If F1 and F2 are irreducible factors of x4+x2+1 with real coefficients and (x3-2 x2+3 x-4/x4+x2+1)=(A x+B/F1)+(C x+D/F2), then A+B+C+D=

Question

If F1 and F2 are irreducible factors of x4+x2+1 with real coefficients and (x3-2 x2+3 x-4/x4+x2+1)=(A x+B/F1)+(C x+D/F2), then A+B+C+D= (A) 2 (B) 1 (C) -3 (D) -4

Tardigrade Admin · Accepted Answer

We have, x4+x2+1 =x4+x2+x2-x2+1 =x4+2 x2+1-x2=(x2+1)2-(x)2 =(x2+x+1)(x2-x+1) Now, (x3-2 x2+3 x-4/x4+x2+1)=(A x+B/x2+x+1)+(C x+D/x2-x+1) ⇒ x3-2 x2+3 x-4=(A x+B) (x2-x+1)+(C x+D)(x2+x+1) ⇒ x3-2 x2+3 x-4=(A+C) x3 +x2(B-A+C+D)+x(A-B+C+D)+(B+D) Comparing of all like terms A +C=1 dots(i) B-A+C+D =-2 A-B+C+D =3 B+D =-4 dots(ii) Adding Eqs. (i) and (ii), we get A+B+C+D=1+(-4)=-3

Q. If F1​ and F2​ are irreducible factors of x4+x2+1 with real coefficients and x4+x2+1x3−2x2+3x−4​=F1​Ax+B​+F2​Cx+D​, then A+B+C+D=

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Q. If $F_{1}$ and $F_{2}$ are irreducible factors of $x^{4} + x^{2} + 1$ with real coefficients and $\frac{x ^{3} - 2 x ^{2} + 3 x - 4}{x ^{4} + x ^{2} + 1} = \frac{A x + B}{F _{1}} + \frac{C x + D}{F _{2}}$ , then $A + B + C + D =$