Algebra 2 Tutorial
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).
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.
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.
f(1)= (1-1)(1+3)A(1) +a+b = 5
f(-3)= (-3-1)(-3+3)A(-3) + a(-3) +b =1
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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