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How Do You Factor A 2 Variable Trinomial?
28 Answers
I hate factoring man, specifically these questions (^2=to the power of 2)
w^2 - 3wz - 10z^2
x^4 + 12x^2 + 20
If anybody could help me understand what to do that would be amazing
w^2 - 3wz - 10z^2
x^4 + 12x^2 + 20
If anybody could help me understand what to do that would be amazing
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Everything's easy if you know the answer. Not so easy if you don't. Best not to make a judgement. Also you can argue that factoring is difficult because you can't guarantee that there even is a "nice" way to factor. How, for example, would you factor w^2 - 4wz-10z^2 ? Or w^2 - 3wz - 10z^3 ? In other words, how did you know that you can just "Think of 2 numbers multiplied together to give -10 and added together to give -3"? Just a tiny change and it no longer works, and it wouldn't be clear to me based on the answer above why not.
erm Jim
I think this is a class question where we know there is an answer
and not a real life question, where nothing fits and some variable are given in imperial unit ....
look out of your ivory tower and you will see me as the man with a dog standing in some dogpoo waving back
and isnt your answer abelian groups? _ didnt evariste galois spent his last night on earth before being DDDDYYYYYIIIIIINNNNNGGGG in a duel, wondering about integer answers to polynomials
( if the answe is no it wdnt surprise me)
I think this is a class question where we know there is an answer
and not a real life question, where nothing fits and some variable are given in imperial unit ....
look out of your ivory tower and you will see me as the man with a dog standing in some dogpoo waving back
and isnt your answer abelian groups? _ didnt evariste galois spent his last night on earth before being DDDDYYYYYIIIIIINNNNNGGGG in a duel, wondering about integer answers to polynomials
( if the answe is no it wdnt surprise me)
A useful rule of thumb in maths, I think, is that "the only problems we like to solve are quadratic equations", ie if we can write an expression so it looks like X^2 + bX + c = 0 then that's now a solvable problem. So I think the first step is to check that we can even do that. Ask yourself the following questions:
1. Is it already in this shape? ie, are the powers of my variable 2,1,0?
2. If not, are they [twice something], [something], and 0?
3. If not, are they [positive thing], 0, [negative a thing]?
4. If not, then, if I write the terms in order of biggest power, middle power, smallest power, then does biggest + smallest = twice the middle?
Example:
1. x^2 + 5x + 4 is in the right shape already.
2. x^4 + 5x^2 + 4 is the same as x^[2*2]+ 5 x^[2]+4 .
3. x + 5 + 4/x is the same as x^[1] + 5 x^[0] + 4 x^[-1].
4. x + 5*sqrt(x)+4 : we can write t = t^(2* 1/2) and sqrt(t) = t^(1/2).
5. t^(11/5) + 5 t^(7/5) + 4 t^(3/5) , 11/5 + 3/5 = 14/5 = 2*7/5.
Every single one of these problems factors nicely as a result, and in more or less the same way, looking like (X + 4)(X + 1) for some clever choice of X.
If you have two variables in the problem, like the first example, then the first thing you should try is to divide everything by the highest power of one of the variables. So:
(w^2 - 3 wz - 10 z^2)/z^2 = w^2/z^2 - 3 wz/z^2 - 10 z^2/z^2
Two powers of z cancel in the last term, and one in the second term, so this is the same as w^2/z^2 - 3 w/z - 10 , which we can write as (w/z)^2 - 3 (w/z) - 10. This is now in the right shape, because we have an expression that looks like "thingy squared + thingy + constant".
This still leaves the challenge of factoring, but I'd regard that as a separate problem from knowing that you can even try to factor.
Bottom line: ask yourself if there is anything you can do to make your expression look like X^2 + bX + c. Clever choice of substitution, divide or multiply by a power of X, etc etc. I can't provide an exhaustive list of what the answers might be, but the skill is to ask the question in the first place.
1. Is it already in this shape? ie, are the powers of my variable 2,1,0?
2. If not, are they [twice something], [something], and 0?
3. If not, are they [positive thing], 0, [negative a thing]?
4. If not, then, if I write the terms in order of biggest power, middle power, smallest power, then does biggest + smallest = twice the middle?
Example:
1. x^2 + 5x + 4 is in the right shape already.
2. x^4 + 5x^2 + 4 is the same as x^[2*2]+ 5 x^[2]+4 .
3. x + 5 + 4/x is the same as x^[1] + 5 x^[0] + 4 x^[-1].
4. x + 5*sqrt(x)+4 : we can write t = t^(2* 1/2) and sqrt(t) = t^(1/2).
5. t^(11/5) + 5 t^(7/5) + 4 t^(3/5) , 11/5 + 3/5 = 14/5 = 2*7/5.
Every single one of these problems factors nicely as a result, and in more or less the same way, looking like (X + 4)(X + 1) for some clever choice of X.
If you have two variables in the problem, like the first example, then the first thing you should try is to divide everything by the highest power of one of the variables. So:
(w^2 - 3 wz - 10 z^2)/z^2 = w^2/z^2 - 3 wz/z^2 - 10 z^2/z^2
Two powers of z cancel in the last term, and one in the second term, so this is the same as w^2/z^2 - 3 w/z - 10 , which we can write as (w/z)^2 - 3 (w/z) - 10. This is now in the right shape, because we have an expression that looks like "thingy squared + thingy + constant".
This still leaves the challenge of factoring, but I'd regard that as a separate problem from knowing that you can even try to factor.
Bottom line: ask yourself if there is anything you can do to make your expression look like X^2 + bX + c. Clever choice of substitution, divide or multiply by a power of X, etc etc. I can't provide an exhaustive list of what the answers might be, but the skill is to ask the question in the first place.
oh, OK take the second one
you only have fourth powers and squares
so any factor has to have squares in it and not x power one
( other wise there would be cubes and exes)
if there are whole number answers ( we think there are) then the factors are plus or minus ( 1,2,4,5,10,20) - in pairs 1,20 and 2,10 and 4,5 - so you can try them in turn
remember the factor theorem - if (x-a) is a factor then putting x=a into the expression will result in zero - - her you put in x2 = -10 ( it works) - so x2 + 10 is a factor
well that is what fatty Barrett ( my maff teacher) taught us in 1968
and Jim you can use the factor theorem to verify there are no integer solutions ( I remember! - took me some time) in your expressions which is the usual entry for students to Galois Theory innit?
you only have fourth powers and squares
so any factor has to have squares in it and not x power one
( other wise there would be cubes and exes)
if there are whole number answers ( we think there are) then the factors are plus or minus ( 1,2,4,5,10,20) - in pairs 1,20 and 2,10 and 4,5 - so you can try them in turn
remember the factor theorem - if (x-a) is a factor then putting x=a into the expression will result in zero - - her you put in x2 = -10 ( it works) - so x2 + 10 is a factor
well that is what fatty Barrett ( my maff teacher) taught us in 1968
and Jim you can use the factor theorem to verify there are no integer solutions ( I remember! - took me some time) in your expressions which is the usual entry for students to Galois Theory innit?
// "the only problems we like to solve are quadratic equations",// hi Jim in my M 203 course er 1985 - we had in the exam with notice a 3x3 matrix to normalise
eigen values and all that and there would be a cubic carefully crafted to give three real integer roots.
( one student made a mistake and factored a different cubic which was so wonderful the examiners used it the next year ....)
hence my observation - these probs have an answer
eigen values and all that and there would be a cubic carefully crafted to give three real integer roots.
( one student made a mistake and factored a different cubic which was so wonderful the examiners used it the next year ....)
hence my observation - these probs have an answer
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