# Question Video: Factorizing the Difference of Two Squares Involving Perfect Squares

Factorise fully 1 − 𝑥² + 14𝑥𝑦 − 49𝑦².

04:09

### Video Transcript

Factorise fully one minus 𝑥 squared plus 14𝑥𝑦 minus 49𝑦 squared.

Our first step is to split the expression into two parts. If we factorise out negative one from the second, third, and fourth term, we are left with one minus 𝑥 squared minus 14𝑥𝑦 plus 49𝑦 squared. The second part of our expression can be factorised into two brackets, or parentheses, 𝑥 minus seven 𝑦 and 𝑥 minus seven 𝑦. We can check this by expanding the two parentheses using the F.O.I.L. method.

Multiplying the first terms gives us 𝑥 squared, as 𝑥 multiplied by 𝑥 is equal to 𝑥 squared. Multiplying the outside terms gives us negative seven 𝑥𝑦. Multiplying the inside terms also gives us negative seven 𝑥𝑦. Finally, multiplying the last terms gives us 49𝑦 squared. Multiplying negative seven 𝑦 by negative seven 𝑦 gives 49𝑦 squared. This simplifies to 𝑥 squared minus 14𝑥𝑦 plus 49𝑦 squared. Therefore, our factorisation is correct.

As the two parentheses are identical, this can be rewritten as one minus 𝑥 minus seven 𝑦 all squared. We now have an expression which is the difference of two squares, as one is a square number and 𝑥 minus seven 𝑦 all squared is also a square number. The difference of any two squares, 𝑎 squared minus 𝑏 squared, can be factorised to give 𝑎 minus 𝑏 multiplied by 𝑎 plus 𝑏.

In our question, 𝑎 is equal to one and 𝑏 is equal to 𝑥 minus seven 𝑦. Our two parentheses, therefore, become one minus 𝑥 minus seven 𝑦 and one plus 𝑥 minus seven 𝑦. The first parenthesis simplifies to one minus 𝑥 plus seven 𝑦, as subtracting negative seven 𝑦 gives positive seven 𝑦. The second parenthesis simplifies to one plus 𝑥 minus seven 𝑦. This means that our fully factorised expression is one minus 𝑥 plus seven 𝑦 multiplied by one plus 𝑥 minus seven 𝑦.

We can check this answer by expanding the two parentheses using the grid method. We have to ensure that we multiply every term in the first parenthesis by every term in the second parenthesis. One multiplied by one is equal to one. One multiplied by 𝑥 is equal to 𝑥. And one multiplied by negative seven 𝑦 is equal to negative seven 𝑦. Multiplying one, 𝑥, and negative seven 𝑦 by negative 𝑥 gives us negative 𝑥, negative 𝑥 squared, and positive seven 𝑥𝑦, respectively. Finally, multiplying one, 𝑥, and negative seven 𝑦 by seven 𝑦 gives us seven 𝑦, seven 𝑥𝑦, and negative 49𝑦 squared.

The two 𝑥 terms and the two seven 𝑦 terms cancel. We are, therefore, left with one, negative 𝑥 squared, 14𝑥𝑦, and finally negative 49𝑦 squared. As this is equivalent to our initial expression, our factorisation is correct.