ScotlandMathsSyllabus dot point

How do you complete the square, read the discriminant, and use the factor theorem to factorise and solve polynomial equations?

Completing the square and the properties of the quadratic, the discriminant and the nature of the roots, the condition for a quadratic to be always positive or always negative, the factor and remainder theorems, and solving and sketching polynomials.

A focused answer to the SQA Higher Mathematics polynomials and quadratics content, covering completing the square, the discriminant and the nature of the roots, the always positive or always negative condition, the factor and remainder theorems, and solving and sketching polynomials.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this dot point is asking
Completing the square
The discriminant
The factor and remainder theorems
Examples in context
Try this

What this dot point is asking

The SQA wants you to complete the square on a quadratic, use the discriminant to decide the nature of the roots, state when a quadratic is always positive or always negative, apply the factor and remainder theorems, and solve and sketch polynomial functions.

Completing the square

Completing the square rewrites a quadratic so the variable appears only once, inside a bracket. This exposes the turning point and the maximum or minimum value immediately, which is why it underlies optimisation and "always positive" arguments.

Complete the square on

2x^2 - 12x + 23

Rewrite $2x^2 - 12x + 23$ in completed-square form and find the turning point.

Step 1: Factor the leading coefficient out of the quadratic and linear terms

The method only works cleanly on a monic expression (coefficient of $x^2$ equal to $1$ ), so factor out $2$ from the first two terms only:

2(x^2 - 6x) + 23

The constant $23$ is left outside because it does not affect the completing-the-square step.

Step 2: Complete the square inside the bracket

Take half the coefficient of $x$ inside the bracket: half of $-6$ is $-3$ , and $(-3)^2 = 9$ . Write $x^2 - 6x = (x - 3)^2 - 9$ (subtracting $9$ corrects the introduced value), then expand the factor of $2$ :

2\big((x - 3)^2 - 9\big) + 23 = 2(x - 3)^2 - 18 + 23

Step 3: Simplify and read off the turning point

2(x - 3)^2 + 5

The minimum of $(x - 3)^2$ is $0$ , so the minimum of the whole expression is $5$ , attained at $x = 3$ .

Final answer: $2x^2 - 12x + 23 = 2(x - 3)^2 + 5$ . The turning point is at $(3, 5)$ , a minimum because the leading coefficient $2$ is positive.

The discriminant

The discriminant is the quantity under the square root in the quadratic formula, so its sign decides how many real roots exist before you solve anything. Examiners frequently set the discriminant equal to zero (for equal roots) or make it positive or negative (for two roots or none) to find an unknown coefficient.

The factor and remainder theorems

The factor theorem turns root-finding for cubics into a search: if substituting $x = h$ gives $f(h) = 0$ , then $(x - h)$ divides $f(x)$ exactly. After finding one factor by trial of the divisors of the constant term, you divide to reduce the cubic to a quadratic you can factorise.

Factorise

f(x) = x^3 - 6x^2 + 11x - 6

Factorise the cubic fully and hence solve $f(x) = 0$ .

Step 1: Find a root using the factor theorem

By the factor theorem, if $f(h) = 0$ then $(x - h)$ is a factor. Sensible candidates to try are the integer divisors of the constant term $-6$ : try $x = 1$ :

f(1) = 1 - 6 + 11 - 6 = 0

Since $f(1) = 0$ , the factor theorem tells us $(x - 1)$ divides $f(x)$ exactly.

Step 2: Divide the cubic by $(x - 1)$ to obtain a quadratic

Using synthetic division (or long division) with $x = 1$ on coefficients $1, -6, 11, -6$ gives quotient $x^2 - 5x + 6$ .

Step 3: Factorise the quadratic and state the roots

x^2 - 5x + 6 = (x - 2)(x - 3)

So $f(x) = (x - 1)(x - 2)(x - 3)$ , and the roots of $f(x) = 0$ are $x = 1, 2, 3$ .

Final answer: $f(x) = (x - 1)(x - 2)(x - 3)$ and the solutions are $x = 1$ , $x = 2$ and $x = 3$ .

The remainder theorem says the remainder when $f(x)$ is divided by $(x - h)$ is $f(h)$ , so a non-zero $f(h)$ gives the remainder directly without doing the division.

Use the discriminant to find a condition on

k

Find the values of $k$ for which $x^2 + kx + 4 = 0$ has two distinct real roots.

Step 1: Identify the coefficients and state the condition needed

Compare with $ax^2 + bx + c = 0$ : here $a = 1$ , $b = k$ and $c = 4$ . Two distinct real roots require the discriminant to be strictly positive, so the condition is $b^2 - 4ac > 0$ .

Step 2: Form the inequality in terms of $k$

Substitute the coefficients:

k^2 - 4(1)(4) > 0 \implies k^2 - 16 > 0

Step 3: Solve the inequality

$k^2 > 16$ means the distance from zero is more than $4$ , giving $k > 4$ or $k < -4$ .

Final answer: The equation has two distinct real roots when $k > 4$ or $k < -4$ . For $-4 < k < 4$ the discriminant is negative (no real roots); at $k = \pm 4$ there is exactly one repeated root.

Examples in context

Completing the square models the path of a projectile. A ball's height is $h = -5t^2 + 20t + 2$ . Completing the square gives $h = -5(t - 2)^2 + 22$ , so the greatest height is $22$ m, reached at $t = 2$ s. Reading the maximum straight off the completed-square form avoids any calculus and is exactly how the vertex form is used in modelling.

Try this

Q1. Find the discriminant of $2x^2 - 4x + 5$ and state the nature of the roots. [2 marks]

Cue. $b^2 - 4ac = 16 - 40 = -24 < 0$ , so no real roots.

Q2. Show that $(x + 2)$ is a factor of $x^3 + 3x^2 - 4$ . [2 marks]

Cue. $f(-2) = -8 + 12 - 4 = 0$ , so $(x + 2)$ is a factor.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20205 marksThe equation

x^2 + (k - 2)x + 9 = 0

has equal roots. Determine the possible values of the constant

k

Show worked answer →

Equal roots means the discriminant is zero: $b^2 - 4ac = 0$ (1 mark).

Here $a = 1$ , $b = k - 2$ , $c = 9$ , so $(k - 2)^2 - 4(1)(9) = 0$ , that is $(k - 2)^2 - 36 = 0$ (2 marks).

Solve: $(k - 2)^2 = 36$ , so $k - 2 = \pm 6$ , giving $k = 8$ or $k = -4$ (2 marks). Markers reward setting the discriminant to zero, substituting the coefficients correctly, and both values of $k$ .

SQA Higher 20235 marks(a) Show that

(x - 3)

is a factor of

f(x) = x^3 - 4x^2 - 3x + 18

. (b) Hence factorise

f(x)

fully and solve

f(x) = 0

Show worked answer →

(a) By the factor theorem, evaluate $f(3) = 27 - 36 - 9 + 18 = 0$ , so $(x - 3)$ is a factor (1 mark).

(b) Divide $f(x)$ by $(x - 3)$ (synthetic division with $3$ on coefficients $1, -4, -3, 18$ gives $1, -1, -6, 0$ ), so the quotient is $x^2 - x - 6$ (2 marks).

Factorise the quadratic: $x^2 - x - 6 = (x - 3)(x + 2)$ , so $f(x) = (x - 3)^2(x + 2)$ (1 mark).

Solve $f(x) = 0$ : $x = 3$ (repeated) or $x = -2$ (1 mark). Markers reward the factor-theorem check, the division to a quadratic, the full factorisation, and the roots.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)