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9x + 2/ 3x- 2x - 8 + 7/3x2 + x - 4   

 

I tried to factor the denominator ending with -8 with (3x+2)(x-4) but it ends because 2 X -4 = -8 so I thought maybe that would work but it ends up being 3x^2 - 10x - 8 and with the denominator ending with -4 I tried (3x+2)(x-2) but it's wrong too.. what am I doing wrong??

 
Guest Jan 12, 2018

Best Answer 

 #5
avatar+1602 
+1

 

Hello again! I see that you are having difficulty finishing the complete simplification of \(\frac{9x+2}{3x^2-2x-8}+\frac{7}{3x^2+x-4}\)

 

I am glad to see that you have factored both quadratics correctly. Rewriting the denominators into its factored form results in \(\frac{9x+2}{(3x+4)(x-2)}+\frac{7}{(3x+4)(x-1)}\). Just like the one I solved over here at https://web2.0calc.com/questions/subtracting-rational-expressions#r1, your goal at the moment is to create a common denominator. Unlike the one I linked to above, finding it is a little bit more difficult. In that example, I noted how you want the LCM (least common multiple); however, you really only need a common multiple. It does not have to be the least, but the computation is easier if you do. 

 

So, what is a common factor of \((3x+4)(x-2)\) and \((3x+4)(x-1)\)? The common factor of 3x+4 is relevant, but it alone will not be enough to figure out a common factor. To figure out a common multiple, just multiply both denominators together. This will always give you a common multiple. It may not be the least one, however. Multiplying both the denominators together yields \((3x+4)^2(x-2)(x-1)\). There is no need to expand yet. Notice, however, that both denominators have a factor of \(3x+4\), so we can divide that from the multiple we have. This gives us the least common multiple of \((x-2)(x-1)\).

 

Now, let's convert each fraction individually. I'll start with \(\frac{9x+2}{(3x+4)(x-2)}\).

 

\(\frac{9x+2}{(3x+4)(x-2)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)I'm not necessarily concerned about expanding just yet. For now, let's just worry about getting the common denominator.
\(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}\) 
  

 

Now, let's convert the next fraction \(\frac{7}{(3x+4)(x-1)}\):

 

\(\frac{7}{(3x+4)(x-1)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)Let's do the multiplication!
\(\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2}\) 
  

 

You might notice that there is an issue, though, because the expression \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}+\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2} \Rightarrow \frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\). You'll see that I factored out some unnecessary factors to create that common denominator. Now, let's add those fractions together!

 

\(\frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\)Now that the denominators are the same, we can combine the fractions.
\(\frac{(9x+2)(x-1)+7(x-2)}{(3x+4)(x-2)(x-1)}\)

Now, let's multiply the numerator. Find the product of the binomials and distribute the 7.

\(\frac{9x^2-7x-2+7x-14}{(3x+4)(x-2)(x-1)}\)Let's combine those like terms!
\(\frac{9x^2-16}{(3x+4)(x-2)(x-1)}\)This is NOT the final answer. This is not fully simplified because the numerator is a difference of squares. Be attentive!
\(\frac{(3x+4)(3x-4)}{(3x+4)(x-2)(x-1)}\)There is a common factor in the numerator and denominator that cancel out.
\(\frac{3x-4}{(x-2)(x-1)}\)Now, expand the denominator again.
\(\frac{3x-4}{x^2-3x+2}\)This is fully simplified; there is nothing else to do here.
  

 

Now, let's consider the restrictions. All the factors of the original problem \((3x+4),(x-2),\text{and}(x-1)\) cannot be divided by zero. Set them equal to zero to figure out the forbidden values. 

 

\(\frac{3x-4}{x^2-3x+2},x\neq -\frac{4}{3},1,2\)

 
TheXSquaredFactor  Jan 12, 2018
edited by TheXSquaredFactor  Jan 12, 2018
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5+0 Answers

 #1
avatar+1602 
+1

You are so close!

 

Let's start with the quadratic \(3x^2-2x-8\). The factors of 8 are 1,2,4, and 8. One of the factors would have to be negative. Since the coefficient of the quadratic term is 3, a prime, it is easy to see that this quadratic will be factored into \((3x+\text{__})(x+\text{__})\). You saw that 2 and -4 are factors of 8, so you filled in the the blanks as follows. \((3x+2)(x-4)\). This is the right idea. Sometimes these require some trial and error. Keep trying until you get the right binomial. 

 

Yet again, you have the right idea with the second binomial. The factors of 4 are 1,2, and 4. Keep trying the factors and changing the signs and changing which binomials those factors belong in. Remember that one must be negative! With time, you'll figure it out. 

 

If you have any further questions, please ask!

