Question

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

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

Answer 1

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

$\frac{x^4}{(x-a)(x-b)(x-c)}=\frac{x^4}{x^3-(a+b+c)}$
$x^2+{\sum }_1^{1}abx$
division of polynomids (- $\pi$a
$x^3-\left({\sum }_1^1\right)x^2+\left({\sum }_1^{1}ab\right)x$
$-\left(\pi a\right)/(x+{\sum }_1^1)a$
$-\left({\sum }_1^{1}a\right)x^3+\left({\sum }_1^{1}ab\right)x^2$
$-\left(\pi a\right)x$
${\sum }_1^{1}a{x}^3-\left({\sum }_1^{1}ab\right)x^2+\left(\pi a\right)x$
${\sum }_1^{1}a{x}^3-({\sum }_1^{1}a{)}^2{x}^2+({\sum }_1^{1}ab\left)\right({\sum }_1^{1}a)$
$+\left(\pi a\right)\left({\sum }_1^{1}a\right)$
$\left[\right({\sum }_1^{1}a{)}^2-\left({\sum }_1^{1}ab\right)]x^2$
$+\left[\pi a\right({\sum }_1^{1}ab\left)\right({\sum }_1^{1}a\left)\right]x$
$-\left(\pi a\right)\left({\sum }_1^{1}a\right)$
$=(x+{\sum }_1^{1}a)+\frac{\theta \left(x\right)}{(x-a)(x-b)(x-c)}$
So, $p\left(x\right)=x+{\sum }_1^{1}a=x+a+b+c$