Posted: 8/17/2011 3:49:12 AM EDT
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I have an example problem that I am stuck on and was wondering if there is anyone who remembers enough algebra to help me with this problem. I'm trying to go from polynomial factored form to standard form and then back to factored form (checking the work).
The problem is: (x + 3)(x + 4)(x + 5) and so far I have: (x + 3)(x + 4)(x + 5) = (x + 3)(x² + 4x + 5x + 20) - Multiplied (x + 4) and (x + 5) = (x + 3)(x² + 9x + 20) - Simplified = x(x² + 9x + 20) + 3(x² + 9x + 20) - Distribute = x³ + 12x² + 47x + 60 Now I think I messed something up along the way because when I try to factor this polynomial (factor by grouping); I can't find a way to get back to the original factored form. (x³ + 12x²) + (47x + 60) = x(x² + 12x) + 1(47x + 60) - Took greatest common factor (GCF) out of each group. And I am stuck here. There is no GCF between the pairs. I have tried grouping differently as well: (x³ + 47x) + (12x² + 60) = x(x² + 47) + 12(x² + 5) I think I could use a second set of eyes (and brains) on this, any help is appreciated. |
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(x + 3)(x + 4)(x + 5)
Do the first pair first (FOIL) = (x2 + 3x + 4x + 12)(x+5) = (x2 + 7x + 12)(x+5) Then normal binomial * trinomial (a+b+c)(d+e) = (ad + bd + cd + ae + be + ce) = x3 + 7x2 + 12x + 5x3 + 35x + 60 = 6x3 + 7x2 + 47x + 60 (ETA: Somehow I do the whole thing and not notice I multipled 3x5 instead of 3x4 in the first step) |
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Quoted:
(x + 3)(x + 4)(x + 5) Do the first pair first (FOIL) = (x2 + 3x + 4x + 12)(x+5) = (x2 + 7x + 12)(x+5) Then normal binomial * trinomial (a+b+c)(d+e) = (ad + bd + cd + ae + be + ce) = x3 + 7x2 + 12x + 5x3 + 35x + 60 = 6x3 + 7x2 + 47x + 60 (ETA: Somehow I do the whole thing and not notice I multipled 3x5 instead of 3x4 in the first step) Thanks for the reply, I learned another way to come about this problem; but: You cubed the 5 on A*E (x² * 5); so we'd still come to the same conclusion of x³ + 12x² + 47x + 60 otherwise. How to get x³ + 12x² + 47x + 60 back to factored form (x+3)(x+4)(x+5) is my question; I want to be able to "check" my work by being able to factor it correctly. |
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I think I figured it out using Google, though I'm not certain it helped me learn anything...
If I divide the problem out, it returns me to the original factored form (duh...I multiplied it to get it into standard form). It's just the textbook was teaching you to multiply to check your work (I think). I am taking an online course to recover a credit and I think thumbing through books and being forced to learn it yourself does not work out very well for someone like me. It is one of the reasons I failed; our teacher would walk in and write the assignments and pages on the board and leave us to our own devices. Anyway...if I divide the polynomial of 4 terms by (x + 3), I should get: (x³ + 12x² + 47x + 60)/(x + 3) -without going into too much long division of polynomials =(x² + 9x + 20) then (x² + 9x + 20)/(x + 4) =(x + 5) (x³ + 12x² + 47x + 60) = (x + 3)(x + 4)(x + 5) ETA: I guess what bothers me is if I was handed the problem x³ + 12x² + 47x + 60 and was told to factor it...I wouldn't really know where to start. |
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Quoted:
Quoted:
(x + 3)(x + 4)(x + 5) Do the first pair first (FOIL) = (x2 + 3x + 4x + 12)(x+5) = (x2 + 7x + 12)(x+5) Then normal binomial * trinomial (a+b+c)(d+e) = (ad + bd + cd + ae + be + ce) = x3 + 7x2 + 12x + 5x3 + 35x + 60 = 6x3 + 7x2 + 47x + 60 (ETA: Somehow I do the whole thing and not notice I multipled 3x5 instead of 3x4 in the first step) Thanks for the reply, I learned another way to come about this problem; but: You cubed the 5 on A*E (x² * 5); so we'd still come to the same conclusion of x³ + 12x² + 47x + 60 otherwise. Oops.  Should be:
= x3 + 7x2 + 12x + 5x2 + 35x + 60 = x3 + 12x2 + 47x + 60 How to get x³ + 12x² + 47x + 60 back to factored form (x+3)(x+4)(x+5) is my question; I want to be able to "check" my work by being able to factor it correctly. How is it x8? That looks like an 8 to me, should be a 3. Starting at x3 + 12x2 + 47x + 60. Getting it back to factored is more difficult, I don't think there's an easy way for these other than breaking up multiplications where you find them and trying 'at random'. Since it has 4 terms you know you have to break it up into at least a (a + b + c)(d + e) format. Knowing that, you know you need to turn those four terms into 6 (to get ad, bd, cd, ae, be, ce). so for 12x2 you need to find two numbers that add to 12 (6 and 6, 5 and 7, etc.), same thing with 47.. I really don't know an easy way to go about that, maybe someone else does. |
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Quoted: I think I figured it out using Google, though I'm not certain it helped me learn anything... If I divide the problem out, it returns me to the original factored form (duh...I multiplied it to get it into standard form). It's just the textbook was teaching you to multiply to check your work (I think). I am taking an online course to recover a credit and I think thumbing through books and being forced to learn it yourself does not work out very well for someone like me. It is one of the reasons I failed; our teacher would walk in and write the assignments and pages on the board and leave us to our own devices. Anyway...if I divide the polynomial of 4 terms by (x + 3), I should get: (x³ + 12x² + 47x + 60)/(x + 3) -without going into too much long division of polynomials =(x² + 9x + 20) then (x² + 9x + 20)/(x + 4) =(x + 5) (x³ + 12x² + 47x + 60) = (x + 3)(x + 4)(x + 5) ETA: I guess what bothers me is if I was handed the problem x³ + 12x² + 47x + 60 and was told to factor it...I wouldn't really know where to start. The best place to start is with the 60. You can break that up into its multiples (e.g. 1 and 60, 2 and 30). Since we have forehand knowledge of the answer we recognize that the multiples 3 and 20 (which is subsequently 4 and 5) will work here. Therefore, I would divide the third order equation by (x+3) as you did above and continue to break it down. Sorry, but I don't remember any quicker way. |
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Quoted:
Oops. Should be:
= x3 + 7x2 + 12x + 5x2 + 35x + 60 = x3 + 12x2 + 47x + 60 How to get x³ + 12x² + 47x + 60 back to factored form (x+3)(x+4)(x+5) is my question; I want to be able to "check" my work by being able to factor it correctly. How is it x8? That looks like an 8 to me, should be a 3. Starting at x3 + 12x2 + 47x + 60. Getting it back to factored is more difficult, I don't think there's an easy way for these other than breaking up multiplications where you find them and trying 'at random'. Since it has 4 terms you know you have to break it up into at least a (a + b + c)(d + e) format. Knowing that, you know you need to turn those four terms into 6 (to get ad, bd, cd, ae, be, ce). so for 12x2 you need to find two numbers that add to 12 (6 and 6, 5 and 7, etc.), same thing with 47.. I really don't know an easy way to go about that, maybe someone else does. It is cubed, I used alt + 0179 and it seems a little smaller than using [sup] bbcode. I'm not even sure if my assignment requires me to go from standard > factored > standard again but this is one thing that bugs me about all math classes...I actually want to learn why and not just learn what I need to in order to pass the class. Probably just causing myself a lot of grief over it. Quoted:
The best place to start is with the 60. You can break that up into its multiples (e.g. 1 and 60, 2 and 30). Since we have forehand knowledge of the answer we recognize that the multiples 3 and 20 (which is subsequently 4 and 5) will work here. Therefore, I would divide the third order equation by (x+3) as you did above and continue to break it down. Sorry, but I don't remember any quicker way. This seems familiar, must be from another lesson I learned...it sorta makes sense...I just can't put it in words. Thanks for the help y'all, I'm gonna stop beating my brain over this and just go from factored to standard form if that is all the assignment is calling for. |
