Is a + b a Factor of a²(b + c) + b²(c + a) + c²(a + b) + 2abc?

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In summary, the conversation discusses proving that a + b is a factor of a²(b + c) + b²(c + a) + c²(a + b) + 2abc and finding the other two factors. The solution involves rearranging the expression and factoring it into (a + b) times another factor. The conversation also discusses the concept of a factor and how it relates to setting a or b equal to negative values.
  • #1
MorallyObtuse
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I tried, doubt I'm even close to correct. Show me where I went wrong or just guide me with the problem please.

Homework Statement


1.) Prove that a + b is a factor of a²(b + c) + b²(c + a) + c²(a + b) + 2abc and write down the other two factors.

2. The attempt at a solution

[tex]a^2(b - a) + b^2(-a + a) + (-a)^2(a + b) + 2ab(-a) = 0[/tex]
[tex]a^2b - a^3 - b^2a + b^2a + a^2 + ab - 2ab = 0[/tex]
[tex]a^2b - a^3 + a^2 - ab = 0[/tex][tex]a^2(-c + c) + (-c)^2(c +a) + c^2(a - c) + 2a(-c) = 0[/tex]
[tex]-a^2c + a^2c + c^2 + ac + c^2a - c^3 - 2ac = 0[/tex]
[tex]c^2 + c^2a - c^3 - ac = 0[/tex]

[tex](-b)^2(b + c) + b^2(c - b) + c^2(-b + b) + 2(-b)bc = 0[/tex]
[tex]b^2 + bc + b^2c - b^3 - c^2b + c^2b - 2bc = 0[/tex]
[tex]b^2 + b^2c - b^3 - bc = 0[/tex]
 
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  • #2
You're not given that a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc is equal to zero. All you're supposed to be doing is to rearrange what you have to put it into a factored form.

Having said that, how does the following expression relate to the expression above?
[tex]a^2(b - a) + b^2(-a + a) + (-a)^2(a + b) + 2ab(-a)[/tex]

Also, are you sure that you have typed the problem exactly as it was given to you? I was able to factor the given expression into (b + c) times another factor, but I haven't been able to write it yet as (a + b) times another factor.
 
  • #3
No, a+ b is a factor. Think of this as a polynomial in a with b and c constants. a+ b= a-(-b) will be a factor if and only if setting a= -b makes the polynomial equal to 0. And, of course, you can find the other factor by dividing by a+b.
 
  • #4
Hi!:smile:
Solution to this is quite simple..
1st open the brackets
:. you have
a2b+ab 2+b2c+c2b+a2c+c2a+2abc
=ab(a+b) +b2c+a2c+c2b+c2a+2abc=ab(a+b)+c 2(a+b)+b2c+a2c+2abc
= (ab+c 2)(a+b)+c(b 2+a 2+2ab)
=(ab+c 2)(a+b)+c(a+b)2
=(a+b)(c2+ab+ac+bc)
That's what I think.
(there could be some mistakes while typing in powers because I kinda get confused while typing them)
I hope this helps!:smile:
 
  • #5
Yes, I typed the question exactly as it was given.
I'm not sure how to 'prove'. The teacher keeps giving these proofs and I get baffled by them.
 
  • #6
HallsofIvy said:
a+ b= a-(-b) will be a factor if and only if setting a= -b makes the polynomial equal to 0.

a + b = 0, and so
a = -b or b = -a
 
Last edited:
  • #7
Hey what's the big deal then .
You "PROVE" it by giving the definition of factor and as it suits this condition :. You can say it is a factor of it.
See if it helps(?)
 

FAQ: Is a + b a Factor of a²(b + c) + b²(c + a) + c²(a + b) + 2abc?

What is the Factor Theorem?

The Factor Theorem is a theorem in algebra that helps to determine whether a given polynomial has a certain factor. It is used to factorize polynomials and solve polynomial equations.

How does the Factor Theorem work?

The Factor Theorem states that if a polynomial P(x) has a factor (x-a), then P(a)=0. This means that if we substitute the value of a into the polynomial and the result is 0, then (x-a) is a factor of P(x).

What is the proof of the Factor Theorem?

The proof of the Factor Theorem involves using polynomial division to show that if (x-a) is a factor of P(x), then the remainder when P(x) is divided by (x-a) is equal to 0. This can be shown using the long division method or synthetic division.

How is the Factor Theorem used in real-world applications?

The Factor Theorem is used in various real-world applications, such as in engineering and physics, to solve polynomial equations. It is also used in finance and economics to analyze and predict trends in data sets.

Are there any limitations to the Factor Theorem?

One limitation of the Factor Theorem is that it only works for polynomials with real coefficients. It also does not work for polynomials with complex roots. Additionally, it only applies to polynomials with integer exponents, not fractional or negative exponents.

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