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

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$

If $\frac{ax^2 + bx + c}{(x - 1)(x + 2)(2x + 3)} = \frac{3}{x - 1} + \frac{2}{x + 2} - \frac{5}{2x + 3}$,then:

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

The partial fractions of $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4}$ are:

Resolve $\frac{2x}{x^4 + x^2 + 1}$ into partial fractions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module