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Quoted:
ETA: I guess what bothers me is if I was handed the problem x³ + 12x² + 47x + 60 and was told to factor it...I wouldn't really know where to start. Easiest way is as follows: Graph x³ + 12x² + 47x + 60. It intersects at -3, -4, and -5. These are the roots. Flip the polarity and add those numbers to X. (x+3)(x+4)(x+5). If you only have one real root (the other roots don't intersect the x axis), you can calculate that root, and divide by it. So if it intersected at 5, divide the equation by (x-5), and (x-5) is one of the factors. Then you can solve the last two variables quadratically. |
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Quoted: Quoted: The best place to start is with the 60. You can break that up into its multiples (e.g. 1 and 60, 2 and 30). Since we have forehand knowledge of the answer we recognize that the multiples 3 and 20 (which is subsequently 4 and 5) will work here. Therefore, I would divide the third order equation by (x+3) as you did above and continue to break it down. Sorry, but I don't remember any quicker way. This seems familiar, must be from another lesson I learned...it sorta makes sense...I just can't put it in words. Thanks for the help y'all, I'm gonna stop beating my brain over this and just go from factored to standard form if that is all the assignment is calling for. Oh don't give up so easy.  Just kidding. I agree that you should focus on the task at hand (simplifying the polynomial) for the moment.After some google, I've found what I think is a good way of explaing this in words that I can't (it's been forever since I've taken algebra, or at least feels that way). http://oakroadsystems.com/math/polysol.htm What you'll be interested in is Rational Roots (in Step 4) and Synthetic Division (in Step 5), but it can't hurt to review the whole page. Good luck! |
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Starting with X3 + 12X2 + 47X + 60 break up the 2 middle terms to re-write the expression as follows:
X3 +7X2 + 5X2 + 12X + 35X + 60 then rearrange the terms as such (have to use factors of 60 as a clue): (X3 + 7X2 + 12X) + (5X2 + 35X + 60) then factor out an "X" from the 1st term and a "5" out of the 2nd term to get: X(X2 +7X + 12) + 5(X2 + 7x + 12) Then factor out "(X2 + 7x + 12)" from both terms to get: (X2 + 7x + 12)(X + 5) Then factor the trinomial X2 + 7x + 12 to get the 3 binomial factors as: (X+4)(X+3)(X+5) That BBcode is a pain, but works. |
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Quoted:
Starting with X3 + 12X2 + 47X + 60 break up the 2 middle terms to re-write the expression as follows: X3 +7X2 + 5X2 + 12X + 35X + 60 then rearrange the terms as such (have to use factors of 60 as a clue): (X3 + 7X2 + 12X) + (5X2 + 35X + 60) then factor out an "X" from the 1st term and a "5" out of the 2nd term to get: X(X2 +7X + 12) + 5(X2 + 7x + 12) Then factor out "(X2 + 7x + 12)" from both terms to get: (X2 + 7x + 12)(X + 5) Then factor the trinomial X2 + 7x + 12 to get the 3 binomial factors as: (X+4)(X+3)(X+5) That BBcode is a pain, but works. The WYSIWYG editor might have an easier way, but I hate it too much for other reasons. I didn't 'check your work' but that looks like the right way to go about it. |
| I listed all of the factors of 60 and then did some trial and error. I first tried to factor it by re-grouping the original terms (which sometimes works) into 2 binomials, but when that dead-ended, I knew it must involve breaking up the middle terms and forming trinomials that would factor.. |
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Quoted:
I listed all of the factors of 60 and then did some trial and error. I first tried to factor it by re-grouping the original terms (which sometimes works) into 2 binomials, but when that dead-ended, I knew it must involve breaking up the middle terms and forming trinomials that would factor.. Ah ok. I was just wondering if you had a special method instead of guess and check, 
Honestly, once you get above degree 3, factoring can be a futile endeavor. 
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