Algebra 2, Algebraic fractions..help

spidermonkey · #1 $Old$ June 27th, 2009, 11:41 AM

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If anyone could help me to show me how to solve these that would be great cause im a little stuck at the moment

masters · #2 $Old$ June 27th, 2009, 11:59 AM

Quote:

Originally Posted by spidermonkey $View Post$

4. $\frac{12xy}{z^3w} \div \frac{20x^2y^3}{6z^2w^4}$

Hi spidermonkey,

Here's #4

$\frac{12xy}{z^3w} \div \frac{20x^2y^3}{6z^2w^4}$

$\frac{12xy}{z^3w} \cdot \frac{3z^2w^4}{10x^2y^3}=\frac{36w^4xyz^2}{10wx^2y^3z^3}=\frac{18w^3}{5xy^2z}$

..

masters · #3 $Old$ June 27th, 2009, 12:22 PM

And #5

$\frac{9x^2-1}{12x^2-12x} \div \frac{9x^2+12x+3}{3x^2-6x+3}=$

$\frac{(3x-1)(3x+1)}{12x(x-1)} \cdot \frac{3(x^2-2x+1)}{3(3x^2+4x+1)}=$

$\frac{(3x-1)(3x+1)}{12x(x-1)} \cdot \frac{3(x-1)(x-1)}{3(3x+1)(x+1)}=$

Are you able to simplify from here?

yeongil · #4 $Old$ June 27th, 2009, 04:22 PM

2)
$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$

Begin by factoring (note that the numerator of the 2nd fraction is a sum of 2 cubes):
$= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{(2a + 1)(4a^2 - 2a + 1)}{2a(2a + 1)}$

Cancel things out:
$\begin{aligned}&= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{{\color{red}(2a + 1)}(4a^2 - 2a + 1)}{2a{\color{red}(2a + 1)}} \\&= \frac{{\color{red}a^4}}{-2{\color{red}a^2}(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2{\color{red}a}} \\&= \frac{a}{-2(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2} \\\end{aligned}$

Multiply:
$= -\frac{a(4a^2 - 2a + 1)}{4(2a^2 - 1)}$

01

beardedoneder · #5 $Old$ June 28th, 2009, 05:40 PM

Quote:

Originally Posted by spidermonkey $View Post$

1.

Quote:

Originally Posted by spidermonkey $View Post$

If anyone could help me to show me how to solve these that would be great cause im a little stuck at the moment

IMO, you can learn more with a demonstration of the concepts necessary to solve the problems, than if you're given the solutions.

Solving problem #1 involves five concepts you should already be familiar with: 1. simplifying and multiplying fractions; 2. the rules of signs; 3. multiplying terms with variables; 4. factoring polynomial terms; and 5. multiplying binomials (unless you're required to present your answer in simplified form).

If you recall, multiplying fractions is a simple matter of multiplying the numerators together, and the denominators together. If the factors were numbers instead of variables, you would simply do this:

To multiply $\tt{\frac{3}{5} \cdot \frac{6}{8}}$ , we first want to see if either fraction can be reduced. $\tt{6\ and\ 8}$ have a greatest common factor of $\tt{2}$ .

So, $\tt{\frac{6}{8} \div \frac{2}{2} = \frac{3}{4}}$ . Therefore, $\tt{\frac{3}{5} \cdot \frac{3}{4}\ =\ \frac{9}{20}}$ .

If we write $\tt{\frac{2}{2} \cdot \frac{3}{4} = \frac{6}{8}}$ symbolically, we would say, $\tt{a=2; b=3; c=4}$ , then $\tt{\frac{a}{a} \cdot \frac{b}{c} = \frac{a \cdot b}{a \cdot c}\ or\ \frac{ab}{ac}}$ .

So, $\tt{ab}$ and $\tt{ac}$ share the common factor $\tt{a}$ .

Multiplying the denominators together in problem #1 is "distributing multiplication over subtraction" by the Distributive Property, and multiplying the numerators together is the multiplication of two binomials. Also, remember from your rules of signs that

$\begin{tabular}{lllll} \tt{ -} & \tt{ times} & \tt{ -} & \tt{ is} & \tt{ + ,}\tabularnewline\tt{ -} & \tt{ times} & \tt{ +} & \tt{ is} & \tt{ - ,}\tabularnewline\tt{ -} & \tt{ plus} & \tt{ -} & \tt{ is} & \tt{ - , and}\tabularnewline\tt{ -} & \tt{ plus} & \tt{ +} & \tt{ is} & \tt{ the sign of the higher number.}\tabularnewline\end{tabular}$

An example of the Distributive Property is $\tt{a(b-c)}$ in: $\tt{\frac{(a-b)(a-b)}{a(b-c)}}$ .

Distributing (multiplying) the $\tt{a}$ across $\tt{b}$ and $\tt{-c}$ produces $\tt{ab-ac}$ .
Reversing the process, you can factor the greatest common factor $\tt{a}$ out of $\tt{ab-ac}$ to produce $\tt{a(b-c)}$ .

