Factorising and completing the square

Q: What is the standard algebraic form of a quadratic expression and what are its components?

The standard form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants with \(a \neq 0\).

Q: How do you factorise an expression using the difference of two squares identity?

Use the identity \(A^2 - B^2 = (A - B)(A + B)\).

Q: How do you factorise a monic quadratic of the form \(x^2 + bx + c\)?

Find two numbers \(p\) and \(q\) such that \(p \times q = c\) and \(p + q = b\), giving \((x + p)(x + q)\).

Q: What are the steps to factorise a non-monic quadratic \(ax^2 + bx + c\) using the "ac" method?

Find two numbers that multiply to give \(ac\) and add to give \(b\), split the middle term, and factorise by grouping.

Q: How do you complete the square on a non-monic quadratic \(ax^2 + bx + c\) where \(a \neq 1\)?

Factor out \(a\) from the \(x^2\) and \(x\) terms first: \(a\left[x^2 + \frac{b}{a}x\right] + c\), then complete the square inside the brackets.

Q: What extra step is essential when completing the square if the \(x^2\) coefficient \(a\) is negative?

Factor out the negative coefficient from the \(x\)-terms first, taking care to reverse the sign of the middle \(x\)-term.

Q: How do you find the vertex (turning point) and optimum value from \(y = a(x + p)^2 + q\)?

The turning point has coordinates \((-p, q)\); the maximum or minimum value is \(q\), occurring at \(x = -p\).

Q: How do you solve a quadratic equation \(ax^2 + bx + c = 0\) by completing the square?

Rewrite the equation into the form \((x + p)^2 = k\), then take the square root of both sides to get \(x + p = \pm\sqrt{k}\).

Q: What common mistake occurs when expanding the outer coefficient in \(a\left[(x + p)^2 - d\right] + c\)?

Forgetting to multiply the subtracted term \(-d\) by the leading coefficient \(a\).

Practice

Further Pure Mathematics · Pearson Edexcel IGCSE · 10 min read · Updated: 23 Oct 2026

Welcome to Quadratic Manipulation

Quadratic expressions appear everywhere in Pearson Edexcel International GCSE Further Pure Mathematics. A quadratic expression is an algebraic expression where the highest power of the variable is \(2\). In its standard form, it is written as:

\(ax^2 + bx + c\)

Here, \(a\), \(b\), and \(c\) are constants (numbers), and \(a \neq 0\). The number \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

Think of algebraic expressions like mechanical objects. In standard form, you can read off the constant term easily. But if you take the expression apart and rebuild it into different forms, you unlock different superpowers:

1. Factorised Form: \(a(x - p)(x - q)\) reveals roots, intercepts, and factors immediately.
2. Completed Square Form: \(a(x + p)^2 + q\) reveals the exact maximum or minimum value and the coordinates of the turning point without needing calculus.

Don't worry if manipulating these expressions feels demanding at first. We will break every technique down into bite-sized, reliable steps that work every single time.

Note: Related quadratic topics—such as the discriminant and functions of roots \(\alpha\) and \(\beta\)—are explored in detail in companion chapters within this section.

Part 1: Factorising Quadratic Expressions

Factorising is the exact reverse of expanding brackets. You are turning a sum of terms into a product of factors.

Type A: The Difference of Two Squares

This is the quickest pattern to spot. Whenever you have two squared terms separated by a minus sign, it factorises using the identity:

\(A^2 - B^2 = (A - B)(A + B)\)

Example 1: Factorise \(4x^2 - 49\).
Notice that \(4x^2 = (2x)^2\) and \(49 = 7^2\).
Applying the identity: \(4x^2 - 49 = (2x - 7)(2x + 7)\).

Example 2: Factorise \(3x^2 - 75\).
Always look for a common factor first! Both terms divide by \(3\):
\(3(x^2 - 25) = 3(x - 5)(x + 5)\).

Type B: Monic Quadratics (where \(a = 1\))

A monic quadratic has the form \(x^2 + bx + c\). To factorise this into \((x + p)(x + q)\), we need two numbers \(p\) and \(q\) such that:

\(p \times q = c\) (they multiply to give the constant term)
\(p + q = b\) (they add to give the middle coefficient)

Memory Trick: "Multiply to the back, Add to the middle."

