# PolynomialsDetermine P(x) if P(3)=2 and if P(x) is divided by x-3 the quotient is x^2+3x-5

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You need to use reminder theorem such that:

`P(x) = (x - 3)(x^2 + 3x - 5) + R(x)`

`R(x)` represents the reminder and its degree needs to be smaller than degree of polynomial `x - 3` , hence R(x) = constant = a

`P(x) = (x - 3)(x^2 + 3x - 5) + a`

The problem provides the information that `P(3) = 2` , hence, susbtituting 2 for x in equation above yields:

`P(3) = (3 - 3)(3^2 + 9 - 5) + a`

`P(3) = 0 + a => P(3) = a => a = 2`

**Hence, evaluating the polynomial `P(x)` , under the given conditions, yields **`P(x) = (x - 3)(x^2 + 3x - 5) + 2.`

Since the quotient is of 2nd order and the divisor is of 1st order, then the order of the polynomial P(x) is :

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

P(x)'s order = 2 + 1

P(x)'s order = 3 order

We'll write the rule of division with reminder:

P(x) = quotient * Divisor + remainder

We know that if P(3) = 2, then the reminder is R = 2. We could write P(x) as:

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

We'll remove the brackets:

P(x)= x^3 - 3x^2 + 3x^2 - 9x - 5x + 15 + 2

We'll eliminate and combine like terms:

**P(x) = x^3 - 14x + 17**