An example of binomial multiplication is $\tt{(a-b)(a-b)}$ in: $\tt{\frac{(a-b)(a-b)}{a(b-c)}}$

Most simple binomials can be multiplied with the F.O.I.L. method, which multiplies the First Terms, then the Outer Terms, then the Inner Terms, then the Last Terms. Finally, you simplify the expression:
$\begin{tabular}{ccccccc} \tt{ $\underbrace{\tt{(a\cdot a)}}$} & \tt{ +} & \tt{ $\underbrace{\tt{(a\cdot-b)}}$} & \tt{ +} & \tt{ $\underbrace{\tt{(-b\cdot a)}}$} & \tt{ +} & \tt{ $\underbrace{\tt{(-b\cdot-b)}}$}\tabularnewline \tt{ F} & & \tt{ O} & & \tt{ I} & & \tt{ L}\tabularnewline \end{tabular}\$ = $\ \tt{\ a^2\ +\ (-ab)\ +\ (-ba)\ +\ b^2}$ ;

Let's simplify the two center terms. 1 is a silent or understood coefficient and since, by the Commutative Property $\tt{-ba = -ab}$ , then $\tt{-1ab+(-1ab) = -2ab}$ .
So, $\tt{(a-b)(a-b)}$ = $\tt{a^2-2ab+b^2}$
Therefore, $\tt{\frac{(a-b)(a-b)}{a(b-c)} = \frac{a^2-2ab+b^2}{ab-ac}}$

I find it less error prone to multiply binomials vertically like in standard multiplication, instead of using F.O.I.L., especially with complex binomials. Consider the non-complex (trivial) example, $\tt{(a-b)(c-d)}$ . :

$\begin{tabular}{ccccc} & & & ${\tt a}$ & ${\tt -b}$\tabularnewline ${\tt x}$ & & & ${\tt c}$ & ${\tt -d}$\tabularnewline \cline{2-5} & & & ${\tt -ad}$ & ${\tt +bd}$\tabularnewline ${\tt +}$ & ${\tt ac}$ & ${\tt -bc}$ & & \tabularnewline \cline{2-5} & ${\tt ac}$ & ${\tt -bc}$ & ${\tt -ad}$ & ${\tt +bd}$\tabularnewline \end{tabular}$

Note: that in the addition rows, the variable terms $\tt{-bc}$ and $\tt{-ad}$ are in separate columns, unlike numerical multiplication where they would be stacked and added together. This, because you can only add terms that have identical variables of the same power.

Now, apply these concepts to solve your given problem. Enjoy.

Hint: If you will first simplify $\tt{\frac{xy-zy}{y}}$ by cancellation, and factor the numerator and denominator of $\tt{\frac{xw-xy}{yz-yw}}$ , it will be easier.

spidermonkey · #6 $Old$ July 9th, 2009, 07:47 PM

Quote:

Originally Posted by yeongil $View Post$

2)
$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$

Begin by factoring (note that the numerator of the 2nd fraction is a sum of 2 cubes):
$= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{(2a + 1)(4a^2 - 2a + 1)}{2a(2a + 1)}$

Cancel things out:
$\begin{aligned}&= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{{\color{red}(2a + 1)}(4a^2 - 2a + 1)}{2a{\color{red}(2a + 1)}} \\&= \frac{{\color{red}a^4}}{-2{\color{red}a^2}(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2{\color{red}a}} \\&= \frac{a}{-2(2a^2 - 1)} \cdot \frac{4a^2 - 2a + 1}{2} \\\end{aligned}$

Multiply:
$= -\frac{a(4a^2 - 2a + 1)}{4(2a^2 - 1)}$

01

thank you for your explanation but what im having problems with is the actual factoring of the problem i just dont get how to do it.

yeongil · #7 $Old$ July 9th, 2009, 09:20 PM

Quote:

Originally Posted by spidermonkey $View Post$

thank you for your explanation but what im having problems with is the actual factoring of the problem i just dont get how to do it.

Okay...

Quote:

$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$

Let's look at the denominator of the first fraction:
$2a^2 - 4a^4$
I don't like how it's not in standard form (ie. descending powers), so I change it to
$-4a^4 + 2a^2$
I then factor out the greatest common monomial factor, which is $-2a^2$ :
${\color{red}-2a^2(2a^2 - 1)}$

Now, look at the numerator of the second fraction:
$8a^3 + 1$
This is a sum of cubes. There is a formula that you should know:
$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
So the sum of cubes can be factored this way:
$8a^3 + 1 = (2a)^3 + 1^3 = (2a + 1)[(2a)^2 - 2a(1) + 1^2] = {\color{red}(2a + 1)(4a^2 - 2a + 1)}$

Finally, look at the denominator of the second fraction:
$2a + 4a^2$
Again, it's not in standard form, so I change to
$4a^2 + 2a$
Factor out the greatest common monomial factor, which is 2a:
${\color{red}2a(2a + 1)}$

So, putting it all together,
$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$
$= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{(2a + 1)(4a^2 - 2a + 1)}{2a(2a + 1)}$

01