Example 3: Factorise \(x^2 - 7x + 12\).
Step 1: Look at the sign of the constant term (\(+12\)). A positive product means both numbers have the same sign. Since the middle term (\(-7\)) is negative, both numbers must be negative.
Step 2: List pairs of negative factors of \(12\):
\(-1\) and \(-12\) (sum = \(-13\))
\(-2\) and \(-6\) (sum = \(-8\))
\(-3\) and \(-4\) (sum = \(-7\)) — Found them!
Step 3: Write out the brackets: \((x - 3)(x - 4)\).

Type C: Non-Monic Quadratics (where \(a \neq 1\))

When the coefficient of \(x^2\) is not \(1\), guessing brackets can take too long. Instead, use the structured "ac" method (also called factorising by grouping).

Step-by-step method for \(ax^2 + bx + c\):
Step 1: Multiply \(a\) by \(c\) to find the target product \(ac\).
Step 2: Find two numbers that multiply to give \(ac\) and add together to give \(b\).
Step 3: Split the middle term \(bx\) into two terms using these two numbers.
Step 4: Factorise in pairs (group the first two terms and the last two terms).

Example 4: Factorise \(6x^2 + 7x - 5\).
Step 1: Calculate \(ac = 6 \times (-5) = -30\). We also need a sum of \(b = +7\).
Step 2: Look for factor pairs of \(-30\) that add to \(+7\):
\(10 \times (-3) = -30\) and \(10 + (-3) = +7\). The numbers are \(+10\) and \(-3\).
Step 3: Split the middle term \(+7x\) into \(+10x - 3x\):
\(6x^2 + 10x - 3x - 5\)
Step 4: Factorise the first two terms and the second two terms separately:
\(2x(3x + 5) - 1(3x + 5)\)
Notice that \((3x + 5)\) is common to both halves! Pull it out as a common factor:
\((3x + 5)(2x - 1)\).

Key Takeaway for Factorising: Always check for a common numerical or variable factor before doing anything else. If \(a = 1\), look for factors of \(c\) summing to \(b\). If \(a \neq 1\), use the \(ac\) grouping method.

Part 2: Completing the Square

Not every quadratic expression factorises nicely with whole numbers. Completing the square is a universal method that works on every single quadratic expression.

Did you know? The name comes from geometry. Ancient mathematicians literally drew a square of area \(x^2\) and two rectangles of combined area \(bx\), and then worked out the missing geometric corner needed to "complete the square"!

Case 1: Monic Quadratics (\(a = 1\))

When expanding \((x + d)^2\), you get \(x^2 + 2dx + d^2\). Notice that the middle term is \(2d\), which is twice \(d\).
Therefore, if you start with \(x^2 + bx\), you can write it as:

\(x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\)

Adding the constant term \(c\) gives the complete rule:

\(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\)

The Golden Rule for \(a = 1\):
1. Half the coefficient of \(x\) and place it inside the bracket: \(\left(x + \frac{b}{2}\right)^2\).
2. Always subtract the square of that number: \(-\left(\frac{b}{2}\right)^2\).
3. Add the original constant term \(c\).

Example 5: Express \(x^2 + 8x - 5\) in the form \((x + p)^2 + q\).
Step 1: Identify \(b = 8\). Half of \(8\) is \(4\).
Step 2: Write \((x + 4)^2 - 4^2 - 5\).
Step 3: Simplify the numbers: \((x + 4)^2 - 16 - 5 = (x + 4)^2 - 21\).
Here, \(p = 4\) and \(q = -21\).

Example 6 (Fractions): Express \(x^2 - 5x + 2\) in completed square form.
Step 1: Identify \(b = -5\). Half of \(-5\) is \(-\frac{5}{2}\).
Step 2: Place inside the bracket and subtract the square:
\(\left(x - \frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + 2\)
Step 3: Calculate the square: \(\left(-\frac{5}{2}\right)^2 = \frac{25}{4}\).
\(\left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{8}{4} = \left(x - \frac{5}{2}\right)^2 - \frac{17}{4}\).

Case 2: Non-Monic Quadratics (\(a \neq 1\))

When the coefficient of \(x^2\) is not \(1\), we must factor out \(a\) from the terms containing \(x\) first. Do not try to do it all in your head in one step—use square brackets to keep your work clean!

Step-by-step procedure:
Step 1: Factor out \(a\) from the \(x^2\) and \(x\) terms.
Step 2: Complete the square inside the bracket on the monic quadratic.
Step 3: Multiply \(a\) back through the square brackets.
Step 4: Combine the numerical constants.

