Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: ralphdspam on April 09, 2011, 08:35:13 pm
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I really need help with simplifying this polynomial. :banghead: (More specifically, I don't know how to solve the parts with the addition/subtraction) (I would like the process, not just the answer.)
((X^2-x-12)/x)/(x/(x-4))
Thanks. :)
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((x2-x-12))/x)/(x/(x-4))
/me puts his thinking hat on.
Okay, so let's start piece by piece and see where we go.
First of all, you can factor (x2-x-12) to (x-4)(x+3). This gives ((x-4)(x+3))/x)/(x/(x-4)).
Now, remember how A/B is the same as A times (1/B)? Well, the same thing works with A/(B/C), however, 1/(B/C) is the same as C/B. So A/(B/C) is the same as A times (C/B). This means that ((x-4)(x+3))/x) which is A in this case, divided by (x/(x-4)) which is B/C in this case, gives ((x-4)(x+3))/x) times ((x-4)/x). Combining these gives ((x-4)(x-4)(x+3)/x2).
Now, I don't know exactly where you want to go with this, but that is pretty simplified there. You could multiply the numerator terms together and get (x2-x-12)(x-4) or (x3-5x2-16x+48).
So (x3-5x2-16x+48)/x2 is an answer.
EDIT: Also, if you set it equal to zero, you could disregard that x2 and use (x-4)(x-4)(x+3)=0. So you have x=4, 4, and -3.
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Thanks, graphmastur. :D
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I'm also learning polynomial factorization at school now, and tried it too, but why can you do this:
First of all, you can factor (x^2-x-12) to (x-4)(x+3). This gives ((x-4)(x+3))/x)/(x/(x-4)).
I have no idea.
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You need to factor x^2-x-12. Find two numbers that multiply to equal A*C and add up to B. -4 and 3 fit this requirement, so you can now write it as (x-4)(x+3). Then you just plug it into the full equation to get ((x-4)(x+3)/x)/(x/(x-4)).
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First of all, the A, B, and C are different from my post above. The general form of a second degree polynomial (ie, one with it's highest term being x2) is Ax2+Bx+C. In the case above, A=1, B=-1, C=-12.
Now, to multiply something of the form (d*x+p)(e*x+q), we use a process called FOIL. This stands for First, Outer, Inner, Last. This means you first multiply the first terms together, which is d*x and e*x in this case, to get d*e*x2. The Outer refers to the first term, and the last term being multiplied together. In this case, d*x and q are multiplied together to get e*x*q. Inner refers to the two inside terms, e*x and p. Multiplying them together gives px. Also note that p and q are usually constants, so you can combine them by (e*p+d*q)x. Lastly, or Last as it were, we have p and q being multiplied together, to get pq.
Now, remember the A, B, and C in the Ax2+Bx+C. To find the d, e, p, and q, that fits a certain A, B, and C, we have to figure these equations out:
A=d*e
B=e*p+d*q
C=p*q
Now, A is 1 in this case, so it simplifies to:
A=1
B=p+q
C=p*q
B is -1 and C is -12. In this case, -4 and 3 work because -3*4=-12=C and -3+4=-1=B.
Does that all make sense?
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Isnt there a formula to factor second degree equations of the type: ax^2+bx+c. I think there is, what is it?
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Isnt there a formula to factor second degree equations of the type: ax^2+bx+c. I think there is, what is it?
Yes, of course, the quadratic formula. Let me link to wolframalpha (http://www.wolframalpha.com/input/?i=quadratic+formula&a=*C.quadratic+formula-_*MathWorld-&a=*FS-_**QuadraticEquation.x-.*QuadraticEquation.a-.*QuadraticEquation.b-.*QuadraticEquation.c--&f3=1&f=QuadraticEquation.a_1&f4=-2&f=QuadraticEquation.b_-2&f5=1&f=QuadraticEquation.c_1)
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All I found here (http://www.wolframalpha.com/input/?i=quadratic+formula&a=*C.quadratic+formula-_*MathWorld-&a=*FS-_**QuadraticEquation.x-.*QuadraticEquation.a-.*QuadraticEquation.b-.*QuadraticEquation.c--&f3=1&f=QuadraticEquation.a_1&f4=-2&f=QuadraticEquation.b_-2&f5=1&f=QuadraticEquation.c_1) was the quadratic formula, not the one to factorize quadratic expressions.
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***BUMP***
How to factorize:
ax^2+bx+c
Using a formula, not the hand method.
Any ideas? Thanks.
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If you find the solutions, ie, x=4 and x=-3, then it is simply x-4=0 and x+3=0 or (x+3)(x-4)=0. That's only if A=1 though. I don't think there is a formula, considering the complexity is basically that of a quadratic formula. There are methods like completing the square that help, though.
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If you find the solutions, ie, x=4 and x=-3, then it is simply x-4=0 and x+3=0 or (x+3)(x-4)=0. That's only if A=1 though. I don't think there is a formula, considering the complexity is basically that of a quadratic formula. There are methods like completing the square that help, though.
Weird, my teacher gave us a formula, if only I had copied it to my notebook as he asked...
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you mean stuffs like this?
Difference of Squares = (A2-B2) = (A-B)(A+B)
Sum of Squares = (A2+B2) = (A-Bi)(A+Bi)
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you mean stuffs like this?
Difference of Squares = (A2-B2) = (A-B)(A+B)
Sum of Squares = (A2+B2) = (A-Bi)(A+Bi)
Nah, those are the 'notable cases', special cases.
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Ok: suppose the roots that you get by running the quadratic formula are r and s (note that there is a plus/minus sign; hence, there are two roots). The quadratic factorized is a(x-r)(x-s), where a is the coefficient of the quadratic (leading) term.
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Thanks a lot phenomist!
I can now fully factorize 3rd and 2nd degree expressions!