When the polynomial p(x) = x3  + 3x2  -2ax + 3, where a is a constant, is divided by x2  + 1 we get a remainder equal to -5x. find a. 

Interesting problem!

P(x)=x^3+3*x^2-2Ax+3

is a third degree polynomial.

The divisor,

D(x)=x^2+1

is a second degree polynomial.

We are given

P(x)/D(x) has a remainder of R(x)= -5x, note that the constant term of the remainder is zero.

The fact that the the quotient Q(x) is a linear expression, and constant term of R(x) equals zero allows us to deduce that the quotient equals Q(x)=x^3/x^2 + 3/1 = x+1, i.e. the sum of the quotients of the leading coefficient divided by that of D(x), or x^2, and the constant term divided by the constant term of D(x), or 1.

Using Q(x)=x+1, the product of the divisor D(x) and the quotient Q(x) added to the remainder R(x) should be equal to the original polynomial P(x), i.e.

P(x) = D(x)*Q(x)+R(x)

=(x^2+1)(x+3)-5x

=x^3+3x^2-4x+3

(To check, divide (using synthetic division) P(x) by D(x) and we should get R(x))

Now solve for A

-2A = -4

A = (-4)/(-2) = 2

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step-by-step explanation: