# Lesson Video: Factoring by Completing the Square Mathematics

In this video, we will learn how to factor an expression by completing the square.

17:27

### Video Transcript

In this video, we will learn how to factor an expression by completing the square. We will begin by recalling the form of a perfect square trinomial. A perfect square trinomial is a polynomial with three terms which can be represented in the form π squared plus or minus two ππ plus π squared. They can be factored as shown. π squared plus two ππ plus π squared is equal to π plus π all squared. And π squared minus two ππ plus π squared is equal to π minus π all squared. In these trinomials, π and π may be variables, constants, or products of variables and constants.

Letβs consider the trinomial π₯ to the fourth power plus four π₯ squared plus four. In order to factor this, we begin by letting the first and last terms be π squared and π squared, respectively. If π squared is equal to π₯ to the fourth power, then π is equal to the square root of this, which is equal to π₯ squared. Likewise, since π squared is equal to four, π is equal to two. And itβs worth noting at this point that π equals negative π₯ squared and π equals negative two would also have been solutions to π squared equals π₯ to the fourth power and π squared equals four. However, since we want π plus π all squared to be equal to π squared plus two ππ plus π squared, we canβt mix the positive and negative root solutions. And in fact, we only use the positive root solutions.

The middle term of our perfect square trinomial is equal to two ππ. Substituting our expressions for π and π, we have two multiplied by π₯ squared multiplied by two. This is equal to the middle term of the initial expression four π₯ squared. The perfect square trinomial π₯ to the fourth power plus four π₯ squared plus four in its factored form is π₯ squared plus two all squared. It is worth noting from our original definition that the expression π₯ to the fourth power minus four π₯ squared plus four can be factored in the same way. This is equal to π₯ squared minus two all squared.

In this video, we will use this to help factor binomials that cannot be factored using other methods. For example, letβs consider the expression π₯ to the fourth power plus four. In this case, our usual methods of factoring do not work. However, we can factor this expression using a method known as completing the square. From the previous trinomial we factored, we know that when the middle term was four π₯ squared, we were able to factor this as a perfect square trinomial. As such, if we artificially introduce a term of four π₯ squared into this polynomial, we would create a perfect square trinomial. It is this method that helps us factor many polynomials of the form π squared plus π squared.

If we have two terms that are both perfect squares, that is, in the form π squared plus π squared, we can create a perfect square trinomial by completing the square as follows. Firstly, we determine the values of π and π. In our example, we already know that π is equal to π₯ squared and π is equal to two. Secondly, we calculate two ππ. Once again, we already know that in our example, this is equal to four π₯ squared. Our next step is to add two ππ and negative two ππ to the expression. In our example, we have π₯ to the fourth power plus four plus four π₯ squared minus four π₯ squared. Since weβre adding the positive and negative of the same term, weβre effectively adding zero, which does not change the polynomial. We can then rewrite the expression as π₯ to the fourth power plus four π₯ squared plus four minus four π₯ squared.

The fourth and final step of completing the square is to factor the perfect square trinomial. π₯ to the fourth power plus four π₯ squared plus four factors to π₯ squared plus two all squared. And the entire expression is as shown. In this example, we can actually go one stage further by noting that four π₯ squared is also a perfect square. It is equal to two π₯ all squared. Rewriting our expression as π₯ squared plus two all squared minus two π₯ all squared, we note it is in the form π squared minus π squared. This is known as the difference of squares. And we know that any expression of this type can be factored as π plus π multiplied by π minus π, where the values of π and π in this example are π₯ squared plus two and two π₯, respectively. Our expression can therefore be factored to π₯ squared plus two plus two π₯ multiplied by π₯ squared plus two minus two π₯. Finally, rewriting the order within the parentheses, we have π₯ to the fourth power plus four is equal to π₯ squared plus two π₯ plus two multiplied by π₯ squared minus two π₯ plus two. It is worth noting that we can check this answer by expanding or distributing the parentheses.

We will now consider a more complicated example of this type.

Factor 625π₯ to the fourth power plus 64π¦ to the fourth power fully by completing the square.

In this question, we are asked to factor the expression by completing the square. As such, we need to manipulate the given binomial to include a perfect square trinomial in the form π squared plus two ππ plus π squared, as this factors to π plus π all squared. We begin by letting the two terms in our expression be π squared and π squared, respectively. This means that π is equal to the square root of 625π₯ to the fourth power. And since the square root of 625 is 25, π is equal to 25π₯ squared. In the same way, π is equal to the square root of 64π¦ to the fourth power. And this is equal to eight π¦ squared.

Our next step is to find an expression for two ππ. This is equal to two multiplied by 25π₯ squared multiplied by eight π¦ squared. This is equal to 400π₯ squared π¦ squared. It is this term we need to add to our expression to create a perfect square trinomial. Since we are adding 400π₯ squared π¦ squared, we also need to subtract this from the initial expression in order for the expression to remain the same. We are now in a position where the first three terms form a perfect square trinomial. And this factors to 25π₯ squared plus eight π¦ squared all squared. And the entire expression can be written as shown.

