x+c\end{aligned}

The missing $x^2$ coefficient is zero, so $c-2=0$ . Therefore $c=2$ , $a=-3$ , $b=2$ .

Your turnPractice

If $(x+1)^2$ is a factor of $x^3+2x^2+ax+b$ , find $a,b$ .

Show worked answer

\begin{aligned}&(x+1)^2(x+c)\\&=x^3+(c+2)x^2\\&\quad+(2c+1)x+c\end{aligned}

Answer

\begin{gathered}c=0\\a=1,\quad b=0\end{gathered}

The input itself can contain the unknown

Worked example

x − a is a factor of x² + ax − 8. Find a.

Use the root $x=a$ without changing the other coefficient $a$ .

\begin{gathered}P(a)=a^2+a\cdot a-8\\2a^2-8=0\\a^2=4\end{gathered}

Answer

a=2\quad\text{or}\quad a=-2

There can be more than one possible polynomial. Both values satisfy the factor condition.

Your turnPractice

$x-a$ is a factor of $x^3+ax-12$ , where $a$ is real. Find $a$ , given $a^3+a^2-12=(a-2)(a^2+3a+6)$ .

Show worked answer

Substitute $x=a$ to get $a^3+a^2-12=0$ . The quadratic has discriminant $9-24=-15<0$ , so no real roots.

Answer

a=2

A whole quadratic factor gives more information

If a cubic has a known quadratic factor, its quotient is linear. You can write that factorisation as an identity and match coefficients. This also works when the quadratic has no real roots.

Worked example

x² + 1 is a factor of 2x³ + 3x² + ax + b. Find a and b.

The leading term forces the quotient’s first term to be $2x$ . Write its constant as $c$ .

\begin{gathered}(x^2+1)(2x+c)\\=2x^3+cx^2+2x+c\end{gathered}

Matching $x^2$ gives $c=3$ . Matching the remaining terms gives $a=2$ , $b=3$ .

If the quadratic factors into distinct linear factors, their roots give two separate zero-value conditions. For example, a factor $(x-1)(x+2)$ requires both $P(1)=0$ and $P(-2)=0$ .

Your turnPractice

$x^2+x-2$ is a factor of $x^3+2x^2+ax+b$ . Find $a,b$ and factorise the cubic completely.

Show worked answer

$x^2+x-2=(x-1)(x+2)$ . Substitution gives $a+b=-3$ and $-2a+b=0$ ; hence $a=-1$ , $b=-2$ .

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

Repeating one root does not give two equations

A repeated factor such as $(x-1)^2$ contains $x-1$ twice. Writing $P(1)=0$ twice still gives only one condition. Instead use the entire quadratic factor.

Worked example

(x − 1)² is a factor of x³ + ax + b. Find a and b.

\begin{aligned}&(x-1)^2(x+c)\\&=(x^2-2x+1)(x+c)\\&=x^3+(c-2)x^2\\&\quad+(1-2c)x+c\end{aligned}

The missing $x^2$ coefficient is zero, so $c-2=0$ . Therefore $c=2$ , $a=-3$ , $b=2$ .

Your turnPractice

If $(x+1)^2$ is a factor of $x^3+2x^2+ax+b$ , find $a,b$ .

Show worked answer

\begin{aligned}&(x+1)^2(x+c)\\&=x^3+(c+2)x^2\\&\quad+(2c+1)x+c\end{aligned}

Answer

\begin{gathered}c=0\\a=1,\quad b=0\end{gathered}

The input itself can contain the unknown

Worked example

x − a is a factor of x² + ax − 8. Find a.

Use the root $x=a$ without changing the other coefficient $a$ .

\begin{gathered}P(a)=a^2+a\cdot a-8\\2a^2-8=0\\a^2=4\end{gathered}

Answer

a=2\quad\text{or}\quad a=-2

There can be more than one possible polynomial. Both values satisfy the factor condition.

Your turnPractice

$x-a$ is a factor of $x^3+ax-12$ , where $a$ is real. Find $a$ , given $a^3+a^2-12=(a-2)(a^2+3a+6)$ .

Show worked answer

Substitute $x=a$ to get $a^3+a^2-12=0$ . The quadratic has discriminant $9-24=-15<0$ , so no real roots.

Answer

a=2