# What is f(2) if dividing f(x) by (x-1) the quotient is x^2-2x+3 and the reminder is 1?

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When we divide f(x) by (x - 1) the quotient is x^2 - 2x + 3 and the remainder is 1.

f(x) / (x - 1) = (x^2 - 2x + 3) + 1/(x - 1)

=> f(x) = (x^2 - 2x + 3)(x - 1) + 1

f(2) = (2^2 - 2*2 + 3)/(2 - 1) + 1

=> f(2) = 4 - 4 + 3 + 1

=> f(2) = 4

**The required value of f(2) = 4**

if dividing f(x) by (x-1) and the quotient is x^2-2x+3 and the reminder is 1

f(2)will be:

f(x) / (x - 1) = (x^2 - 2x + 3) + 1/(x - 1)

f(x) = (x^2 - 2x + 3)(x - 1) + 1

f(2) = (2^2 - 2*2 + 3)/(2 - 1) + 1

=4.

To determine the value of the polynomial f(x), when x=2, we'll have to find out the expression of f(x).

We'll determine the order of polynomial f(x), based on the given orders of, divisor, quotient and reminder.

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 reminder theorem:

f(x) = divisor*quotient + remainder

f(x) = (x-1)*(x^2 -2x+3)+ 1

We'll replace x by 3:

f(2) = (2-1)*(2^2 -4+3)+ 1

**The value of the polynomial f(x), if x = 2, is: f(2) = 4.**