Problem with "Rational Algebraic Expressions" (Factoring)

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MaxRabbit

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Jan 25, 2008
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Hi guys-I need to find (a) the domain and (b) the zeros of each function. Here is an example of a problem that is completely stumping me:

h(t)=t^3+4t^2-t-4 over t^3-t^2+t-1
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I don't understand either parts, but I'm starting with part (a) where I make t^3-t^2+t-1 (the denominator) equal to zero. However, I'm having a heck of a time factoring this. I tried everything I know how to do and can't seem to get a good answer. The answer according to this sites calc should be \(\displaystyle \left(t-1\right)\,\left(t^2+1\right)$$\) which would make the answer 1. But I don't understand at all how to get this-so I'm just looking for a solution and instructions on how you got it.

By the way-I may be able to figure this out if you walk me through (a), but it would help even more if you could help with (b) too.

Thanks a ton for any help.
 
Your first step should be to factor the numerator and denominator completely and reduce. That will produce \(\displaystyle \frac{(t+4)(t+1)}{t^2+1}\). Can you take it from there?
 
I'm afraid not-my entire problem is with the factoring part. I just don't know how it's getting from the original to that factored answer. Also, that denominator doesn't appear to be the same answer the calculator gave-why's that?

So anyway, I'd really like a step-by-step on factoring so I can figure out and continue on my own from there.
 
MaxRabbit said:
I just don't know how it's getting from the original to that factored answer.... I'd really like a step-by-step on factoring so I can figure out and continue on my own from there.
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Actually, just giving you the specific steps for these specific polynomials, without any explanation, won't teach you the general terms or techniques. For that, you need lessons! :wink:

To learn how to factor, try the following sequence:

. . . . .Simple Factoring: "In Pairs"

. . . . .Factoring Quadratics

. . . . .Special Factoring

MaxRabbit said:
Also, that denominator doesn't appear to be the same answer the calculator gave-why's that?
Click to expand...
Since we have no idea what you plugged into your calculator (for instance, did you forget the grouping symbols then, as you did here?), nor what result it returned, I'm afraid it is impossible for us to explain what happened. Sorry! :oops:

Once you have studied the above lessons, you should be able to do the factorization. You use "in pairs" factoring first on each polynomial, and then you'll apply "special" factoring to the difference of squares. Then you'll cancel the one common factor, and you'll end up with the simplified form. :idea:

You'll use the regular factoring of quadratics for most of your other exercises; it is the most common factoring you'll be doing. So be sure to study that lesson, too! Have fun! :D

Eliz.
 
Dear MaxRabbit,

Whenever you are considering factoring a 4-term polynomial (just like your stated numerator and denomonator), think of factoring by grouping. Look in your books index under 'F' for "factoring by grouping" or something similar, then read the lesson on that topic. Here is an example of 'factoring by grouping':

\(\displaystyle $4x^2+24x-2x-12$\quad\)
\(\displaystyle $4x^2+24x+-2x-12$\quad\) If the third term is negative, change the - to +-. Know why?
\(\displaystyle $(4x^2+24x)+(-2x-12)$\quad\) Group the terms just by adding parenthesis.
\(\displaystyle $4x(x+6)-2(x+6)$\quad\) Factor out the GCF in each group. If this will work, the ``parethesis factors'' must match, i.e., the \(\displaystyle $(x+6)$\).
\(\displaystyle $(x+6)(4x-2)$\quad\) Factor out the ``parethesis factor'', \(\displaystyle $(x+6)$\). It is the GCF of the terms \(\displaystyle $4x(x+6)$\) and \(\displaystyle $-2(x+6)$\). You are done.
 
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