Next, we notice that 400π₯ squared π¦ squared is a perfect square. It can be written as 20π₯π¦ all squared. Our expression simplifies to 25π₯ squared plus eight π¦ squared all squared minus 20π₯π¦ all squared. This is written in the form π squared minus π squared, which is known as the difference of squares. And we know that this can be factored to π plus π multiplied by π minus π. Our expression can therefore be written as 25π₯ squared plus eight π¦ squared plus 20π₯π¦ multiplied by 25π₯ squared plus eight π¦ squared minus 20π₯π¦. Rearranging the order of the terms, we have that 625π₯ to the fourth power plus 64π¦ to the fourth power is equal to 25π₯ squared plus 20π₯π¦ plus eight π¦ squared multiplied by 25π₯ squared minus 20π₯π¦ plus eight π¦ squared. We could check this answer by distributing the parentheses, where all terms apart from the first and last will cancel.

We will now look at one more example where other factoring methods are required first. But in the resulting expression, one of the factors can be factored using completing the square.

Factor π₯ to the eighth power minus 16π¦ to the eighth power by completing the square.

In this question, it may not immediately be obvious how we can solve the problem by completing the square. However, we do notice that our expression is of the form π squared minus π squared. This is known as the difference of squares and can be factored as π plus π multiplied by π minus π. Letting π squared equal π₯ to the eighth power, we know that π is equal to π₯ to the fourth power. And letting π squared equals 16π¦ to the eighth power, we know that π is the square root of this, which is equal to four π¦ to the fourth power. The original expression can therefore be rewritten as π₯ to the fourth power all squared minus four π¦ to the fourth power all squared. And factoring this using the difference of squares, we have π₯ to the fourth power plus four π¦ to the fourth power multiplied by π₯ to the fourth power minus four π¦ to the fourth power.

The second part of our expression is once again written in the form π squared minus π squared. π₯ to the fourth power minus four π¦ to the fourth power is equal to π₯ squared plus two π¦ squared multiplied by π₯ squared minus two π¦ squared. These two parentheses cannot be factored any further. So we now need to consider the expression π₯ to the fourth power plus four π¦ to the fourth power. It is this expression that weβll be able to factor by completing the square. This is in the form π squared plus π squared. And we need to manipulate this so it is in the form of a perfect square trinomial.

We know that any perfect square trinomial of the form π squared plus two ππ plus π squared can be factored into the form π plus π all squared. Since π squared is equal to π₯ to the fourth power, π is equal to π₯ squared. Likewise, since π squared is equal to four π¦ to the fourth power, π is equal to two π¦ squared. The term two ππ is therefore equal to two multiplied by π₯ squared multiplied by two π¦ squared. And this is equal to four π₯ squared π¦ squared.

We need to add the positive and negative of this to the expression in our first set of parentheses to create our perfect square trinomial. The full expression can be written as shown. And we can now factor the perfect square trinomial contained in the first set of parentheses. This is equal to π₯ squared plus two π¦ squared all squared. And rewriting four π₯ squared π¦ squared as two π₯π¦ all squared, we have π₯ squared plus two π¦ squared all squared minus two π₯π¦ all squared multiplied by π₯ squared plus two π¦ squared multiplied by π₯ squared minus two π¦ squared.

We note that the expression within the brackets is the difference of squares. And using the rule we saw earlier, this factors to π₯ squared plus two π¦ squared plus two π₯π¦ multiplied by π₯ squared plus two π¦ squared minus two π₯π¦. We can then reorder the terms in these parentheses as shown, giving us a final answer of π₯ squared plus two π₯π¦ plus two π¦ squared multiplied by π₯ squared minus two π₯π¦ plus two π¦ squared multiplied by π₯ squared plus two π¦ squared multiplied by π₯ squared minus two π¦ squared. This is the fully factored form of π₯ to the eighth power minus 16π¦ to the eighth power.

We will now finish this video by recapping the key points. We saw in this video that a perfect square trinomial is of the form π squared plus or minus two ππ plus π squared and can be factored as π plus or minus π all squared. If weβre asked to factor a polynomial that we cannot factor by other means and the polynomial contains two terms of the form π squared plus π squared, we can complete the square to form a perfect square trinomial. The term that must be added to create this is positive or negative two ππ. It is important to note that both two ππ and negative two ππ must be included so that we are effectively adding zero. Whilst we did not see an example of this type in this video, if possible, we need to factor out the greatest common factor first, as this reduces the complexity of the polynomial that needs to be factored.

Often after completing the square, factoring the resulting expression or parts of the expression using the difference of squares will be possible. As with our second example, it may also be the case that we need to do this prior to completing the square. In all cases, once the polynomial has been factored in one way, it is important to check to see if the resulting expressions can be factored further.

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