 
TheXSquaredFactor  Jan 12, 2018
edited by TheXSquaredFactor  Jan 12, 2018
 #3
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0

Thank you so much

 
Guest Jan 12, 2018
 #4
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0

Okay I figured out the one ending with  -8 is (3x+4)(x-2) and the one ending in -4 is (3x+4)(x-1) and I see the common factor is 3x+4 and I've seen many examples but I still don't know where I go from there 

 
Guest Jan 12, 2018
 #5
avatar+1602 
+1
Best Answer

 

Hello again! I see that you are having difficulty finishing the complete simplification of \(\frac{9x+2}{3x^2-2x-8}+\frac{7}{3x^2+x-4}\)

 

I am glad to see that you have factored both quadratics correctly. Rewriting the denominators into its factored form results in \(\frac{9x+2}{(3x+4)(x-2)}+\frac{7}{(3x+4)(x-1)}\). Just like the one I solved over here at https://web2.0calc.com/questions/subtracting-rational-expressions#r1, your goal at the moment is to create a common denominator. Unlike the one I linked to above, finding it is a little bit more difficult. In that example, I noted how you want the LCM (least common multiple); however, you really only need a common multiple. It does not have to be the least, but the computation is easier if you do. 

 

So, what is a common factor of \((3x+4)(x-2)\) and \((3x+4)(x-1)\)? The common factor of 3x+4 is relevant, but it alone will not be enough to figure out a common factor. To figure out a common multiple, just multiply both denominators together. This will always give you a common multiple. It may not be the least one, however. Multiplying both the denominators together yields \((3x+4)^2(x-2)(x-1)\). There is no need to expand yet. Notice, however, that both denominators have a factor of \(3x+4\), so we can divide that from the multiple we have. This gives us the least common multiple of \((x-2)(x-1)\).

 

Now, let's convert each fraction individually. I'll start with \(\frac{9x+2}{(3x+4)(x-2)}\).

 

\(\frac{9x+2}{(3x+4)(x-2)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)I'm not necessarily concerned about expanding just yet. For now, let's just worry about getting the common denominator.
\(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}\) 
  

 

Now, let's convert the next fraction \(\frac{7}{(3x+4)(x-1)}\):

 

\(\frac{7}{(3x+4)(x-1)}*\frac{(x-2)(x-1)}{(x-2)(x-1)}\)Let's do the multiplication!
\(\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2}\) 
  

 

You might notice that there is an issue, though, because the expression \(\frac{(9x+2)(x-2)(x-1)}{(3x+4)(x-2)^2(x-1)}+\frac{7(x-2)(x-1)}{(3x+4)(x-2)(x-1)^2} \Rightarrow \frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\). You'll see that I factored out some unnecessary factors to create that common denominator. Now, let's add those fractions together!

 

\(\frac{(9x+2)(x-1)}{(3x+4)(x-2)(x-1)}+\frac{7(x-2)}{(3x+4)(x-2)(x-1)}\)Now that the denominators are the same, we can combine the fractions.
\(\frac{(9x+2)(x-1)+7(x-2)}{(3x+4)(x-2)(x-1)}\)

Now, let's multiply the numerator. Find the product of the binomials and distribute the 7.

\(\frac{9x^2-7x-2+7x-14}{(3x+4)(x-2)(x-1)}\)Let's combine those like terms!
\(\frac{9x^2-16}{(3x+4)(x-2)(x-1)}\)This is NOT the final answer. This is not fully simplified because the numerator is a difference of squares. Be attentive!
\(\frac{(3x+4)(3x-4)}{(3x+4)(x-2)(x-1)}\)There is a common factor in the numerator and denominator that cancel out.
\(\frac{3x-4}{(x-2)(x-1)}\)Now, expand the denominator again.
\(\frac{3x-4}{x^2-3x+2}\)This is fully simplified; there is nothing else to do here.
  

 

Now, let's consider the restrictions. All the factors of the original problem \((3x+4),(x-2),\text{and}(x-1)\) cannot be divided by zero. Set them equal to zero to figure out the forbidden values. 

 

\(\frac{3x-4}{x^2-3x+2},x\neq -\frac{4}{3},1,2\)

 
TheXSquaredFactor  Jan 12, 2018
edited by TheXSquaredFactor  Jan 12, 2018
 #2
avatar+80978 
+1

3x^2  -  2x  - 8

 

Note....write  -2x   as   -6x  + 4x   and we have that

 

3x^2 - 6x +  4x  - 8       factor by grouping

 

3x ( x - 2) + 4 ( x - 2)             x - 2  is common  ....so....

 

(x - 2) (3x + 4)      is the factorization we are looking for

 

 

Similarly...

 

3x^2  +  x  -  4

 

Write x  as     -3x  + 4x       and we have

 

3x^2 - 3x  + 4x - 4       factor by grouping

 

3x ( x - 1)  + 4 ( x - 1)      so....the factorization is

 

(x - 1)  ( 3x + 4)

 

 

So we have

 

[ 9x + 2 ] / [ (x - 2) (3x + 4) ]   +  7 / [ (x - 1) (3x + 4) ]

 

[ (9x + 2)(x - 1)  + 7 ( x -2)  ]  /  [ (x - 2)(x - 1) (3x + 4) ]

 

[ 9x^2  + 2x - 9x - 2  + 7x - 14 ]  /  [ (x - 2) (x - 1) (3x + 4 ]

 

[ 9x^2 - 16 ]  /  [ (x - 2) (x - 1) (3x + 4) ]

 

[ (3x + 4) (3x - 4) ]  /  [ (x -2) ( x - 1) (3x + 4) ]

 

(3x - 4)  / [ (x - 2) (x - 1) ]   =      (3x - 4)  / ( x^2 - 3x + 2 )

 

 

 

cool cool cool

 
CPhill  Jan 12, 2018

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