If F_1 and F_2 are irreducible factors of x^4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We have,$x^4+x^2+1 = x^4+2x^2+1-x^2 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$.Let $F_1 = x^2+x+1$ and $F_2 = x^2-x+1$.Then,$\frac{x^3-2x^2+3x-4}{x^4

Vedclass · Accepted Answer

-$3$

Vedclass · Answer

(C) We have,$x^4+x^2+1 = x^4+2x^2+1-x^2 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$.Let $F_1 = x^2+x+1$ and $F_2 = x^2-x+1$.Then,$\frac{x^3-2x^2+3x-4}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.Multiplying both sides by $x^4+x^2+1$,we get:$x^3-2x^2+3x-4 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$.Expanding the right side:$x^3-2x^2+3x-4 = (A+C)x^3 + (B-A+C+D)x^2 + (A-B+C+D)x + (B+D)$.Comparing coefficients of like powers of $x$:$1) A+C = 1$$2) B-A+C+D = -2$$3) A-B+C+D = 3$$4) B+D = -4$We want to find $A+B+C+D$. From equation $(1)$,$A+C=1$. From equation $(4)$,$B+D=-4$.Therefore,$A+B+C+D = (A+C) + (B+D) = 1 + (-4) = -3$.

If $F_1$ and $F_2$ are irreducible factors of $x^4+x^2+1$ with real coefficients and $\frac{x^3-2x^2+3x-4}{x^4+x^2+1}=\frac{Ax+B}{F_1}+\frac{Cx+D}{F_2}$,then $A+B+C+D=$

Similar Questions

The partial fractions of $\frac{x^2 - 5}{x^2 - 3x + 2}$ are

If $\frac{1}{(3x+1)(x-2)}=\frac{A}{3x+1}+\frac{B}{x-2}$ and $\frac{x+1}{(3x+1)(x-2)}=\frac{C}{3x+1}+\frac{D}{x-2}$,then

$\frac{x^4}{x^3-3x+2}$ is a

If $\frac{x-4}{x^2-5x+6}$ can be expanded in the ascending powers of $x$,then the coefficient of $x^3$ is

$\text{If } \frac{3x^2+1}{(x^2+1)(x^2+2)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+2} + \frac{Ex+F}{(x^2+2)^2}, \text{ then } A+C+E = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module