Reduction to Lowest Terms

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BlBl

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Howdy folks, I've got a number of example problems that are giving me trouble in the radical and rational expressions chapter of this book. Here's the first one that's bugging me:

1. \(\displaystyle \dfrac{x^3+x^2+x+1}{x^3+3x^2+3x+1}\)

I thought of trying to break it down by factoring and cancellation, but it doesn't seem to be the right approach:

\(\displaystyle \dfrac{x^2(x+1)+(x+1)}{x^2(x+3)+(3x+1)}\)

Maybe if I change the order of the terms to get the 3s in one place? I'll keep trying, but feel free to offer some direction.

2. \(\displaystyle \dfrac{(x^2+1)^23x^2-x^3(2x)(x^2+1)2}{(x^2+1)^4}\)

Ok, just from looking at it I'm guessing that it's a common factor thing. I probably need to rearrange some things so it's in the proper order.

\(\displaystyle \dfrac{(x^2+1)(x^2+1)x^2[3-4x^2]}{(x^2+1)^4}\)

Maybe I've gone in the wrong direction here. Should I have thrown in a negative exponent and reduced the denominator or something? Help?

3. \(\displaystyle \dfrac{(x+h)^3-x^3}{h}\)

The first thing I did with this was to multiply it by \(\displaystyle \frac{h}{h}\) to get

\(\displaystyle h[(x+h)^3-x^3]\)

Would it have been wiser to just solve the numerator and then divide it by the denominator?

There are two more problems I'll list but these are enough for now. Any direction on them would be much obliged.
 
Reduction to Lowest Terms

\(\displaystyle \dfrac{x^3+x^2+x+1}{x^3+3x^2+3x+1}\)
Click to expand...

\(\displaystyle \dfrac{x^2(x+1)+(x+1)}{(x+1)^3}\)

\(\displaystyle \dfrac{(x+1)(x^2+1)}{(x+1)^3}\)

\(\displaystyle \dfrac{(x^2+1)}{(x+1)^2}\)
 
2.

\(\displaystyle \dfrac{(x^2+1)[(x^2+1)(3x^2)-(4x^4)]}{(x^2+1)^4}

\)

\(\displaystyle \dfrac{(x^2+1)(3x^2)-(4x^4)}{(x^2+1)^3}\)

You could factor an x^2 from the terms on top, but they would not cancel anything in the denominator.
 
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3. Expand (x + h)^3. Simplify. Factor an h from the numerator. Cancel with h in the denominator.
 
Reduce Algebraic Expression to Lowest Terms

Question (1) below.


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


Factoring by groups in the numerator, I got:


x^2(x + 1) + (x + 1)


Factoring by groups in the denominator, I got:


x^2(x + 3) + (3x + 1)


Together:


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


Cancel out the x^2 term.


[(x + 1) + (x + 1)]/(x + 3) + (3x + 1) =


Combine like terms on the top and bottom.


(2x + 2)/(4x + 4) =


Factor on the top and bottom again.


2(x + 1)/4(x + 1)


Cancel out the quantity (x + 1) on the top and bottom.


We are left with 2/4 = 1/2

















 
Question (1) below.


[(x^3 + x^2) + (x + 1)]/(x^3 + 3x^2) + (3x + 1) = 1/2
Click to expand...

If the above left hand expression simplifies to 1/2, then this must be true for all values of x. A good way to check this is to simply plug in some values for x and see if it works. If we let x = 0, our expression simplifies to 1/1, which does not equal 1/2. Therefore we know the above simplification is not true.

So, where is the problem?

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


Cancel out the x^2 term.
Click to expand...

The x^2 term is not a factor of ALL the terms in both the numerator and denominator, so we cannot cancel it out.
 
wjm11 said:
\(\displaystyle \dfrac{x^2(x+1)+(x+1)}{(x+1)^3}\)

\(\displaystyle \dfrac{(x+1)(x^2+1)}{(x+1)^3}\)

\(\displaystyle \dfrac{(x^2+1)}{(x+1)^2}\)
Click to expand...

Thanks for this wjm. I couldn't really follow your work here, but it was the fact that the you made the denominator \(\displaystyle (x+1)^3\) that really helped me out here. Once I had that I was able to crack it for myself.

When I tried following your work just now I got confused at step 2, as I don't know how you got that numerator without changing the denominator. Mine went like this:

\(\displaystyle \dfrac{x^3+x^2+x+1}{x^3+3x^2+3x+1}=\dfrac{x^2(x+1)+(x+1)}{(x+1)^3}\),

\(\displaystyle \dfrac{x^2(x+1)+(x+1)}{(x+1)^3}\div\dfrac{(x+1)}{(x+1)}=\dfrac{x^2+1}{(x+1)^2}\)

For some reason the answer given in the book gives the denominator in an expanded form: \(\displaystyle x^2+2x+1\) But it's all the same.
 
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nyc_function said:
Question (1) below. … We are left with 2/4 = 1/2
Click to expand...

This was not the answer that the book gave, but thanks anyway nyc_function. Also, please consider writing everything in \(\displaystyle \LaTeX\), as it's much, much easier on the eyes.
 
wjm11 said:
2.

\(\displaystyle \dfrac{(x^2+1)[(x^2+1)(3x^2)-(4x^4)]}{(x^2+1)^4}

\)

\(\displaystyle \dfrac{(x^2+1)(3x^2)-(4x^4)}{(x^2+1)^3}\)

You could factor an x^2 from the terms on top, but they would not cancel anything in the denominator.
Click to expand...

Thank you for this one as well. As with the last one, something in your writing of the problem definitely helped me see what I needed here. I split the \(\displaystyle 3x^2\) in the original numerator needlessly, and that sent me in the wrong direction for forming the expression. So what I did went like this:

\(\displaystyle \dfrac{(x^2+1)^23x^2-x^3(2x)(x^2+1)2}{(x^2+1)^4}=\dfrac{(x^2+1)[(x^2+1)(3x^2)-(4x^4)]}{(x^2+1)^4}\),

\(\displaystyle \dfrac{(x^2+1)[(3x^4+3x^2-4x^4)]}{(x^2+1)^4}=\dfrac{(x^2+1)[(3x^2-x^4)]}{(x^2+1)^4}=\div\dfrac{x^2+1}{x^2+1}\),

\(\displaystyle =\dfrac{3x^2-x^4}{(x^2+1)^3}\)
 
wjm11 said:
3. Expand (x + h)^3. Simplify. Factor an h from the numerator. Cancel with h in the denominator.
Click to expand...

Yes, this is exactly right ;) I should've just gone through with the problem and seen how it worked out. But I got stuck in a "factor-cancel" problem solving mode, and made the wrong turn.

\(\displaystyle \dfrac{(x+h)^3-x^3}{h}=\dfrac{(x^3+3hx^2+3h^2x+h^3)-x^3}{h}=\dfrac{(3hx^2+3h^2x+h^3)}{h}\),

\(\displaystyle =3x^2+3hx+h^2\)
 
Should I put the other problems on here, or should I start a new thread?
 
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