Question

Find the values of x that satisfies the following inequality: 4x+3/x^3+2x^2+3x+6<2/x-3

Answer 1

\[\frac{ 4x+3 }{ x^3+2x^2+3x+6 } <\frac{ 2 }{ x-3 }\]

Answer 2

hw doI begin?

Answer 3

carry out cross multiplication i.e.
(4x+3)(x-3)<2(x^3+2x^2+3x+6)

Answer 4

ok then...

Answer 5

cancel the like terms from both sides having same sign and solve the remaining problem

Answer 6

\[\frac{4x+3}{x^3+2x^2+3x+6}< \frac{2}{x-3}\]
i.e. \[(4x+3)(x-3)<2(x^3+2x^2+3x+6)\]
\[4x(x-3)+3(x-3)<2x^3+4x^2+6x+12\]
\[\rightarrow 4x^2-12x+3x-9<2x^3+4x^2+6x+12\]
in the next step we will transfer variables on LHS and constants on RHS
\[4x^2-12x+3x-2x^3-4x^2-6x<+12 +9\]
\[\rightarrow -15x-2x^3<21 \rightarrow 2x^3+15x<-21 \rightarrow 2x^3+15x+21<0\]

Answer 7

I am not able to see the fraction in proper order. What shld I do?

Answer 8

[\frac{ -15x-2^3-21 }/{ (x+2)(x^2+3)(x+3)} < 0\]\]

Answer 9

where are you stuck?

Answer 10

nw, hw to solve for x values?
is it x = -3,-2,3 not be included

Answer 11

the inequality you should end up with is:\[2x^3+15x+21<0\]solving this depends on what you have been taught in class regarding cubic equations.

Answer 12

is it use factor theorem?

Answer 13

I need help...

Answer 14

use whatever theorem(s) you have been taught in class.