Quadratic Equations

Lessons

Senior 3 · Mathematics

Quadratic Equations

5 lessons

Lesson 4 of 5

Solving equations with synthetic division

22 min

One root opens the door to the others

Suppose you notice that $x = 5$ makes $x^{2} - 2 x - 15$ equal to zero. Then $x - 5$ must be a factor, and dividing by it leaves a simpler factor that gives the other root. In S2 you met a quick layout for this division, synthetic division, also called Horner's method. In this lesson you use it to solve quadratic equations when one root can be guessed, and to solve cubic equations, which reduce to a quadratic after one division.

Important

The factor theorem. For a polynomial $P (x)$ , the number $r$ is a root of $P (x) = 0$ exactly when $x - r$ is a factor of $P (x)$ .

Why: dividing $P (x)$ by $x - r$ gives a quotient $Q (x)$ and a number remainder $R$ : $P (x) = (x - r) Q (x) + R$ . Put $x = r$ : $P(r) = 0 \times Q(r) + R = R$ . So the remainder is $P (r)$ , and it is zero exactly when $r$ is a root.

Definition

Synthetic division (Horner's method): a short layout for dividing a polynomial by $x - r$ using only its coefficients.

Notation: write $r$ on the left and the coefficients in order of falling powers (a 0 for any missing power). Bring the first coefficient down; then repeatedly multiply the last result by $r$ , write the product under the next coefficient, and add. The last number is the remainder; the others are the coefficients of the quotient, one degree lower.

Example: dividing $x^{2} - 2 x - 15$ by $x - 5$ gives the bottom row $1,\ 3,\ 0$ : quotient $x + 3$ , remainder 0 (Worked example 1).

Important

Where to look for a root. If all the coefficients are integers, any integer root divides the constant term, and any fraction root $\frac{p}{q}$ (in lowest terms) has $p$ dividing the constant term and $q$ dividing the leading coefficient.

Why (integer case): if $r$ is an integer root of $x^{2} + b x + c = 0$ , then $c = - r^{2} - b r = r (- r - b)$ , a multiple of $r$ . The same argument works for any degree.

Worked example 1: the second root of a quadratic

Show that $x = 5$ is a root of $x^{2} - 2 x - 15 = 0$ , and use synthetic division to find the other root.

Solution $5^{2} - 2 (5) - 15 = 25 - 10 - 15 = 0$ , so 5 is a root. Divide by $x - 5$ :

$r$	$x^{2}$	$x$	constant
$5$	$1$	$- 2$	$- 15$
		$5$	$15$
	$1$	$3$	$0$

Bring down 1. Then $1 \times 5 = 5$ and $- 2 + 5 = 3$ ; then $3 \times 5 = 15$ and $- 15 + 15 = 0$ . The remainder is 0 (as the factor theorem predicts) and the quotient is $x + 3$ . So

x^2 - 2x - 15 = (x - 5)(x + 3) = 0, \quad\text{giving}\quad x = 5 \ \text{ or } \ x = -3.

Worked example 2: a leading coefficient that is not 1

Solve $2 x^{2} + x - 6 = 0$ by finding one root by trial.

Solution Integer roots must divide $- 6$ : try $\pm 1, \pm 2, \pm 3, \pm 6$ . $x = 1$ gives $2 + 1 - 6 = - 3$ , not 0. $x = - 2$ gives $8 - 2 - 6 = 0$ . ✓ Divide by $x + 2$ , that is $x - (- 2)$ :

$r$	$x^{2}$	$x$	constant
$- 2$	$2$	$1$	$- 6$
		$- 4$	$6$
	$2$	$- 3$	$0$

The quotient is $2 x - 3$ , so $2 x^{2} + x - 6 = (x + 2) (2 x - 3)$ , and the roots are $x = - 2$ and $x = \frac{3}{2}$ .

Worked example 3: a cubic equation

Solve $2 x^{3} - 3 x^{2} - 11 x + 6 = 0$ .

Solution Write $P (x) = 2 x^{3} - 3 x^{2} - 11 x + 6$ . Test divisors of 6: $P (1) = 2 - 3 - 11 + 6 = - 6$ ; $P (- 1) = - 2 - 3 + 11 + 6 = 12$ ; $P (2) = 16 - 12 - 22 + 6 = - 12$ ; $P (3) = 54 - 27 - 33 + 6 = 0$ . ✓ Divide by $x - 3$ :

$r$	$x^{3}$	$x^{2}$	$x$	constant
$3$	$2$	$- 3$	$- 11$	$6$
		$6$	$9$	$- 6$
	$2$	$3$	$- 2$	$0$

The quotient is the quadratic $2 x^{2} + 3 x - 2$ , which factorises as $(2 x - 1) (x + 2)$ . So

2 x^{3} - 3 x^{2} - 11 x + 6 = (x - 3) (2 x - 1) (x + 2) = 0,

and the roots are $x = 3$ , $x = \frac{1}{2}$ and $x = - 2$ .

