Solving Polynomial Equations | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Find a useful root, then show the factorisation

A root is an input making the polynomial zero. When a calculator is permitted and supports the degree, its polynomial solver is a fast first method for an ordinary solve question. Enter all coefficients, including zeros for missing powers.

A simple root such as $2$ suggests $x-2$ ; a fraction such as $1/2$ suggests $2x-1$ . Verify the exact substitution, then divide or match coefficients. If a question specifies a theorem or algebraic method, show that method.

Worked example

Solve x³ − 4x² + x + 6 = 0.

The solver gives −1, 2 and 3. These are convenient exact roots, so write the factorisation, then expand it to check the original coefficients:

\begin{gathered}x^3-4x^2+x+6\\=(x+1)(x-2)(x-3)\\x=-1,\ 2,\ 3\end{gathered}

Alternatively, test $P(2)=8-16+2+6=0$ and divide by $x-2$ to obtain $x^2-2x-3=(x-3)(x+1)$ .

Without a solver, try small positive and negative factors of the constant term for a monic integer polynomial. For non-monic integer polynomials, rational candidates may also have denominators dividing the leading coefficient. Testing a candidate is not proof until its substitution is zero.

Your turnPractice

Solve $2x^3+x^2-5x+2=0$ , giving a checked factorisation.

Show worked answer

$P(1)=0$ . Division gives $2x^2+3x-2=(2x-1)(x+2)$ .

Answer

\begin{gathered}(x-1)(2x-1)(x+2)\\=0\end{gathered}

Thus $x=1,\ 1/2,\ -2$ . The factors expand back to the given cubic.

Awkward roots: solve the remaining quadratic exactly

A decimal is not an exact surd. After removing a simple factor, use the quadratic formula from MF19 on the remaining quadratic. Keep the leading coefficient in a factorisation; roots alone do not determine its scale.

Worked example

Solve x³ − 2x² − 2x + 4 = 0 exactly.

$P(2)=0$ . Division gives $P(x)=(x-2)(x^2-2)$ .

x^2-2=0\Rightarrow x=\pm\sqrt2

Answer

x=2,\quad x=\pm\sqrt2

Here taking a square root is shortest. For $ax^2+bx+c=0$ with $a\ne0$ , the general exact route is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

For a quartic, remove two known distinct linear factors or a known quadratic factor to leave a quadratic. For example:

\begin{gathered}x^4-x^3-3x^2+x+2\\=(x^2-1)(x^2-x-2)\\=(x-1)(x+1)^2(x-2)\end{gathered}

This has three distinct real roots: −1, 1 and 2. The repeated −1 is one value, although its factor occurs twice.

Your turnPractice

Solve $x^4-5x^2+6=0$ exactly. You may treat it as a quadratic in $u=x^2$ .

Show worked answer

\begin{gathered}u^2-5u+6=0\\(u-2)(u-3)=0\end{gathered}

Answer

x=\pm\sqrt2,\qquad x=\pm\sqrt3

Both positive values of $u$ give two real square roots.

Prove when there are no further real roots

The discriminant of $ax^2+bx+c$ is $b^2-4ac$ , the quantity under the square root in the formula. Positive gives two real roots; zero gives one repeated root; negative gives no real roots.

Your turn9709/22/O/N/24 Q5(b)

Factorise $P(x)=2x^3+x^2-2x+8$ and show that $P(x)=0$ has exactly one real root. You already know $x+2$ is a factor.

Show worked answer

Divide by $x+2$ ; successive quotient terms are $2x^2$ , $-3x$ , 4.

\begin{gathered}P(x)\\=(x+2)(2x^2-3x+4)\end{gathered}

The quadratic discriminant is $(-3)^2-4(2)(4)=-23$ , so it contributes no real roots.

Answer

x=-2

Find a useful root, then show the factorisation

Worked example

Solve x³ − 4x² + x + 6 = 0.

The solver gives −1, 2 and 3. These are convenient exact roots, so write the factorisation, then expand it to check the original coefficients:

\begin{gathered}x^3-4x^2+x+6\\=(x+1)(x-2)(x-3)\\x=-1,\ 2,\ 3\end{gathered}

Alternatively, test $P(2)=8-16+2+6=0$ and divide by $x-2$ to obtain $x^2-2x-3=(x-3)(x+1)$ .

Your turnPractice

Solve $2x^3+x^2-5x+2=0$ , giving a checked factorisation.

Show worked answer

$P(1)=0$ . Division gives $2x^2+3x-2=(2x-1)(x+2)$ .

Answer

\begin{gathered}(x-1)(2x-1)(x+2)\\=0\end{gathered}

Thus $x=1,\ 1/2,\ -2$ . The factors expand back to the given cubic.

Awkward roots: solve the remaining quadratic exactly

Worked example

Solve x³ − 2x² − 2x + 4 = 0 exactly.

$P(2)=0$ . Division gives $P(x)=(x-2)(x^2-2)$ .

x^2-2=0\Rightarrow x=\pm\sqrt2

Answer

x=2,\quad x=\pm\sqrt2

Here taking a square root is shortest. For $ax^2+bx+c=0$ with $a\ne0$ , the general exact route is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

For a quartic, remove two known distinct linear factors or a known quadratic factor to leave a quadratic. For example:

\begin{gathered}x^4-x^3-3x^2+x+2\\=(x^2-1)(x^2-x-2)\\=(x-1)(x+1)^2(x-2)\end{gathered}

This has three distinct real roots: −1, 1 and 2. The repeated −1 is one value, although its factor occurs twice.

Your turnPractice

Solve $x^4-5x^2+6=0$ exactly. You may treat it as a quadratic in $u=x^2$ .

Show worked answer

\begin{gathered}u^2-5u+6=0\\(u-2)(u-3)=0\end{gathered}

Answer

x=\pm\sqrt2,\qquad x=\pm\sqrt3

Both positive values of $u$ give two real square roots.

Prove when there are no further real roots

The discriminant of $ax^2+bx+c$ is $b^2-4ac$ , the quantity under the square root in the formula. Positive gives two real roots; zero gives one repeated root; negative gives no real roots.

Your turn9709/22/O/N/24 Q5(b)

Factorise $P(x)=2x^3+x^2-2x+8$ and show that $P(x)=0$ has exactly one real root. You already know $x+2$ is a factor.

Show worked answer

Divide by $x+2$ ; successive quotient terms are $2x^2$ , $-3x$ , 4.

\begin{gathered}P(x)\\=(x+2)(2x^2-3x+4)\end{gathered}

The quadratic discriminant is $(-3)^2-4(2)(4)=-23$ , so it contributes no real roots.

Answer

x=-2