# The polinomial f(x) divided x-3 results in a quotient of x^2+3x-5 and reminder of two. Find f(3).

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Given that the quotient of dividing f(x) by (x-3) , the results will be x^2 + 3x -5 and remainder of 2:

Then f(x) is a product of two functions, including (x-3), plus two.

==> f(x) = (x-3)*R(x) + 2

Let us write :

f(x) / (x-3) = (x^+ 3x -5) + 2

We will multiply by (x-3).

==> f(x) = (x-3)(x^2 +3x -5) + 2

Now to find f(3) we will substitute with x= 3 :

f(3) = (3-3)( 3^2 + 3&3 -5) + 2

==> f(3) = 0 + 2 = 2

Then, we know that:

**f(3) = 2**

By division algorithm if a/ b = q and remainder is r, then a = bq+r.

Given that the f(x)/(x-2) = x^2+3x-5 is the quotient and remainder is 2.

Therefore f(x) = (x-3)(x^2+3x-5) +2.

So to find the value of f(3), we just put x= 3 in f(x) = (x-3)(x^2+3x-5).

Therefore f(3) = (3-3){3^2+3*3-5} +2

f(3) = (0){....}+2

f(3) = 2.

Therefore f(3) = 2.

To determine f(3), first we need to find out the expression of f(x). Since the quotient is of 2nd order and the divisor is of 1st order, then the order of f(x) is :

f(x)'s order = quotient's order + divisor's order

f(x)'s order = 2 + 1

f(x)'s order = 3 order

We'll write the rule of division with reminder:

f(x) = quotient *Divisor + remainder

f(x)= (x^2 +3x-5)(x-3) + 2

We'll substitute x by 3:

f(3) = (9 + 9 - 5)(3 - 3) + 2

**f(3) = 2**