The graph of $y = 2 x^{3} - 3 x^{2} - 11 x + 6$ crosses the $x$ -axis at the three roots. A cubic can have up to three real roots.

Worked example 4: when the quadratic factor needs the formula

Solve $x^{3} - 3 x^{2} - 2 x + 4 = 0$ . Give exact roots.

Solution The coefficients add up to $1 - 3 - 2 + 4 = 0$ , so $x = 1$ is a root. Divide by $x - 1$ :

$r$	$x^{3}$	$x^{2}$	$x$	constant
$1$	$1$	$- 3$	$- 2$	$4$
		$1$	$- 2$	$- 4$
	$1$	$- 2$	$- 4$	$0$

The quotient $x^{2} - 2 x - 4$ does not factorise ( $\Delta = 4 + 16 = 20$ is not a perfect square), so use the formula:

x = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}.

The roots are $x = 1$ , $x = 1 + \sqrt{5} \approx 3.24$ and $x = 1 - \sqrt{5} \approx -1.24$ .

Warning

A missing power still needs its place. For $x^{3} - 7 x + 6$ the coefficients are $1,\ 0,\ -7,\ 6$ , not $1,\ -7,\ 6$ . Leaving out the 0 shifts every later number into the wrong column and gives a wrong quotient.

Divide by $x - r$ , write $r$ . To divide by $x + 2$ , write $- 2$ in the corner.

Tip

Two free tests. If the coefficients add up to 0, then $x = 1$ is a root. If the sum of the coefficients of the even powers equals the sum for the odd powers, then $x = - 1$ is a root. Try these before any other number.

PreMath: Solve Cubic Equations – Synthetic Division Method (10 min). Watch how one root is found by testing the divisors of the constant term, how synthetic division reduces the cubic to a quadratic, and how the quadratic is then solved.

✦

Practice

Use synthetic division in every question. The last three are harder.

Show that $x = 4$ is a root of $x^{2} - x - 12 = 0$ and find the other root.
Given that $x = 2$ is a root of $3 x^{2} - 5 x - 2 = 0$ , find the other root.
Solve $x^{3} - 7 x + 6 = 0$ .
Solve $x^{3} + 2 x^{2} - 5 x - 6 = 0$ .
Solve $2 x^{3} + x^{2} - 13 x + 6 = 0$ .
Solve $x^{3} - 4 x^{2} + 2 x + 1 = 0$ , giving exact roots.
$x = - 2$ is a root of $x^{3} + k x^{2} - x - 2 = 0$ . Find $k$ and the other two roots.

Show answersHide answers

$16 - 4 - 12 = 0$ . Row: $1,\ 3,\ 0$ ; quotient $x + 3$ ; other root $x = - 3$ .
Row: $3,\ 1,\ 0$ ; quotient $3 x + 1$ ; other root $x = -\frac{1}{3}$ .
$x = 1$ (coefficients $1 + 0 - 7 + 6 = 0$ ). Row: $1,\ 1,\ -6,\ 0$ ; $x^{2} + x - 6 = (x + 3) (x - 2)$ . Roots $1,\ 2,\ -3$ .
$x = - 1$ : $- 1 + 2 + 5 - 6 = 0$ . Row: $1,\ 1,\ -6,\ 0$ ; $(x + 3) (x - 2)$ . Roots $-1,\ 2,\ -3$ .
$x = 2$ : $16 + 4 - 26 + 6 = 0$ . Row: $2,\ 5,\ -3,\ 0$ ; $2 x^{2} + 5 x - 3 = (2 x - 1) (x + 3)$ . Roots $2,\ \frac{1}{2},\ -3$ .
$x = 1$ (coefficients add to 0). Row: $1,\ -3,\ -1,\ 0$ ; $x^{2} - 3 x - 1 = 0$ gives $x = \frac{3 \pm \sqrt{13}}{2}$ . Roots $1,\ \frac{3 + \sqrt{13}}{2} \approx 3.30,\ \frac{3 - \sqrt{13}}{2} \approx -0.30$ .
$- 8 + 4 k + 2 - 2 = 0$ , so $k = 2$ . Row for $1,\ 2,\ -1,\ -2$ with $- 2$ : $1,\ 0,\ -1,\ 0$ ; $x^{2} - 1 = 0$ . Other roots $1$ and $- 1$ .

Sources and credits

5.5: Dividing Polynomials (synthetic division and the factor theorem), from Intermediate Algebra by Lynn Marecek and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The rule for where to look for roots, the two free tests and the worked examples. Written by Ganzaa.
Video: Solve Cubic Equations - Synthetic Division Method, PreMath, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1One root opens the door to the others

One root opens the door to the others