How do you solve the polynomial inequality and state the answer in interval notation given #(x^3+20x)/8>=x^2+2#?

The steps to solving any nonlinear inequality are:

• Rewrite the inequality so that there is a zero on the right side, in the form #f(x) lt 0# or #f(x) gt 0# (or #f(x) le 0# or #f(x) ge 0#).
• Factorise the function, #f(x)# if possible.
• Find the critical values where a sign change can occur by finding the roots of #f(x)=0#, (or any factor of #f(x)=0#).
• Determine the sign of #f(x)# in the intervals formed by the critical values.

The solution will be those intervals in which the function has the correct signs satisfying the inequality.

We can rearrange the equation as follows:

# (x^3+20x)/8 ge x^2+2 #

# :. (x^3+20x)/8 - x^2+2 ge 0 #

# :. (x^3+20x) - 8(x^2+2) ge 0 #

# :. x^3+20x - 8x^2-16 ge 0 #

By inspection we see that #x-2# is a factor of the cubic, and so we can use algebraic long division to get

# (x-2)(x^2-6x+8) ge 0 #

So, now we can fully factorise to get:

# (x-2)(x-2)(x-4) ge 0 #
# :. (x-2)^2(x-4) ge 0 #

Which we can write as:

# f(x) ge 0 # where #f(x)=(x-2)^2(x-4) #

So, the critical points where a sign change can occur, are:

# x -2 = 0 => x = 2 #
# x-4 = 0 => x = 4 #

We must now examine the sign of #f(x)# in each interval, partitioned by these critical points i.e:

# x le 2; \ \ 2 le x le 4; \ \ x ge 4 #

The easiest way to do this is to look at the sign of each factor (or component) of #f(x)# via a sign chart

# {: ( ul("factor"), ul(x lt 2), ul(2 lt x le 4), ul(x gt 4) ), ( x-2, -, +, + ), ( (x-2)^2, +, +, + ), ( x-4, -, -, + ), ( ul(" "), ul(" "), ul(" "), ul(" ") ), ( f(x), -,- ,+ ) :} #

So, overall, we have established that the function, #f(x) lt 0# if #x gt 4#,

In addition for the solutions #f(x)=0#, we have:

# (x-2)^2(x-4) = 0 => x=2,4#

Hence, the solution of the inequality is:

#x=2, x=4, x gt 4 #

Or:

#x=2, x ge 4 #