Question

If $\frac{x^4}{(x-a)(x-b)(x-c)}=P\left(x\right)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$ then P (x) =

Answer 1

$(x-a)(x-b)(x-c)=[x^2-(a+b)x+ab](x-c)$
$=x^2-c{x}^2-(a+b)x^2+c(a+b)x+abx-abc$
$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$

$\frac{x^4}{(x-a)(x-b)(x-c)}=P\left(x\right)+\frac{A}{x-a}+\frac{A}{x-b}+\frac{C}{x-c}\therefore x^4=P\left(x\right)(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-5)+C(x+a)(x-b)$
Dividend = Quotient $\times$ Divisor + Remainder
P(x) = Quotient
$\therefore$ P(x) = x+a+b+c