Difference or Sum of Squares with Complex Numbers

Sum and Difference of Squares

The difference of squares is a method for factoring certain kinds of binomials. Let's do a quick review of this method before we dive a little bit deeper.

Any binomial with the form $(a^2-b^2)$ can be factored into $(a+b)(a-b)$. Here's an example: $$(x^2-4)=(x+2)(x-2)$$ Try one for yourself to check your understanding.

$(9x^2-25)$ can be factored into $(3x+5)(3x-5)$.

When you originally learned this concept you were probably told that this only works when you are taking a difference of two squares. Now that you have an understanding of complex numbers and their conjugates, we can actually use a new technique.

When you incorporate complex numbers, you can factor the sum of two squares. Here is the generalized form: $$(a^2+b^2)=(a+bi)(a-bi)$$ This may look familiar to you as the two factors are conjugates. They are normally used to remove the imaginary component from complex numbers. Let's look at an example. $$(x^2+9)=(x+3i)(x-3i)$$ This uses the exact same process as the difference of squares, but we are also using the imaginary $i$.

Try one for yourself:

The binomial $25x^2+36$ can be factored into $(5x+6i)(5x-6i)$.

Wrap Up

The big takeaway here is that you can factor out the sum of squares using the exact same process as factoring out the difference of squares. $$\begin{align}(a^2-b^2)&=(a+b)(a-b)\\ (a^2+b^2)&=(a+bi)(a-bi)\end{align}$$

So as to not get a headache and deal with complex numbers, we shall let $c\geq 0$ for the equation $x^{2}-c =0$. Now we define a new variable $d = \sqrt{c}$. Because $c\geq 0$, $\sqrt{c}$ is a real non-negative number and by definition $d\geq 0$. Replacing $c$ with $d$, we now have: $$x^{2}-c=x^{2}-d^{2}$$ Looking at the latter expression, we have a quadratic expression which can be easily factored, termed difference of squares. The factored form is going to be $(x+d)(x-d)$. Expanding: $$(x+d)(x-d)$$ $$= x^{2}-xd+xd-d^{2}$$ $$= x^{2} - d^{2}$$

Replacing $d$ with $\sqrt{c}$, we have $(x+\sqrt{c})(x-\sqrt{c})$. So next time, if you have an expression of the form: $x^{2}-c =0, c\geq 0$, the factored form is $(x+\sqrt{c})(x-\sqrt{c})$.

Note, this factoring method is also valid if only an expression, but recognize that $x^{2}+1 =4$ can be simplified into $x^{2}-3 =0$ and thereby factored into $(x+\sqrt{3})(x-\sqrt{3})=0$. But if only presented with $x^{2}-7$, the factoring method still applies, resulting in $(x+\sqrt{7})(x-\sqrt{7})$. Additionally, considering the equation $x^{2}-c =0$, the factored form tells us that $\pm\sqrt{c}$ are our roots, values for $x$ that satisfy the above equation.

If $c<0$, then $\sqrt{c}$ would not be real, which is not preferred. But we will address this question later here.

We can use the difference of two squares method of factoring to multiply numbers that have an even difference.

This works because of the following algebraic equation:

$x^2-y^2 = (x-y) (x+y)$

The difference of two squares method outlines that, if we have two numbers (let's say $a$ and $b$) whose difference is even, we can find their product by a) finding the midpoint between them, and b) plugging in that midpoint and the distance of one of the numbers from that midpoint into our difference of two squares formula: $x^2-y^2$.

If that was hard to follow, take a look at the following example, and then read that previous paragraph again. It might take a few tries until you really get the gist of it. You might also want to try some practice problems.

Let's say we want to multiply the numbers 14 and 6. We first want to check that they have an even difference (this is the only way we can verify that our difference of squares method of multiplication will work cleanly). When we subtract 6 from 14, we get 8. Is 8 an even number? Yes!

Now that we know our two numbers have an even difference, let's first find the midpoint between them. We do this by adding the numbers and then dividing that sum by two (you can think of finding the midpoint between two numbers as taking their average):

$$\frac{14 + 6}{2} = 10$$

Now that we have our midpoint, we want to find the distance between that midpoint and either of our numbers. We'll be using 6 to calculate this value for the purposes of this example, but just be aware that we could also use 14:

$$6 - 10 = -4$$

So, now that we have a midpoint of 10 and a distance value of -4, we can plug those numbers into our difference of two squares formula: we'll let $x= 10$ and we'll let $y= -4$ (note that we always want to plug our midpoint into the first term of our difference of two squares formula and that we always want to plug our distance value into the formula's second term).

When we plug our specific values into our difference of two squares formula, we get the following result:

$\\x^2-y^2 = 10^2-(-4)^2 = 100 - 16 = 84$

If we check our work by using the normal method of multiplication, we should find that our two results match.

Also, notice that if we plug our values for $x$ and $y$ into the right-hand side of our difference of two squares equation, we get:

$\\(x+y) (x-y) = (10+(-4)) (10-(-4)) = (6)*(14)$

Recall that 6 and 14 are the exact two numbers that we set out to multiply in the first place!

(Video) Difference of Squares Multiplication

by Mr. T's Math Videos

This video by YouTube channel Mr T's Math Videos shows how to multiply two numbers using the difference of squares method and explains why this method works.

(Video) Multiplying Surds

by HEGARTYMATHS

Here's a good video by HEGARTYMATHS showing you how to quickly multiply two binomials together if they are of the form $(a+b)(a-b)$. In order to do this, you just need to pick one of those two binomials, square its second term and then subtract the result from the square of its first term. In other words, $(a+b)(a-b)=a^2-b^2$. This is called the difference of two squares equation.

(Video) Factoring Differences of Squares

by yaymath

Here's a video by yaymath that shows you how to factor a difference of squares (a polynomial that is in the form of $A^2-B^2$) by using the following method:

1. If able, factor out any greatest common factors.
2. Write down two sets of parentheses with one containing a plus sign and the other containing a minus sign.
3. Take the square root of the polynomial's first term and write the result inside both sets of parentheses, to the left of each sign.
4. Take the square root of the polynomial's second term and write the result inside both sets of parentheses, to the right of each sign.

Our instructor then shows us how to solve an equation that contains a difference of squares.

Note: if the polynomial you're working with isn't a binomial or contains addition, the method above cannot be used.

(Video) The Difference of Squares

by mathman1024

This by mathman1024 that shows how to employ the difference of squares method in factoring binomials.

$\large{\text{Summary}}$

This method makes use of the difference of squares equation, which states the following:

$$a^2-b^2=(a+b)(a-b)$$

The first binomial that our instructor factors is:

$$x^2-25$$

However, you might notice that we can rewrite this binomial like so:

$$(x)^2-(5)^2$$

As our instructor points out, we can let $x$ be our $a$ in our difference of squares equation, and we can let $5$ be our $b$ in our difference of squares equation. Then, we just have to substitute those terms into the right-hand side of our equation:

$$(x+5)(x-5)$$

He then goes over some additional problems, and some of them are more complex than others. See how many you can figure out on your own, and then watch the video above to see if you were correct.

1. $x^2-49$
2. $4x^2-9$
3. $x^2-\frac{121}{64}$
4. $x^{10}-16$
5. $49x^4-81y^{12}$
6. $100x^7-25x^3y^4$ (Hint: do the terms in this binomial have a GCF?)
7. $x^4-81$ (Hint: can either of this binomial's factors be factored again?)