Finding Unknown Polynomial Coefficients | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Each condition gives an equation

If a coefficient is unknown, a factor or remainder statement still tells you the value of the polynomial at one input. Substitute carefully, then solve the resulting equations for the unknown coefficients.

Worked example

Find a and b when x + 2 is a factor of ax³ + bx² − ax + 8, and division by x − 2 leaves 24.

These are the conditions in 9709/22/O/N/24 Q5(a). The first means $P(-2)=0$ ; the second means $P(2)=24$ .

\begin{gathered}-8a+4b+2a+8=0\\-6a+4b=-8\\[4pt]8a+4b-2a+8=24\\6a+4b=16\end{gathered}

Add the equations to remove $a$ : $8b=8$ , so $b=1$ . Substitute back: $6a+4=16$ , so $a=2$ .

Check in both original conditions, not just the simplified system: $P(-2)=-16+4+4+8=0$ and $P(2)=16+4-4+8=24$ .

Your turnPractice

$P(x)=2x^3+ax^2+bx+3$ . Division by $x-1$ leaves 5, and $x+1$ is a factor. Find $a,b$ .

Show worked answer

\begin{gathered}P(1)=5\\a+b=0\\[4pt]P(-1)=0\\a-b=-1\end{gathered}

Answer

a=-\tfrac12,\qquad b=\tfrac12

Add to get $2a=-1$ , then substitute. These coefficients return 5 and 0 at the required inputs.

A non-monic divisor may give a fraction

Worked example

For P(x) = 2x³ + ax² + bx + 1, P(1) = 5 and 2x + 1 is a factor. Find a and b.

The factor gives $P(-1/2)=0$ .

\begin{aligned}a+b&=2,\\-\tfrac14+\tfrac a4-\tfrac b2+1&=0\\\Rightarrow a-2b&=-3.\end{aligned}

Multiply the second condition by 4 to remove fractions. Subtract it from the first: $3b=5$ . Thus $b=5/3$ and $a=1/3$ .

Your turnPractice

$P(x)=4x^3+ax+b$ leaves remainder 7 when divided by $x-1$ , and remainder 0 when divided by $2x-1$ . Find $a,b$ .

Show worked answer

\begin{gathered}a+b=3\\\tfrac12+\tfrac a2+b=0\end{gathered}

Double the second equation: $a+2b=-1$ . Subtract the first to get $b=-4$ , then $a=7$ .

Answer

a=7,\qquad b=-4

Each condition gives an equation

Worked example

Find a and b when x + 2 is a factor of ax³ + bx² − ax + 8, and division by x − 2 leaves 24.

These are the conditions in 9709/22/O/N/24 Q5(a). The first means $P(-2)=0$ ; the second means $P(2)=24$ .

\begin{gathered}-8a+4b+2a+8=0\\-6a+4b=-8\\[4pt]8a+4b-2a+8=24\\6a+4b=16\end{gathered}

Add the equations to remove $a$ : $8b=8$ , so $b=1$ . Substitute back: $6a+4=16$ , so $a=2$ .

Check in both original conditions, not just the simplified system: $P(-2)=-16+4+4+8=0$ and $P(2)=16+4-4+8=24$ .

Your turnPractice

$P(x)=2x^3+ax^2+bx+3$ . Division by $x-1$ leaves 5, and $x+1$ is a factor. Find $a,b$ .

Show worked answer

\begin{gathered}P(1)=5\\a+b=0\\[4pt]P(-1)=0\\a-b=-1\end{gathered}

Answer

a=-\tfrac12,\qquad b=\tfrac12

Add to get $2a=-1$ , then substitute. These coefficients return 5 and 0 at the required inputs.

A non-monic divisor may give a fraction

Worked example

For P(x) = 2x³ + ax² + bx + 1, P(1) = 5 and 2x + 1 is a factor. Find a and b.

The factor gives $P(-1/2)=0$ .

\begin{aligned}a+b&=2,\\-\tfrac14+\tfrac a4-\tfrac b2+1&=0\\\Rightarrow a-2b&=-3.\end{aligned}

Multiply the second condition by 4 to remove fractions. Subtract it from the first: $3b=5$ . Thus $b=5/3$ and $a=1/3$ .

Your turnPractice

$P(x)=4x^3+ax+b$ leaves remainder 7 when divided by $x-1$ , and remainder 0 when divided by $2x-1$ . Find $a,b$ .

Show worked answer

\begin{gathered}a+b=3\\\tfrac12+\tfrac a2+b=0\end{gathered}

Double the second equation: $a+2b=-1$ . Subtract the first to get $b=-4$ , then $a=7$ .

Answer

a=7,\qquad b=-4