# Quadratic concepts -- multiplying binomials  (Page 4/1)

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This module teaches about multiplying binomials. Specifically about common patterns that can be memorized and using the "FOIL" method.

The following three formulae should be memorized.

${\left(x+a\right)}^{2}={x}^{2}+2\text{ax}+{a}^{2}$
${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$
$\left(x+a\right)\left(x-a\right)={x}^{2}-{a}^{2}$

It is important to have these three formulae on the top of your head. It is also nice to be able to show why these formulae work, for instance by using FOIL. But the most important thing of all is knowing what these three formulae mean, and how to use them .

These three are all “algebraic generalizations,” as discussed in the first unit on functions. That is, they are equations that hold true for any values of $x$ and $a$ . It may help if you think of the second equation above as standing for:

${\left(\text{Anthing}-\text{Anything Else}\right)}^{2}={\text{Anything}}^{2}-2{\left(\text{Anything Else}\right)}^{2}$ For instance, suppose the Anything (or $x$ ) is 5, and the Anything Else (or $a$ ) is 3.

${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$ , when $x=5$ , $a=3$ .

• $5-{3}^{2}\stackrel{?}{=}{5}^{2}-2\left(3\right)\left(5\right)+{3}^{2}$
• ${2}^{2}\stackrel{?}{=}\text{25}-\text{30}+9$
• $4=4$

It worked! Now, let’s leave the Anything as $x$ , but play with different values of $a$ .

More examples of ${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$

$\begin{array}{cccc}a=1:& \left(x-1{\right)}^{2}& =& {x}^{2}-2x+1\\ a=2:& \left(x-2{\right)}^{2}& =& {x}^{2}-4x+4\\ a=3:& \left(x-3{\right)}^{2}& =& {x}^{2}-6x+9\\ a=5:& \left(x-5{\right)}^{2}& =& {x}^{2}-10x+25\\ a=10:& \left(x-10{\right)}^{2}& =& {x}^{2}-20x+100\end{array}$

Once you’ve seen a few of these, the pattern becomes evident: the number doubles to create the middle term (the coefficient of $x$ ), and squares to create the final term (the number).

${\left(2y-6\right)}^{2}$

There are three ways you can approach this.

${\left(2y-6\right)}^{2}$ , computed three different ways
Square each term FOIL Using the formula above
$\begin{array}{cc}& {\left(2y-6\right)}^{2}\\ & {\left(2y\right)}^{2}-2\left(6\right)\left(2y\right)+{6}^{2}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$ $\begin{array}{cc}& \left(2y-6\right)\left(2y-6\right)\\ & \left(2y\right)\left(2y\right)-\left(2y\right)6-\left(2y\right)6+\text{36}\\ & {4y}^{2}-\text{12}y-\text{12}y+\text{36}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$ $\begin{array}{cc}& {\left(2y-6\right)}^{2}\\ & {\left(2y\right)}^{2}-2\left(6\right)\left(2y\right)+{6}^{2}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$

Did it work? If a formula is true, it should work for any $y$ -value; let’s test each one with $y=5$ . (Note that the second two methods got the same answer, so we only need to test that once.)

 $\begin{array}{ccc}{\left(2y-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{36}\\ {\left(2\cdot 5-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{36}\\ {\left(\text{10}-6\right)}^{2}& \stackrel{?}{=}& \text{100}-\text{36}\\ {4}^{2}& \stackrel{?}{=}& \text{64}✗\end{array}$ $\begin{array}{ccc}{\left(2y-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{24}y+\text{36}\\ {\left(2\cdot 5-6\right)}^{2}& \stackrel{?}{=}& 4{\left(5\right)}^{2}-\text{24}\cdot 5+\text{36}\\ {\left(\text{10}-6\right)}^{2}& \stackrel{?}{=}& \text{100}-\text{120}+\text{36}\\ {4}^{2}& \stackrel{?}{=}& \text{16}✓\end{array}$

We conclude that squaring each term individually does not work. The other two methods both give the same answer, which works.

The first method is the easiest, of course. And it looks good. ${\left(2y\right)}^{2}$ is indeed ${4y}^{2}$ . And ${6}^{2}$ is indeed 36. But as you can see, it led us to a false answer —an algebraic generalization that did not hold up.

I just can’t stress this point enough. It sounds like a detail, but it causes errors all through Algebra II and beyond. When you’re adding or subtracting things, and then squaring them, you can’t just square them one at a time. Mathematically, ${\left(x+a\right)}^{2}\ne {x}^{2}+{a}^{2}$ . You can confirm this with numbers all day. ${\left(7+3\right)}^{2}=\text{100}$ , but ${7}^{2}+{3}^{2}=\text{58}$ . They’re not the same.

So that leaves the other two methods. FOIL will never lead you astray. But the third approach, the formula, has three distinct advantages.

1. The formula is faster than FOIL.
2. Using these formulae is a specific case of the vital mathematical skill of using any formula—learning how to plug numbers and variables into some equation that you’ve been given, and therefore understanding the abstraction that formulae represent.
3. Before this unit is done, we will be completing the square, which requires running that particular formula backward —which you cannot do with FOIL.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I got X =-6
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ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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