Factorising By Completing The Square
Every question with a fully worked solution.
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Theory
Completing the square rewrites \(x^2 + bx + c\) as \(\left(x + \dfrac{b}{2}\right)^2\) minus or plus a number. When the result is a square minus a positive number, the difference of two squares factorises it, with surds if needed. This Year 10 maths page covers odd and fractional coefficients, a leading coefficient other than \(1\), a negative \(x^2\) term, smallest values, and how to show a quadratic has no real factors.
A perfect square \((x + h)^2 = x^2 + 2hx + h^2\) has a constant equal to the square of half the coefficient of \(x\). So \(x^2 + bx\) needs \(\left(\dfrac{b}{2}\right)^2\) added to become a perfect square: that is completing the square.
To keep the expression equal, add and subtract the same number: \(x^2 + bx + c\) \(= \left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c\).
If the leftover number is negative, as in \((x - 1)^2 - 6\), write \(6 = (\sqrt{6})^2\) and use the difference of two squares. If it is positive, as in \((x - 1)^2 + 6\), the expression is always positive and has no real linear factors.
The form \(a(x + h)^2 + k\) is called the completed-square form. It also shows the smallest (or largest) value of the quadratic, \(k\), when \(x = -h\).
How to factorise by completing the square
- Prepare: if the \(x^2\) coefficient is not \(1\), take it out of the \(x\) terms (or of everything). If it is negative, take out \(-1\).
- Halve the coefficient of \(x\), square it, and add and subtract that number.
- Write the perfect square and collect the constants: \((x + h)^2 - d^2\).
- Factorise as a difference of two squares, write any surd in simplest form, and simplify each bracket.
- Check by expanding, or by substituting a value of \(x\).
Using the completed square
- Finding unknowns: if the factors are \((x + h + d)(x + h - d)\), they differ by \(2d\). A given difference of factors, or a given perfect square, fixes \(d\) and hence the constant.
- Areas: factorising an area expression gives possible length and width; their difference is \(2d\).
- No real factors: if the completed square is \((x + h)^2 + k\) with \(k > 0\), the expression is always positive, so it has no real linear factors (Example 4).
- Two letters: in \(x^2 - y^2 + 6y - 9\), the last three terms are \(-(y - 3)^2\), so the expression is \(x^2 - (y - 3)^2 = (x - y + 3)(x + y - 3)\).
(i) Fill the gaps: \(x^2 + 10x + \underline{\quad} = (x + \underline{\quad})^2\) and \(x^2 - 5x + \underline{\quad} = (x - \underline{\quad})^2\).
(ii) By completing the square, factorise \(x^2 + 10x + 21\).
(i) Halve the coefficient of \(x\) and square it.
| \(x^2 + 10x + 25\) | \(=\) | \(\boldsymbol{(x + 5)^2}\) |
| \(x^2 - 5x + \dfrac{25}{4}\) | \(=\) | \(\boldsymbol{\left(x - \dfrac{5}{2}\right)^2}\) |
(ii) Add and subtract \(25\), then use the difference of two squares.
\(x^2 + 10x + 21\)
| \(\) | \(=\) | \(x^2 + 10x + 25 - 25 + 21\) |
| \(\) | \(=\) | \((x + 5)^2 - 4\) |
| \(\) | \(=\) | \((x + 5)^2 - 2^2\) |
| \(\) | \(=\) | \((x + 5 + 2)(x + 5 - 2)\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 7)(x + 3)}\) |
This trinomial could also be factorised with product and sum. Completing the square works even when that fails, as Example 2 shows.
(i) \(x^2 + 10x + 18\)
(ii) \(x^2 + 3x - 1\)
(i) No integers multiply to \(18\) and add to \(10\), so complete the square.
\(x^2 + 10x + 18\)
| \(\) | \(=\) | \((x + 5)^2 - 25 + 18\) |
| \(\) | \(=\) | \((x + 5)^2 - 7\) |
| \(\) | \(=\) | \((x + 5)^2 - (\sqrt{7})^2\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 5 - \sqrt{7})(x + 5 + \sqrt{7})}\) |
(ii) Half of \(3\) is \(\dfrac{3}{2}\), and \(\left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}\).
\(x^2 + 3x - 1\)
| \(\) | \(=\) | \(\left(x + \dfrac{3}{2}\right)^2 - \dfrac{9}{4} - \dfrac{4}{4}\) |
| \(\) | \(=\) | \(\left(x + \dfrac{3}{2}\right)^2 - \left(\dfrac{\sqrt{13}}{2}\right)^2\) |
| \(\) | \(=\) | \(\boldsymbol{\left(x + \dfrac{3 - \sqrt{13}}{2}\right)\left(x + \dfrac{3 + \sqrt{13}}{2}\right)}\) |
\(\dfrac{13}{4} = \left(\dfrac{\sqrt{13}}{2}\right)^2\), because \(\sqrt{\dfrac{13}{4}} = \dfrac{\sqrt{13}}{\sqrt{4}}\).
(i) Write \(2x^2 + 12x + 7\) in the form \(a(x + h)^2 + k\), then factorise it fully with surds.
(ii) Write \(8 - 2x - x^2\) in the form \(c^2 - (x + h)^2\) and factorise it.
(i) Take out \(2\) from the \(x\) terms only, and complete the square inside.
| \(2x^2 + 12x + 7\) | \(=\) | \(2(x^2 + 6x) + 7\) |
| \(\) | \(=\) | \(2[(x + 3)^2 - 9] + 7\) |
| \(\) | \(=\) | \(\boldsymbol{2(x + 3)^2 - 11}\) |
To factorise, take out \(2\) from both terms: \(\sqrt{\dfrac{11}{2}} = \dfrac{\sqrt{11}}{\sqrt{2}} = \dfrac{\sqrt{22}}{2}\).
| \(\) | \(=\) | \(2\left[(x + 3)^2 - \dfrac{11}{2}\right]\) |
| \(\) | \(=\) | \(\boldsymbol{2\left(x + 3 - \dfrac{\sqrt{22}}{2}\right)\left(x + 3 + \dfrac{\sqrt{22}}{2}\right)}\) |
(ii) Take out \(-1\) from the \(x\) terms.
| \(8 - 2x - x^2\) | \(=\) | \(8 - (x^2 + 2x)\) |
| \(\) | \(=\) | \(8 - [(x + 1)^2 - 1]\) |
| \(\) | \(=\) | \(\boldsymbol{9 - (x + 1)^2}\) |
| \(\) | \(=\) | \((3 - x - 1)(3 + x + 1)\) |
| \(\) | \(=\) | \(\boldsymbol{(2 - x)(4 + x)}\) |
(ii) Find the smallest value of \(x^2 + 6x + 14\), and the value of \(x\) that gives it.
(i) Complete the square.
| \(x^2 - 4x + 7\) | \(=\) | \((x - 2)^2 - 4 + 7\) |
| \(\) | \(=\) | \((x - 2)^2 + 3\) |
A square is never negative, so \((x - 2)^2 + 3 \ge 3\) for every \(x\). The expression is a square plus a number, not a difference of two squares, and it is never \(0\). A product of linear factors \((x - r)(x - s)\) would be \(0\) at \(x = r\), so no real linear factors exist.
(ii) Complete the square: half of \(6\) is \(3\), and \(3^2 = 9\).
| \(x^2 + 6x + 14\) | \(=\) | \((x + 3)^2 - 9 + 14\) |
| \(\) | \(=\) | \((x + 3)^2 + 5\) |
The square is smallest, \(0\), when \(x + 3 = 0\). So the smallest value is \(\boldsymbol{5}\), when \(\boldsymbol{x = -3}\).
Check: \((-3)^2 + 6(-3) + 14 = 9 - 18 + 14 = 5\). ✓
Common pitfalls
Frequently asked questions
How do you complete the square?
Take half of the coefficient of \(x\) and square it. Add and subtract that number, then write the first three terms as a squared bracket: \(x^2 + 8x + 3\) \(= (x + 4)^2 - 16 + 3\) \(= (x + 4)^2 - 13\).
Why do you add and subtract the same number?
Adding a number makes a perfect square, and subtracting the same number keeps the expression equal to the original. Together they add zero.
How do you factorise a quadratic with surds?
Complete the square to get \((x + h)^2 - d\). Write \(d\) as \((\sqrt{d})^2\), then use \(a^2 - b^2 = (a - b)(a + b)\).
How do you complete the square when the coefficient of x squared is not 1?
Take the coefficient out of the x squared and x terms, complete the square inside the bracket, then multiply the subtracted square by that coefficient when you remove the bracket.
How can you tell a quadratic has no real factors?
Complete the square. If you get a bracket squared plus a positive number, the expression is always positive, so it is never zero and has no real linear factors.