Number of real roots of equation
(x+1) (x+2) (x+3) (x+4) -8 =0 is

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Solution

The correct option is A 0
$\begin{matrix} (x + 1) (x + 2) (x + 4) = 8 x^{4} + 10^{3} + 35 x^{2} + 50 x + 16 = 0 F r o m O e s c a n t e s r u l e o f s i g n o f s i g n T h e r e w i l l b e n o p o s i t i v e r o o t s f (- x) = x^{4} - 10 x^{3} + 35 x^{2} - 50 x + 60 = 0 a n d p o s i b i l i t y o f n e g a t i v e r o o t s a n d 0, 2 o r 4 b u t n o n e g a t i v e n u m b e r m a k i n g t h i s e q u a t i o n^{'} 0^{'} s o i t h a s n o r e a l r o o t s \end{matrix}$

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Number of real roots of equation (x+1) (x+2) (x+3) (x+4) -8 =0 is

The correct option is A 0(x+1)(x+2)(x+4)=8x4+103+35x2+50x+16=0FromOescantesruleofsignofsignTherewillbenopositiverootsf(−x)=x4−10x3+35x2−50x+60=0andposibilityofnegativerootsand0,2or4butnonegativenumbermakingthisequation′0′soithasnorealroots

Number of real roots of equation
(x+1) (x+2) (x+3) (x+4) -8 =0 is