The cubic equation x^3 + ax^2 + bx 26 = 0 has 3 positive, distinct, integer roots. Find the values a and b.
I don't understand how to work this out.
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Offline2ReputationRep:What factor would I use?

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If the equation has three distinct roots, you will be able to factorise it into something of the form:
Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, fgh. You know that this must equal 26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full. 
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Offline2ReputationRep:(Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that 
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Offline2ReputationRep:(Original post by porkstein)
If the equation has three distinct roots, you will be able to factorise it into something of the form:
Where f, g and h are your solutions. When you multiply this all out, you're going to get a bunch of different terms. Importantly, you will find that you have only one constant term, fgh. You know that this must equal 26, which tells you what your three solutions must be (here you can apply the restriction that the solutions are distinct and positive integers). Put your solutions back into the factorised form and multiply out in full. 
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Offline2ReputationRep:I got 1,2,13, can't it be 1,2, 13?

Offline1ReputationRep:(Original post by Math12345)
I got 1,2,13, can't it be 1,2, 13?
If 1, 2, and 13 are the roots, then you'll have the cubic equation:
But , so 1, 2 and 13 can't be the three roots (plus they're not even positive in the first place) 
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Offline2ReputationRep:(Original post by Blazy)
Notice the condition upon the question  they are three POSITIVE, distinct integer roots.*
If 1, 2, and 13 are the roots, then you'll have the cubic equation:
But , so 1, 2 and 13 can't be the three roots (plus they're not even positive in the first place) 
Offline2ReputationRep:(Original post by Math12345)
I wish I understood this.
It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):
Means:
a_{4} just represents some general constant that is multiplied by the x^{4} term. You could call it "the x^{4} coefficient". As such the expression above involving that capital sigma (the big Elike symbol) is just a way of expressing some general polynomial (of degree n).
, which should really be written as
Means:
So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a_{0}, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on just how many things you multiply together.
if
This is just a fancy way of saying that the roots are all different. The second root of the equation, r_{2}, is not equal to the fifth root of the equation, r_{5}. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.
This is just a fancy way of saying the roots are integers. 
Offline3ReputationRep:(Original post by porkstein)
TheMagicMan is being almost deliberately confusing, introducing notation that simply isn't necessary for this question. He's essentially just expressing what the question says in words with symbols, presumably because that satisfies him or something. I can't see how he could've thought it would help you in any way.
It really isn't relevant, but in case you'd like to know what he means (it's not as hard as it looks):
Means:
a_{4} just represents some general constant that is multiplied by the x^{4} term. You could call it "the x^{4} coefficient". As such the expression above involving that capital sigma (the big Elike symbol) is just a way of expressing some general polynomial (of degree n).
, which should really be written as
Means:
So looking back at the equation involving this capital pi, what we're trying to say is that the product of all the roots of the equation is equal to a_{0}, the constant term of the polynomial. This is similar to what I said about fgh being equal to 26. The (1)^n bit just comes about because if, say, 5 is a root of the equation, then in the factorised form it's going to appear as (x minus 5). When all these roots are multiplied together you'll potentially get a minus term floating around, depending on how many things you multiply together.
if
This is just a fancy way of saying that the roots are all different. The second root of the equation, r_{2}, is not equal to the fifth root of the equation, r_{5}. Nor is the fifth root equal to the fourth root, and so on. The i just stands in for any number, as does j. When you run through all the combinations of roots, if you ask yourself, say, whether the third root of the equation equals the third root of the equation, the answer is obviously yes. Thus the i =/= j just says that the roots are all different, providing you don't consider the same root twice in your comparisons.
This is just a fancy way of saying the roots are integers.
Also, when there is an 'obvious' range for the summation and product limits, there is really no need to put them on.
Anyhow, like most people here, I'm just trying to help, so... 
Offline2ReputationRep:(Original post by TheMagicMan)
...
And if you're going to leave the limits off, at least be consistent in doing so. 
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Can somebody please post a solution with the answers
edit: nevermind 
Offline0ReputationRep:
I am sorry but why is everone confusing this kid on such a basic question.
If it is a cubic and the constant is 26, then the factors have to make 26.
Now let's take all the factors of 26; they are 1,2,13,26.
A factor can not be 26, since then you would have 1 and 1 as the other two factors which aren't distinct. Therefore, the three factors are 1,2,13.
Therefore we see the equation can be written in this format:
(x1)(x2)(x13)=0
The roots of the above equation are x=1,2,13: Three positive, distinct, integer roots!
Now simply expand the brackets and you get:
x^3โ16x^2+41xโ26 = 0.
Therefore, a =16 and b=41 
Offline0ReputationRep:(Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that 
Offline0ReputationRep:(Original post by Blazy)
Notice the condition upon the question  they are three POSITIVE, distinct integer roots.*
If 1, 2, and 13 are the roots, then you'll have the cubic equation:
But , so 1, 2 and 13 can't be the three roots (plus they're not even positive in the first place) 
Offline3ReputationRep:(Original post by GreenLantern1)
Don't try and act really clever with this. You are being annoying by trying to look so great and confusing the kid, when there is a very easy solution which you have failed to find with your "complex" method. 
Offline0ReputationRep:(Original post by TheMagicMan)
It's not complex...it's exactly the same as whatever else has been suggested 
Online3ReputationRep:(Original post by TheMagicMan)
It's not complex...it's exactly the same as whatever else has been suggested
When you help someone with maths (in TSR or anywhere else), you should first consider the level and ability of the person that you are helping. Sometimes, you may think that you're helping out but really, you're just making them more confused. 
Offline0ReputationRep:(Original post by TheMagicMan)
If are the roots of the polynomial , then . Use the fact that if , and that
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