Example 7: Express \(2x^2 + 12x + 11\) in the form \(a(x + p)^2 + q\).
Step 1: Factor out \(2\) from the first two terms:
\(2\left[x^2 + 6x\right] + 11\)
Step 2: Complete the square inside the square brackets (half of \(6\) is \(3\)):
\(2\left[(x + 3)^2 - 3^2\right] + 11\)
\(2\left[(x + 3)^2 - 9\right] + 11\)
Step 3: Expand the outer bracket by multiplying by \(2\):
\(2(x + 3)^2 - 18 + 11\)
Step 4: Combine the constants:
\(2(x + 3)^2 - 7\).
Here, \(a = 2\), \(p = 3\), and \(q = -7\).

Case 3: When \(a\) is Negative (Crucial Exam Skill)

Questions where \(a < 0\) appear frequently in Pearson Edexcel IGCSE Further Pure Mathematics. Take extra care with minus signs!

Example 8: Write \(7 - 4x - 2x^2\) in the form \(A - B(x + C)^2\).
Step 1: Rearrange into descending powers: \(-2x^2 - 4x + 7\).
Step 2: Factor out \(-2\) from the \(x\)-terms:
\(-2\left[x^2 + 2x\right] + 7\)
Notice that \(-4x\) divided by \(-2\) becomes \(+2x\).
Step 3: Complete the square inside the brackets:
\(-2\left[(x + 1)^2 - 1^2\right] + 7\)
\(-2\left[(x + 1)^2 - 1\right] + 7\)
Step 4: Multiply through by \(-2\):
\(-2(x + 1)^2 + 2 + 7\)
Be careful: \(-2 \times (-1) = +2\).
Step 5: Collect constants and write in the required order:
\(9 - 2(x + 1)^2\).
Here, \(A = 9\), \(B = 2\), and \(C = 1\).

Key Takeaway for Completing the Square: Always isolate \(a\) before completing the square on the bracketed terms. Keep a set of square brackets \([\dots]\) while working inside, and multiply through carefully at the end.

Part 3: Why Do We Complete the Square? (Applications)

1. Finding Turning Points, Maximums, and Minimums

Since any real number squared is always greater than or equal to zero, we know that:

\((x + p)^2 \geq 0\) for all real values of \(x\).

This simple fact allows us to read key features of a quadratic graph directly from the completed square form \(y = a(x + p)^2 + q\):

When \(a > 0\) (a positive, "cup-shaped" parabola \(\cup\)):
- The smallest possible value of \((x + p)^2\) is \(0\), which occurs when \(x = -p\).
- The minimum value of the function is \(q\).
- The coordinates of the minimum point (vertex) are \((-p, q)\).
- The equation of the line of symmetry is \(x = -p\).

When \(a < 0\) (a negative, "cap-shaped" parabola \(\cap\)):
- The largest possible value of the expression occurs when \((x + p)^2 = 0\), which occurs when \(x = -p\).
- The maximum value of the function is \(q\).
- The coordinates of the maximum point (vertex) are \((-p, q)\).

Example 9: For the curve \(y = 3(x - 4)^2 + 10\):
- The minimum value is \(10\).
- This occurs when \(x - 4 = 0 \implies x = 4\).
- The turning point is \((4, 10)\).

Example 10: State the maximum value of \(f(x) = 15 - 2(x + 3)^2\) and the value of \(x\) at which it occurs.
Since \((x + 3)^2 \geq 0\), the term \(-2(x + 3)^2\) is at most \(0\).
- The maximum value is \(15\).
- It occurs when \(x + 3 = 0 \implies x = -3\).

2. Solving Quadratic Equations Exactly

Completing the square gives an algebraic way to solve equations without needing to memorise the quadratic formula (in fact, this is how the quadratic formula itself is proved!).

Example 11: Solve \(x^2 + 6x - 2 = 0\), leaving your answer in exact surd form.
Step 1: Complete the square on the left-hand side:
\((x + 3)^2 - 3^2 - 2 = 0\)
\((x + 3)^2 - 11 = 0\)
Step 2: Rearrange to isolate the squared term:
\((x + 3)^2 = 11\)
Step 3: Take the square root of both sides (remembering the \(\pm\) sign!):
\(x + 3 = \pm\sqrt{11}\)
Step 4: Subtract \(3\) to find \(x\):
\(x = -3 \pm\sqrt{11}\).
The exact roots are \(x = -3 + \sqrt{11}\) and \(x = -3 - \sqrt{11}\).

