• <blockquote id="a6oys"></blockquote>
• <tt id="a6oys"></tt>
• Paul's Online Notes
Home / Algebra Trig Review / Algebra / Multiplying Polynomials
Show General Notice Show Mobile Notice Show All Notes Hide All Notes
General Notice

I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.

Paul
August 27, 2020

Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Multiplying Polynomials

Multiply each of the following. Show All Solutions Hide All Solutions

1. $$\left( {7x - 4} \right)\left( {7x + 4} \right)$$
Show Solution

Most people remember learning the FOIL method of multiplying polynomials from an Algebra class. I’m not very fond of the FOIL method for the simple reason that it only works when you are multiplying two polynomials each of which has exactly two terms (i.e. you’re multiplying two binomials). If you have more than two polynomials or either of them has more, or less than, two terms in it the FOIL method fails.

The FOIL method has its purpose, but you’ve got to remember that it doesn’t always work. The correct way to think about multiplying polynomials is to remember the rule that every term in the second polynomial gets multiplied by every term in the first polynomial.

So, in this case we’ve got.

$\left( {7x - 4} \right)\left( {7x + 4} \right) = 49{x^2} + 28x - 28x - 16 = 49{x^2} - 16$

Always remember to simplify the results if possible and combine like terms.

This problem was to remind you of the formula

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
2. $${\left( {2x - 5} \right)^2}$$
Show Solution

Remember that $${3^2} = \left( 3 \right)\left( 3 \right)$$ and so

${\left( {2x - 5} \right)^2} = \left( {2x - 5} \right)\left( {2x - 5} \right) = 4{x^2} - 20x + 25$

This problem is to remind you that

${\left( {a - b} \right)^n} \ne {a^n} - {b^n}\hspace{0.25in}\,\,\,\,\,\,\,{\rm{and}}\hspace{0.5in}{\left( {a + b} \right)^n} \ne {a^n} + {b^n}$

so do not make that mistake!

${\left( {2x - 5} \right)^2} \ne 4{x^2} - 25$

There are actually a couple of formulas here.

\begin{align*}{\left( {a + b} \right)^2} & = {a^2} + 2ab + {b^2}\\ {\left( {a - b} \right)^2} &= {a^2} - 2ab + {b^2}\end{align*}

You can memorize these if you’d like, but if you don’t remember them you can always just FOIL out the two polynomials and be done with it…

3. $$2{\left( {x + 3} \right)^2}$$
Show Solution

Be careful in dealing with the 2 out in front of everything. Remember that order of operations tells us that we first need to square things out before multiplying the 2 through.

$2{\left( {x + 3} \right)^2} = 2\left( {{x^2} + 6x + 9} \right) = 2{x^2} + 12x + 18$

Do, do not do the following

$2{\left( {x + 3} \right)^2} \ne {\left( {2x + 6} \right)^2} = 4{x^2} + 24x + 36$

It is clear that if you multiply the 2 through before squaring the term out you will get very different answers!

There is a simple rule to remember here. You can only distribute a number through a set of parenthesis if there isn’t any exponent on the term in the parenthesis.

4. $$\left( {2{x^3} - x} \right)\left( {\sqrt x + \frac{2}{x}} \right)$$
Show Solution

While the second term is not a polynomial you do the multiplication in exactly same way. The only thing that you’ve got to do is first convert everything to exponents then multiply.

\begin{align*}\left( {2{x^3} - x} \right)\left( {\sqrt x + \frac{2}{x}} \right) & = \left( {2{x^3} - x} \right)\left( {{x^{\frac{1}{2}}} + 2{x^{ - 1}}} \right)\\ & = 4{x^2} + 2{x^{\frac{7}{2}}} - {x^{\frac{3}{2}}} - 2\end{align*}
5. $$\left( {3x + 2} \right)\left( {{x^2} - 9x + 12} \right)$$
Show Solution

Remember that the FOIL method will not work on this problem. Just multiply every term in the second polynomial by every term in the first polynomial and you’ll be done.

\begin{align*}\left( {3x + 2} \right)\left( {{x^2} - 9x + 12} \right) & = {x^2}\left( {3x + 2} \right) - 9x\left( {3x + 2} \right) + 12\left( {3x + 2} \right)\\ & = 3{x^3} - 25{x^2} + 18x + 24\end{align*}
亚洲欧美中文日韩视频 - 视频 - 在线观看 - 影视资讯 - av网 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>