Doing an SAT practice test and came across this question:

Doing an SAT practice test and came across this question:
"Which of the following is equivalent to [math]\frac{3b^2+9b}{3b+6}[/math]
A ) [math]b-\frac{6}{3b}[/math]
B ) [math]b[/math]
C ) [math]b+1-\frac{6}{3b+6}[/math]
D ) [math]b+3[/math]"

I got frustrated and used a calculator even tho it is on the no-calculator section, and to my confusion the answer is C.
My attempts at reducing it: [math]\frac{3b^2+9b}{3b+6} = \frac{b^2+3b}{b+2} = [/math]???
Can somebody explain this to my babby brain? How can I arrive at the answer C?

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en.wikipedia.org/wiki/Polynomial_long_division#Example
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Go the fuck to sleep Red.

This

This, Red will never learn how to factor neither squares nor women

you got this from the 2017 sat princeton prep book or a site that stole from it
the answer to this question is incorrect
t. tutored out of the book during the summer

there you can go to bed now

thanks, he's a little bit slow in the head; so now he can rest well.

no problem. glad to help ;)

what?
the answer
[math]\frac{3b^2+9b}{3b+6} [/math] is the same as [math]\frac{3b^2+6b+3b}{3b+6} [/math] which is [math] \frac{b(3b+6)+3b} {3b+6} [/math] or [math] b+ \frac{3b}{3b+6} [/math]
and you can add zero as [math] \frac{6}{3b+6}-\frac{6}{3b+6} [/math] or [math] b+ \frac{3b}{3b+6}+\frac{6}{3b+6}-\frac{6}{3b+6} [/math]
which gets you [math] b+ \frac{3b+6}{3b+6}-\frac{6}{3b+6} [/math] for a final answer of [math] b+ 1-\frac{6}{3b+6} [/math] C

You can brute force this in about 10 seconds by choosing an easy number like [eqn]b=1[/eqn] and seeing that the only one the original expression is equal to is C.

http://www.emathhelp.net/calculators/algebra-1/polynomial-long-division-calculator/?numer=3x^2+%2B+9x&denom=3x+%2B+6&steps=on

When in doubt, use long division.

what the fuck is 3''??

Putting b = 0 we get 0. So it must be B or C. But it is easy to see that it is not B, so it must be C.

Is there a name for this method? I never seen it before.

Why not use partial fraction decomposition

Your dick.

Still not clear on how you get from [math]\frac{b(3b+6)=3b}{3b+6}[/math] to [math]b+\frac{3b}{3b+6}[/math]
Can you explain a little more pls, I am slow.

[math]\frac{b(3b+6)+3b}{3b+6}[/math] to [math]b+\frac{3b}{3b+6}[/math] rather.

This is fucked, my tex is correct.
Well you get what I'm saying.

easy.
[eqn]\frac{b(3b+6)+3b}{3b+6}=\frac{b(3b+6)}{3b+6}+\frac{3b}{3b+6}=b\cdot\frac{3b+6}{3b+6}+\frac{3b}{3b+6}=b\cdot1+\frac{3b}{3b+6}=b+\frac{3b}{3b+6}[/eqn]

Treat b and 1 as being simplified versions of some number over the denominator 3b+6.

Multiply them by the denominator to get the simplified version then combine

>equivalent
Under what equivalence relation? Equality? Then why didn't they just say equal?

Learn polynomial long division:
en.wikipedia.org/wiki/Polynomial_long_division#Example
That's the name of an algorithm for the method

If the answer method will take too long a time, refrain from it and try thinking of a faster method. Since we are comparing the four choices to a single fraction, try writing the choices as a fraction and eliminate what are not possible or probable.
"B" is the easiest it will be [math]\frac{b}{1}
then it is not the answer and we eliminate it.
"A" as a single fraction will be [math]\frac{3b^2-6}{3b}
this is also not the answer and we eliminate A.
"D" will also be eliminated and "C" will be the only answer left and most probable.

You can factor out a 3 and do synthetic division leaving you with b + 1 + (-2/(b+2)). Then, put multiply that last term by 3/3 and you get the previous with -6/(3b+6). Then you inject yourself with ketamine to forget that you don't know beta-tier math.

>tutored out of the book this summer
wow, you were getting paid to teach people wrong?

you need long division of polynomials

>not using synthetic division