Part 4: Common Mistakes to Avoid

Mistake 1: Adding instead of subtracting the squared term inside the completed square.
Incorrect: \(x^2 + 6x = (x + 3)^2 + 9\).
Correction: Expanding \((x + 3)^2\) produces \(x^2 + 6x + 9\). To get back to \(x^2 + 6x\), you must subtract \(9\):
\(x^2 + 6x = (x + 3)^2 - 9\). Always subtract \(\left(\frac{b}{2}\right)^2\).

Mistake 2: Sign error when dealing with negative coefficients.
When writing \(-x^2 + 6x\) as \(-\left[x^2 - 6x\right]\), students often forget to change the sign of the \(x\)-term inside the bracket.
Check: \(-1 \times (-6x) = +6x\). Always expand your bracket in your head to check signs!

Mistake 3: Forgetting to multiply the subtracted value by \(a\).
In \(3x^2 + 12x + 1 = 3\left[(x + 2)^2 - 4\right] + 1\), students often write \(3(x + 2)^2 - 4 + 1 = 3(x + 2)^2 - 3\).
Correction: The \(-4\) is inside the bracket! It must be multiplied by \(3\):
\(3 \times (-4) = -12\), giving \(3(x + 2)^2 - 12 + 1 = 3(x + 2)^2 - 11\).

Mistake 4: Writing the vertex coordinates with the wrong signs.
For \(y = (x + 5)^2 + 4\), the minimum point is \((-5, 4)\), not \((5, 4)\).
Remember: set the bracket equal to zero (\(x + 5 = 0 \implies x = -5\)).

Part 5: Quick Revision Checklist

Before moving on to the next topic, ensure you can confidently:
• Recognise and factorise the difference of two squares: \(A^2 - B^2 = (A - B)(A + B)\).
• Factorise non-monic quadratics \(ax^2 + bx + c\) systematically using the \(ac\) grouping method.
• Complete the square for \(x^2 + bx + c\) by writing \(\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\).
• Factor out \(a\) safely when completing the square for \(ax^2 + bx + c\), especially when \(a < 0\).
• State the maximum or minimum value and the turning point \((-p, q)\) directly from \(a(x + p)^2 + q\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the standard algebraic form of a quadratic expression and what are its components?

The standard form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants with \(a \neq 0\).

How do you factorise an expression using the difference of two squares identity?

Use the identity \(A^2 - B^2 = (A - B)(A + B)\).

How do you factorise a monic quadratic of the form \(x^2 + bx + c\)?

Find two numbers \(p\) and \(q\) such that \(p \times q = c\) and \(p + q = b\), giving \((x + p)(x + q)\).

What are the steps to factorise a non-monic quadratic \(ax^2 + bx + c\) using the "ac" method?

Find two numbers that multiply to give \(ac\) and add to give \(b\), split the middle term, and factorise by grouping.

What is the formula for completing the square on a monic quadratic \(x^2 + bx + c\)?

\(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\).

How do you complete the square on a non-monic quadratic \(ax^2 + bx + c\) where \(a \neq 1\)?

Factor out \(a\) from the \(x^2\) and \(x\) terms first: \(a\left[x^2 + \frac{b}{a}x\right] + c\), then complete the square inside the brackets.

What extra step is essential when completing the square if the \(x^2\) coefficient \(a\) is negative?

Factor out the negative coefficient from the \(x\)-terms first, taking care to reverse the sign of the middle \(x\)-term.

How do you find the vertex (turning point) and optimum value from \(y = a(x + p)^2 + q\)?

The turning point has coordinates \((-p, q)\); the maximum or minimum value is \(q\), occurring at \(x = -p\).

How do you solve a quadratic equation \(ax^2 + bx + c = 0\) by completing the square?

Rewrite the equation into the form \((x + p)^2 = k\), then take the square root of both sides to get \(x + p = \pm\sqrt{k}\).

What common mistake occurs when expanding the outer coefficient in \(a\left[(x + p)^2 - d\right] + c\)?

Forgetting to multiply the subtracted term \(-d\) by the leading coefficient \(a\).

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