# Use the remainder theorem to: find the remainder when f(x) is divided by x – 6, and determine whether x – 6 is a factor of f(x). f(x)=x^4 -5x^2 -36; x-6

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You need to remember that the reminder of a polynomial divided by a binomial x-a is a constant value r(a), hence if r(a)=0, then x-a express a factor of polynomial.

Notice that the polynomial `x^4 -5x^2 -36` is divided by binomial x - 6, hence, considering the reminder theorem, all you need to do is to substitute 6 for x in polynomial to evaluate the reminder such that:

`r(6) = 6^4 - 5*(6)^2 - 36 =gt r(6) = 1296 - 180 - 36`

`r(6) = 1080`

You may also use the alternative method of changing the variable such that `x^2 = y` , hence you need to solve for y the equation:

`y^2 - 5y - 36 = 0`

`y_(1,2) = (5+-sqrt(25+144))/2 =gt y_(1,2) = (5+-13)/2`

`y_1 = 9; y_2 = -4`

You need to solve for x the equation `x^2 = 9 =gt x_(1,2)=+-3` `x^2 = -4 =gt x_(3,4) = +-sqrt(-4) =gt x_(3,4) = +-2i`

**Notice that none of the roots of the equation `x^4 -5x^2 -36 = 0` is 6, hence x - 6 is not a factor for `x^4 -5x^2 -36` , and the reminder is the constant value `r(6)=1080` **.

`f(x)= x^4 - 5x^2 - 36`

We will use the remainder theorim :

==> f(x)=(x-6)*Q(x) + R

==> Substitue with x = 6

`==gt f(6) = (6-6)*Q(6)+ R `

`==gt f(6) = R `

`==gt 6^4 - 5(6^2) -36 = R `

`==gt 1080 = R`

`` **==> The remainder is R= 1080**

**Since the remainder is NOT zero, then (x-6) is NOT a factor of f(x).**