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How does the remainder of P(x)/x-a=Q(x)+R/x-a compare to R/x-a

What does the remainder of this division represent?

4 years ago

Answered By Leonardo F

Basically, we need to start from here: if a polynomial P(x) is divided by a divisor D(x), resulting in a quocient Q(x) and a remainder R(x), the mathematical expression is:

 $P\left(x\right)=D\left(x\right)\times Q\left(x\right)+R\left(x\right)$P(x)=D(x)×Q(x)+R(x) 

The result of the division of a polynomial in x, P(x), by a binomial of the form  $x-a,a\text{ ∈ }I,$xa,aI, is  $\frac{P\left(x\right)}{x-a}=Q\left(x\right)+\frac{R}{x-a}$P(x)xa =Q(x)+Rxa  , where Q(x) is the quotient and R is the remainder.

Hence, the remainder of the division stated in the problem is R. It compares to  $\frac{R}{x-a}$Rxa  with the following equation:

Remainder$=R=$=R=  $\frac{R}{x-a}\times\left(x-a\right)$Rxa ×(xa) 

This means that we must multiply the term  $\frac{R}{x-a}$Rxa   by   $x-a$xa to get the remainder of the division stated in the problem.

We can check the division of a polynomial P(x) by multiplying the quotient, Q(x), by the binomial divisor, x - a, and adding the remainder, R. The result should be equivalent to the polynomial dividend, P(x):

 $P\left(x\right)=\left(x-a\right)Q\left(x\right)+R$P(x)=(xa)Q(x)+R 

The remainder of the division is the polynomial "left over" after dividing polynomial P(x) by "x-a". If "a" is a root or zero of the polynomial (P(a)=0), then the remainder (R) will be zero.