If F1 and F2 are irreducible factors of x4+x2+1 with real coefficients and x2-2x2+3x-4x4+x2+1=Ax+BF1+Cx+DF2 then A+B+C+D=

x2-2x2+3x-4x4+x2+1=Ax+Bx2-x+1+Cx+Dx2+x+1x3-2x2+3x-4=(Ax+B)(x2+x+1)+(Cx+D)(x2-x+1)x3 coeff=A+C=1const B+D=-4A+B+C+D=-3

If F1 and F2 are irreducible factors of x4+x2+1 with real coefficients and x2-2x2+3x-4x4+x2+1=Ax+BF1+Cx+DF2 then A+B+C+D=

x2-2x2+3x-4x4+x2+1=Ax+Bx2-x+1+Cx+Dx2+x+1x3-2x2+3x-4=(Ax+B)(x2+x+1)+(Cx+D)(x2-x+1)x3 coeff=A+C=1const B+D=-4A+B+C+D=-3

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$- 2$

$- 3$

$- 4$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} \frac{x^{2} - 2 x^{2} + 3 x - 4}{x^{4} + x^{2} + 1} & = & \frac{Ax + B}{x^{2} - x + 1} + \frac{Cx + D}{x^{2} + x + 1} \\ x^{3} - 2 x^{2} + 3 x - 4 & = & (Ax + B) (x^{2} + x + 1) + (Cx + D) (x^{2} - x + 1) \\ x^{3} coeff & = & A + C = 1 \\ const B + D & = & - 4 \\ A + B + C + D & = & - 3 \end{array}$