Question

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

• A. $x-a$
• B. $x-a-b$
• C. $x-a-b-c$
• D. $x+a+b+c$

Answer 1

The correct option is D $x+a+b+c$

Carefully observing LHS and RHS we can

say that $P\left(x\right)$ has to be a linear

equation, so let $P\left(x\right)=mx+n$

$\frac{x^4}{(x-a)(x-b)(x-c)}=(mx+n)+\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{c}{(x-c)}$

$\frac{x^4}{(x-a)(x-b)(x-c)}=(mx+n)(x-a)(x-b)(x-c)+A(x-b)(x-c)$

$\frac{+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)}$

Comparing cofficients of $x^4,x^3,x^2,x$ in both sides

$1=m=$ (1)

$0=-am-bm-cm+n$

$\Rightarrow 0=-a-b-c+n$

$\Rightarrow a+b+c=n-$ (2)

$\therefore P\left(x\right)=mx+n$

$=x+(a+b+c)$

$\therefore$ Answer option (D)