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itskratika_07

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If a and b are non-zero, and a + b = 0, prove that the polynomials A(x) = x^3 + ax^2 − x + b and
B(x) = x^3 + bx^2 − x + a have a common factor of degree 2 but are not identical polynomials. What
is the common factor?
 

synthesisFR

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If a and b are non-zero, and a + b = 0, prove that the polynomials A(x) = x^3 + ax^2 − x + b and
B(x) = x^3 + bx^2 − x + a have a common factor of degree 2 but are not identical polynomials. What
is the common factor?
I’m in bed so I can’t be bothered doing the question but list all the conditions the question gives first and then see how they can be used
 

KT878

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If a and b are non-zero, and a + b = 0, prove that the polynomials A(x) = x^3 + ax^2 − x + b and
B(x) = x^3 + bx^2 − x + a have a common factor of degree 2 but are not identical polynomials. What
is the common factor?
Could you add A(x) and B(x) to get 2x^3 + (a+b)x^2 - 2x + (a+b)
Then sub a+b = 0
A(x) + B(x) = 2x^3 - 2x = 2x(x^2 -1)
So the factor would be x^2 - 1 since if both A(x) and B(x) are divisible by the factor then their sum must also be divisible by the same factor.
Note that x=0 is not a factor as a cannot equal b which cannot equal 0.
 

gazzaboy

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Could you add A(x) and B(x) to get 2x^3 + (a+b)x^2 - 2x + (a+b)
Then sub a+b = 0
A(x) + B(x) = 2x^3 - 2x = 2x(x^2 -1)
So the factor would be x^2 - 1 since if both A(x) and B(x) are divisible by the factor then their sum must also be divisible by the same factor.
Note that x=0 is not a factor as a cannot equal b which cannot equal 0.
This is a pretty neat idea! If we were being pedantic though, it hasn't been technically proven that A(x) and B(x) have a common factor. You've just shown that if there IS a common factor, it must be . Just because a term is a factor of the sum of two polynomials doesn't mean it is a factor of the original two polynomials. You still need to show that A(x) and B(x) actually have a common factor of degree 2.

A more standard way to do it would be to use the substitution a = -b in A(x) to get . You can then try to factorise this by grouping the right terms to get .

Similarly, , and you can see that x^2 - 1 is a common factor. Since , they are not identical polynomials.
 

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