Solving polynomial equations

Covers AS91577

The factor theorem

The factor theorem says that if $P(a) = 0$ , then $(x - a)$ is a factor of the polynomial $P(x)$ .
Method for a cubic: test small values ( $x = \pm 1, \pm 2, \ldots$ ) to find a value that gives 0, giving one linear factor; then divide to get a quadratic factor and solve that.

Complex roots come in conjugate pairs

If a polynomial has real coefficients, then any complex roots occur in conjugate pairs: if $a + bi$ is a root, so is $a - bi$ .
This is why a cubic with real coefficients always has at least one real root (complex roots pair up, and three roots cannot split into pairs without one left over).

Building factors from roots

A conjugate pair of roots $a \pm bi$ comes from a quadratic factor with real coefficients:

$\big(x - (a + bi)\big)\big(x - (a - bi)\big) = x^2 - 2ax + (a^2 + b^2)$

Worked ExampleSolving a cubic with a complex pair

Solve $x^3 - 5x^2 + 17x - 13 = 0$ , given that $x = 1$ is one root.

Step 1 — Use the known real root to get a factor

Since $x = 1$ is a root, $(x - 1)$ is a factor. Dividing $x^3 - 5x^2 + 17x - 13$ by $(x - 1)$ gives:

$x^3 - 5x^2 + 17x - 13 = (x - 1)(x^2 - 4x + 13)$

Step 2 — Solve the quadratic factor

Use the quadratic formula on $x^2 - 4x + 13 = 0$ :

$x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2}$

Step 3 — Simplify the square root of a negative

$\sqrt{-36} = 6i$ , so:

$x = \frac{4 \pm 6i}{2} = 2 \pm 3i$

The three roots are $x = 1$ , $x = 2 + 3i$ and $x = 2 - 3i$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Solve $x^2 + 9 = 0$ .

Merit

Solve $x^2 - 6x + 25 = 0$ , giving answers in the form $a + bi$ .

Excellence

A cubic with real coefficients has roots $x = 2$ and $x = 1 + i$ . Find the cubic in the form $x^3 + bx^2 + cx + d = 0$ .

Solving polynomial equations

Covers AS91577

The factor theorem

The factor theorem says that if $P(a) = 0$ , then $(x - a)$ is a factor of the polynomial $P(x)$ .
Method for a cubic: test small values ( $x = \pm 1, \pm 2, \ldots$ ) to find a value that gives 0, giving one linear factor; then divide to get a quadratic factor and solve that.

Complex roots come in conjugate pairs

If a polynomial has real coefficients, then any complex roots occur in conjugate pairs: if $a + bi$ is a root, so is $a - bi$ .
This is why a cubic with real coefficients always has at least one real root (complex roots pair up, and three roots cannot split into pairs without one left over).

Building factors from roots

A conjugate pair of roots $a \pm bi$ comes from a quadratic factor with real coefficients:

$\big(x - (a + bi)\big)\big(x - (a - bi)\big) = x^2 - 2ax + (a^2 + b^2)$

Worked ExampleSolving a cubic with a complex pair

Solve $x^3 - 5x^2 + 17x - 13 = 0$ , given that $x = 1$ is one root.

Step 1 — Use the known real root to get a factor

Since $x = 1$ is a root, $(x - 1)$ is a factor. Dividing $x^3 - 5x^2 + 17x - 13$ by $(x - 1)$ gives:

$x^3 - 5x^2 + 17x - 13 = (x - 1)(x^2 - 4x + 13)$

Step 2 — Solve the quadratic factor

Use the quadratic formula on $x^2 - 4x + 13 = 0$ :

$x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2}$

Step 3 — Simplify the square root of a negative

$\sqrt{-36} = 6i$ , so:

$x = \frac{4 \pm 6i}{2} = 2 \pm 3i$

The three roots are $x = 1$ , $x = 2 + 3i$ and $x = 2 - 3i$ .

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

Solve $x^2 + 9 = 0$ .

Merit

Solve $x^2 - 6x + 25 = 0$ , giving answers in the form $a + bi$ .

Excellence

A cubic with real coefficients has roots $x = 2$ and $x = 1 + i$ . Find the cubic in the form $x^3 + bx^2 + cx + d = 0$ .