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Given polynomial f=x^4+ax^3+bx+c and g divide f, g=(x^2-1)(x-t), show that for any t...

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greenbel | Honors

Posted June 28, 2013 at 1:54 PM via web

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Given polynomial f=x^4+ax^3+bx+c and g divide f, g=(x^2-1)(x-t), show that for any t real g divide f?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 28, 2013 at 2:25 PM (Answer #1)

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The problem provides the information that polynomial `g(x)` is divisor of polynomial `f(x)` , hence, each factor of `g(x)` is a divisor of `f(x)` .

Hence, if `(x^2 - 1)` is a divisor of `f(x)` , then `f(+-1) = 0` , thus, replacing `+-1` for x in equation of `f(x)` yields:

`f(1) = 1 + a + b + c = 0 => a + b + c = -1`

`f(-1) = 1 - a - b + c = 0 => -a - b + c = -1`

Adding f(1) and f(-1) yields:

`f(1) + f(-1) = 2c = -2 => c = -1 => a + b = 0 => a = -b`Replacing `-a` for `b` and `-1` for `c` in polynomial `f(x)` yields:

`f(x) = x^4 + ax^3 - ax - 1`

You need to group the terms, such that:

`f(x) = (x^4 - 1) + (ax^3 - ax)`

`f(x) = (x^2 - 1)(x^2 + 1) + ax(x^2 - 1)`

Factoring out `(x^2 - 1)` yields:

`f(x) = (x^2 - 1)(x^2 + ax + 1)`

Since `g(x)` is a divisor of `f(x)` , hence, the factor `(x - t)` is a divisor of `f(x)` , thus `(x - t)` is a divisor of factor `h(x) = x^2 + ax + 1` , such that:

`h(t) = t^2 + at + 1 = 0 => t_(1,2) = (-a+-sqrt(a^2 - 4))/2`

Hence, since the coefficient a is a real number, then t is also a real number.

Hence, testing the validity of the statement that for all `t in R` , the polynomial `g(x)` is a divisor of polynomial `f(x)` , yields that `f(x)` is divided by `g(x)` for all `t in R` .

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