Basic Math - I feel dumb!

  • Thread starter Sanyo
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Hello! I have a major final on Friday and I need to figure out how to do these problems. For the life of me I canot figure out how to simplify those fractions. And to multiple, I was under the impression you factor out first and then cross out. But everything comes out positive?

For the multiplication, it should be x2+12x+35, not 25. sorry

Any help is appreciated!
 

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arildno

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Let's take the first:
[tex]\frac{3x^{2}-27}{18-6x}[/tex]
It is smart to focus on the numerator and denominator separately:
FIRST SIMPLIFICATION:
Numerator: [tex]3x^{2}-27[/tex]
Note that 3 is common factor, since 27=3*9.
Therefore, you can write:
[tex]3x^{2}-27=3*(x^{2}-9)[/tex]
Denominator: [tex]18-6x[/tex]
Here, 6 is a common factor, since 18=6*3
Thus, you can write: [tex]18-6x=6*(3-x)[/tex]

Therefore, we have:
[tex]\frac{3x^{2}-27}{18-6x}=\frac{3*(x^{2}-9)}{6*(3-x)}[/tex]
But, now we can see that because 6=3*2, we have:
[tex]\frac{3*(x^{2}-9)}{6*(3-x)}=\frac{3*(x^{2}-9)}{3*2*(3-x)}=\frac{(x^{2}-9)}{2*(3-x)}[/tex]
We have managed a simplification!!!!

SECOND SIMPLIFICATION:
Numerator:
If you are perceptive, you see that we have: [tex]x^{2}-9=(x+3)*(x-3)[/tex]
Let us write this a bit differently, by extracting a minus sign from the last parenthesis:
[tex]x^{2}-9=(x+3)*(x-3)=-(x+3)*(3-x)[/tex]
But now, we have:
[tex]\frac{(x^{2}-9)}{2*(3-x)}=\frac{-(x+3)*(3-x)}{2*(3-x)}=-\frac{(x+3)}{2}[/tex]
since the factor (3-x) occurs in both numerator and denominator.
Got that?
 

dextercioby

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Can u factor them...?All of them can be factored & easily simplified.Take the first.What can u do to it...?

Daniel.
 

Galileo

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In all those problems the idea is to factorize.
Some useful identities will help you spot how to factor certain polynomials immediately:

[tex](a+b)^2=a^2+2ab+b^2[/tex]
[tex](a-b)^2=a^2-2ab+b^2[/tex]
[tex](a+b)(a-b)=a^2-b^2[/tex]

Identifying the right side will help you factor it as the left side. eg: [itex]x^2-9=x^2-3^2=(x+3)(x-3)[/itex].

Also, be careful when simplifying as follows:

[tex]\frac{x^2-25}{x-5}=\frac{(x+5)(x-5)}{x-5}=x+5[/tex]

In the last step I divided top and bottom by x-5, which is not allowed when x=5. So the last step is wrong. Be sure to add [itex]x\not=5[/itex] or don't simplify further.
 
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arildno

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I would rather say that x isn't allowed to be 5 in the first place..
In my case, Galileo's argument requires that x must be different from 3.
 
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Thanks! That was stupid, all I had to do was factor out.

But for the multiplication, I'm still unsure.

wouldn't x2+12x+35 = (x+7)(x+5) ?

and x2+10x+25 = (x+5)(x+5)

and for x2+7x+10 = (x+5)(x+2)

and for x2+9x+14= (x+7)(x+2)
 
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[tex]\frac{x^2 + 12x + 25}{x^2 + 10x + 25} \times \frac{x^2 + 7x + 10}{x ^2 + 9x + 14} =[/tex]

[tex]\frac{x^4 + 7x^3 + 10x^2 + 12x^3 + 84x^2 + 120x + 25x^2 + 175x + 250}{x^4 + 9x^3 + 14x^2 + 10x^3 + 90x^2 + 140x + 25x^2 + 225x + 350}[/tex] etc. is one way or you can factorise.

The Bob (2004 ©)
 

arildno

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Sure, your factoring looks okay to me.
 

arildno

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The Bob said:
[tex]\frac{x^2 + 12x + 25}{x^2 + 10x + 25} \times \frac{x^2 + 7x + 10}{x ^2 + 9x + 14} =[/tex]

[tex]\frac{x^4 + 7x^3 + 10x^2 + 12x^3 84x^2 + 120x + 25x^2 + 175x + 250}{x^4 + 9x^3 + 14x^2 + 10x^3 + 90x^2 + 140x + 25x^2 + 225x + 350}[/tex] etc.

The Bob (2004 ©)
You have a delightfully original conception of SIMPLIFICATION, THe Bob!
 
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arildno said:
You have a delightfully original conception of SIMPLIFICATION, THe Bob!
The original question says multiply. Not a word about simplification.

arildno said:
Sure, your factoring looks okay to me.
It might be ok but if you see the original question the first:

Sanyo said:
x^2+12x+35 = (x+7)(x+5)
should be x^2 + 12x + 25, hence, the longer way of doing it.

The Bob (2004 ©)
 
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Should be:

[tex]\frac{x^2 + 12x + 35}{x^2 + 10x + 25} \times \frac{x^2 + 7x + 10}{x ^2 + 9x + 14} [/tex]

and then I thought it was:

[tex]\frac{(x+7)(x+5)}{(x+5)(x+5)} \times \frac{(x+5)(x+2)}{(x+7)(x+2)}[/tex]

Would they all cancel out making it 1?
 

arildno

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That seems correct.
 

learningphysics

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Sanyo said:
Should be:

[tex]\frac{x^2 + 12x + 35}{x^2 + 10x + 25} \times \frac{x^2 + 7x + 10}{x ^2 + 9x + 14} [/tex]

and then I thought it was:

[tex]\frac{(x+7)(x+5)}{(x+5)(x+5)} \times \frac{(x+5)(x+2)}{(x+7)(x+2)}[/tex]

Would they all cancel out making it 1?
Exactly! Well done. :smile:

EDIT: Follow Galileo's advice and write that x cannot equal -5,-7 or -2.
 
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Sanyo said:
Should be:

[tex]\frac{x^2 + 12x + 35}{x^2 + 10x + 25} \times \frac{x^2 + 7x + 10}{x ^2 + 9x + 14} [/tex]
Now that this information has come out, I would like to apologise to arildno for an offence I may have caused. With the information given I did not see a simpler way without the quadratic equation. Please accept my apologise. :smile:

The Bob (2004 ©)
 

arildno

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The Bob said:
Now that this information has come out, I would like to apologise to arildno for an offence I may have caused. With the information given I did not see a simpler way without the quadratic equation. Please accept my apologise. :smile:

The Bob (2004 ©)
None needed; I was able to sleep unperturbed last night, and that's the important thing, isn't it? :wink:
 
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arildno said:
None needed; I was able to sleep unperturbed last night, and that's the important thing, isn't it? :wink:
:rofl: Yeap. It is. :biggrin:

The Bob (2004 ©)
 

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