## Algebra 2 Tutorial

*Remainder Theorem*

#### Intro

the problem exploits the fundamental concept of the remainder theorem. if a polynomial f(x) is divided by Q(x), then the remainder R(x) is a polynomial whose degree is less than the degree of Q(x). If Q(x) is a quadratic expression, then R(X) is at most a linear expression.Also, if a polynomial f(x) is divided by x-a, the remainder is f(a).

#### Sample Problem

Given that the polynomial f(x) leaves a remainder of 5 when divided by x-1 and leaves a remainder of 1 when divided by x+3, what is the remainder when f(x) is divided by the product

(x-1)(x+3)?

#### Solution

f(x) is divided by Q(x)=(x-1)(x+3), then the remainder R(x) has the form

R(x) = ax + b, where a and b are constants.

so,

f(x) = (x-1)(x+3)A(x) + R(x) = (x-1)(x+3)A(x) + ax +b, where A(x) is the quotient.

when f(x) is divided by x-1 and x+3, the remainders are f(1)=5 and f(-3)=1 respectively.

Now,

f(1)= (1-1)(1+3)A(1) +a+b = 5

f(-3)= (-3-1)(-3+3)A(-3) + a(-3) +b =1

we have

a +b =5 ----------- (1) -3a + b=1 ----------(2)

solving both equations simultaneously we obtain;

(1) – (2) gives, 4a = 4. Hence a = 1

substituting the value of a, in equation (1) we obtain: 1 + b= 5

so b = 4.

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