produto da soma pela diferença

Difference of Squares with Geometry

Can we understand this with geometry?

Imagine two squares, an blue square with length of $a$ and an orange square with side length of $b$. $a$ is shorter than $b$.

We know the formula for the area of a square, where we square the side length, i.e., $x^2$. Since we know that the area of the blue square is $a^2$, and we know that the area of the orange square is $b^2$, we also know that:

The difference of area between the two squares is $a^2-b^2$.

Alternatively, we can also calculate the difference of area by adding two small rectangles ($1$ and $2$) together.

$$\begin{align} &[(a-b)\times b] + [(a-b)\times a] \\ =&(a-b)\times(a+b) \\ =&(a-b)(a+b) \end{align}$$

We can rewrite any difference of two squares to be the product of two binomials.

We can also use this formula in reverse if we're given a product of binomials and want to expand.

(Video) Multiplying Binomial Conjugates

by mathman1024

This video by mathman1024 covers multiplication with binomial conjugates.

$\large{\text{Summary}}$

A conjugate is when you change the sign in the middle of the two terms. So if you have $3x+1$ then its conjugate would be $3x-1$.

When we multiply two conjugates together the middle terms end up canceling each other out. $$\begin{gather} (a+b)(a-b)\\ a^2-ab+ab+b^2\\ a^2-b^2 \end{gather}$$ So we're left with the difference of two squares. This becomes a new formula for us to quickly multiply binomials. $$(a+b)(a-b)=a^2-b^2$$

He looks at the example: $(x+5)(x-5)$. $$\begin{gather} (x+5)(x-5)\\ (x)^2-(5)^2\\ x^2-25 \end{gather}$$

Try these next ones yourself and then see if you get the same answers.

1. $(x+7)(x-7)$

2. $(x-11)(x+11)$

3. $(5x+8)(5x-8)$

(Video) Difference of Squares

by mathman1024

Here's another video with some more complicated examples. You still use the formula exactly the same. Try doing these on your own first.

1. $(9x^{15}+2x^{11})(9x^{15}-2x^{11})$

2. $(\frac1{10}x^{13}y^4-\frac47)(\frac1{10}x^{13}y^4+\frac47)$

3. $(8a^5b-7x^2y^3)(8a^5b+7x^2y^3)$

Remember the rules for multiplying exponents and monomials!

Here is one other problem that uses this expansion pattern in the denominator.

Example: Find $\frac{1}{a + 4} + \frac{1}{a - 4}$ if $a = \frac{3}{5}.$

Solution: Let's add the fractions together and simplify before plugging in $a:$

$$\frac{1}{a + 4} + \frac{1}{a - 4} = \frac{a - 4}{(a + 4)(a - 4)} + \frac{a + 4}{(a + 4)(a - 4)} = \frac{2a}{a^2 - 16}.$$

Now plug in $a = \frac{3}{5}:$

$$\frac{2\left(\frac{3}{5}\right)}{\left(\frac{3}{5}\right)^2 - 16} = \frac{\frac{6}{5}}{\frac{9}{25} - 16} = \boxed{-\frac{30}{391}}.$$

Multiplying Matching Binomials

What do we do when we want to multiply a pair of "matching" binomials (i.e. two binomials that happen to be the sum and difference of the same two terms), like $2x-5$ and $2x+5$?

We could certainly use FOIL but it turns out that there's an even easier approach available to us when our binomials have "matching" terms.

Let's break it down.

Use FOIL to multiply $\left(2x-5\right)$ and $\left(2x+5\right)$. What is the result?

The result is $4x^2 - 25$.

See how the expression we got from FOILing together our two binomials contained a couple of middle terms that simply canceled out?

Well, this observation sets us up to learn a new shortcut for multiplying matching binomials.

When we've got matching binomials, we can simply perform the following two steps:

1. Square the binomials' first term

2. Square the binomials' second term and subtract the result from the previous step's result

Note that we can only use this shortcut because, since our two binomials have the same terms, and since one binomial uses addition while the other uses subtraction, we know that any middle terms that result from multiplying them together will cancel out.

Let's try this shortcut out on our two matching binomials from before: $\left(2x-5\right)$ and $\left(2x+5\right)$.

1. Square the binomials' first term

2. Square the binomials' second term and subtract the result from the previous step's result

Finally, we subtract this result from the previous step's result. This gives us:

$4x^2-25$

Notice that this is exactly the same answer we got when we used FOIL.

Now, let's practice this shortcut on a few more examples.

Yup! We've taken our binomials' second term, squared it, and subtracted the result from our binomials' first term (which we've also squared):

Again, we've taken the two-step process that we've learned above and applied it to our two binomials:

One last time, let's see our shortcut in action: