Answered By Prabhat K

 $\frac{P\left(x\right)}{x-a}=Q\left(x\right)+\frac{R}{\left(x-a\right)}$P(x)x−a =Q(x)+R(x−a)  (a) Multiply both sides by x-a: 

 $\frac{P\left(x\right)}{\left(x-a\right)}\left(x-a\right)=\left(Q\left(x\right)+\frac{R}{\left(x-a\right)}\right)\left(x-a\right)$P(x)(x−a) (x−a)=(Q(x)+R(x−a) )(x−a) 

On the left hand side, the x-a on top and bottom cancel each other out.On the riight hand side, lets distribute (x-a) inside the bracket by multiplying each term by x-a:  $P\left(x\right)=Q\left(x\right)\left(x-a\right)+\frac{R}{\left(x-a\right)}\left(x-a\right)$P(x)=Q(x)(x−a)+R(x−a) (x−a) On the very last term, x-a cancels out on bottom and top since it is being multipled together. Leaves you with:  $P\left(x\right)=Q\left(x\right)\left(x-a\right)+R$P(x)=Q(x)(x−a)+R (b) P(a) means substitute x=a into the equation we got above:  $P\left(a\right)=Q\left(a\right)\left(a-a\right)+R$P(a)=Q(a)(a−a)‍+R  a-a=0, so that term will be 0: P(a) = R(c) The answer to part b relates to the remainder theorem because it yeilds the remainder of the question. If you are dividing something like x-4 to a quadratic function or a function of higher order, a=4 substituted into the polynomial (dividend) will give you the remainder of the result. Review the labels on what means what in the remainder theorem. P(x) is the dividend, Q(x) is resulting quotient and R is the remainder. (x-a) is the divisor. (d) The "R" is a symbol for remainder. Therefore, if there is no remainder, R=0 and that means the divisor is a FACTOR of the dividend polynomial. When R=0:   $P\left(x\right)=Q\left(x\right)\left(x-a\right)+R$P(x)=Q(x)(x−a)+R  $P\left(x\right)=Q\left(x\right)\left(x-a\right)$P(x)=Q(x)(x−a)    

As can be seen, the remaining quotient and (x-a) are both factors of polynomial P(x), this is the factor theorem.