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Perfect squares

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Image: Pexels/Rolanda de wet

Use this resource to explore perfect squares. Understanding perfect squares will help you with recognise patterns in algebraic expressions, making factorising and expanding algebraic expressions much easier.

A perfect square is an algebraic term that is the square of an integer. In other words, it is a term multiplied by itself. Some examples of perfect squares are:

  • \(5^{2}\)
  • \(x^{2}\)
  • \(a^{2}b^{2}\)
  • \((xy)^{2}\)
  • \((x+1)^{2}\)

Here, we will deal with perfect squares that have the general form:

\[(x\pm y)^{2}=x^{2}\pm2xy+y^{2}\]

It is useful to know this form because if a quadratic is a perfect square, it can be very neatly factorised.

Expanding perfect squares

There are two rules you should remember when it comes to perfect squares. These will make your life easier when it comes to expanding and factorising algebraic expressions.

The first is \((a+b)^{2}\).

\[\begin{align*} (a+b)^{2} & = (a+b)(a+b)\\
& = a^{2}+ab+ba+b^{2}\\
& = a^{2}+2ab+b^{2}
\end{align*}\]

The equation you should remember is:

\[(a+b)^{2}=a^{2}+2ab+b^{2}\]

The second rule is \((a-b)^{2}\).

\[\begin{align*} (a-b)^{2} & = (a-b)(a-b)\\
& = a^{2}-ab-ba+b^{2}\\
& = a^{2}-2ab+b^{2}
\end{align*}\]

The equation you should remember is:

\[(a-b)^{2}=a^{2}-2ab+b^{2}\]

In these two equations:

  • the first and last terms must be positive and must be perfect squares
  • the middle term must be twice the product of the first and last terms and may be positive or negative.

Using perfect squares to factorise

The rules for expanding perfect squares can be used in reverse to factorise algebraic expressions. Let's look at some examples.

Example 1 – using perfect squares to factorise quadratics

Factorise \(x^{2}+14x+49\).

Take the square root of \(x^{2}\) and \(49\). These are \(x\) and \(7\), respectively. We can confirm whether they are correct by checking that the coefficients \(a\) and \(b\) work with the equation.

\[2ab=2\times x\times7=14x\]

This matches with the \(2ab\) term in the quadratic so it is a perfect square. Since the sign in front of the \(2ab\) term is positive, it must be the form \((x+y)^{2}\). Therefore:

\[x^{2}+14x+49=(x+7)^{2}\]

Factorise \(y^{2}-20y+25\).

Take the square root of \(y^{2}\) and \(25\). These are \(y\) and \(5\), respectively. We can confirm whether they are correct by checking that the coefficients \(a\) and \(b\) work with the equation.

\[2ab=2\times y\times5=10y\]

This does not match with the \(2ab\) term in the quadratic, so \(y^{2}-20y+25\) is not a perfect square. We actually cannot factorise it further.

Factorise \(4a^{2}-12a-9\).

We can take the square root of \(4a^{2}\) but not the last term, since it is a negative. Therefore, \(4a^{2}-12a-9\) is not a perfect square. We also cannot factorise it further.

Factorise \(100x^{2}-180x+81\).

Take the square root of \(100x^{2}\) and \(81\). These are \(10x\) and \(9\), respectively. We can confirm whether they are correct by checking that the coefficients \(a\) and \(b\) work with the equation.

\[2ab=2\times10x\times9=180x\]

This matches with the \(2ab\) term in the quadratic so it is a perfect square. Since the sign in front of the \(2ab\) term is negative, it must be the form \((x-y)^{2}\). Therefore:
\[100x^{2}-180x+81=(10x-9)^{2}\]

Factorise \(50x^{2}+80x+32\).

At first glance, this expression does not appear to be a perfect square because the squares of \(50x^{2}\) and \(32\) are not integers. However, all of these terms have a common factor that we can use to factorise.

\[50x^{2}+80x+32=2(25x^{2}+40x+16)\]

We can take the square roots of \(25x^{2}\) and \(16\) to get the integers \(5x\) and \(4\). We can confirm whether they are correct by checking that the coefficients \(a\) and \(b\) work with the equation.

\[2ab=2\times5x\times4=40x\]

This matches with the \(2ab\) term in the quadratic so it is a perfect square. Since the sign in front of the \(2ab\) term is postiive, it must be the form \((x+y)^{2}\). Therefore:

\[50x^{2}+80x+32=2(5x+4)^{2}\]

Note that although the quadratic contains a perfect square after the factor of \(2\) is removed, the original quadratic is not a perfect square.

Your turn – using perfect squares to factorise quadratics

  1. Factorise the following into perfect squares.
    1. \(a^{2}+2a+1\)
    2. \(x^{2}-4x+4\)
    3. \(25x^{2}-10x+1\)
    4. \(4y^{2}-6y+9\)
    5. \(81x^{2}+108x+36\)
    6. \(9a^{2}-24a-16\)
    7. \(16x^{2}-40xy+25y^{2}\)
    8. \(121z^{2}+88z+64\)
    9. \(2x^{2}+8x+8\)

    1. \((a+1)^{2}\)
    2. \((x-2)^{2}\)
    3. \((5x-1)^{2}\)
    4. Not a perfect square
    5. \((9x+6)^{2}\)
    6. Not a perfect square
    7. \((4x-5y)^{2}\)
    8. Not a perfect square
    9. Not a perfect square, but \(2x^{2}+8x+8=2(x+2)^{2}\)

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