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Using Remainder theorem find the remainder for (5x^4+4x^2+1)/(4x^2 - 1)

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xetaalpha2 | Student | (Level 2) Honors

Posted July 21, 2012 at 5:31 PM via web

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Using Remainder theorem find the remainder for (5x^4+4x^2+1)/(4x^2 - 1)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 21, 2012 at 5:55 PM (Answer #1)

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You should remember the reminder theorem such that:

`p(x) = q(x)*d(x) + r(x)`

`p(x)`  represents the polynomial to be divided

`q(x)`  represents the quotient

`d(x)`  represents the divisor

`r(x)`  represents the reminder

You should come up with the following notations such that:

`p(x) = 5x^4+4x^2+1`

`d(x) = 4x^2 - 1`

You need to remember that the order of polynomial that represents the reminder is less than the order of the polynomial that represents the divisor.

Since the polynomial that represents the divisor is a quadratic polynomial, hence, the reminder will be represented by a linear function `r(x) = ax+b` .

You should find the roots of the divisor such that:

`4x^2 - 1 = 0 =gt 4x^2 = 1 =gt x^2 = 1/4 =gt x_(1,2) = +-1/2`

You need to substitute `1/2`  for x in reminder theorem such that:

`p(1/2) = q(1/2)*d(1/2) + r(1/2)`

Since `1/2`  is a root of d(x), hence `d(1/2) = 0`  such that:

`p(1/2) = q(1/2)*0 + r(1/2)`

`p(1/2) = r(1/2)`

`p(1/2) = 5/16 + 1 + 1 =gt p(1/2) = (5 + 32)/16 =gt p(1/2) = 37/16`

`r(1/2) = a/2 + b`

Hence, `a/2 + b= 37/16` .

You need to substitute `-1/2`  for x in reminder theorem such that:

`p(-1/2) = r(-1/2)`

`p(-1/2) = 37/16`

`r(-1/2) = -a/2 + b`

Hence, `-a/2 + b = 37/16` .

You need to add `-a/2 + b = 37/16 ` and `a/2 + b = 37/16`  such that:

`-a/2 + b + a/2 + b= 2*37/16`

`2b = 2*37/16 =gt b = 37/16`

`a/2 = 37/16 - 37/16 =gt a/2 = 0 =gt a = 0`

Hence, evaluating the reminder of the given division, using reminder theorem, yields `r(x) = 37/16` .

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