Functions: Higher degree polynomials
Higher degree inequalities
In the same manner as when solving a quadratic inequality, we can also solve an inequality with higher degree polynomials.
Solving a higher degree inequality
Procedure  Example  
We solve the following inequality \[\blue{f(x)} \gt \green{g(x)}\] in which #\blue{f(x)}# and #\green{g(x)}# are polynomials.  #\blue{x^6+x^3+6}\ (solid)\gt \green{2x^3+10}\ (dashed)# The solution is #x \lt \sqrt[3]{4} \land x \gt 1#. 

Step 1  We solve the equality \[\blue{f(x)} = \green{g(x)}\]  
Step 2  We sketch the graphs #\blue{f(x)}# and #\green{g(x)}#.  
Step 3  With the help of step 1 and 2, determine for which values of #x# the inequality holds. In a coordinate system, the biggest graph is the one above the other. 
Please note that this procedure also holds for the inequality signs #\geq# and #\leq#, only now the #x#values of the intersection points are also part of the solution.
#q\gt 5^{{{1}\over{7}}}\land q\lt 1#
Step 1  We solve the equality #q^{14}+6\cdot q^7+10\cdot q+5=10\cdot q#. This is done like this: \[\begin{array}{rcl} q^{14}+6\cdot q^7+10\cdot q+5&=&10\cdot q \\ &&\phantom{xxx}\blue{\text{original equation}}\\ q^{14}+6\cdot q^7+5&=&0 \\&&\phantom{xxx}\blue{\text{reduced to }0}\\ \left(q^7+1\right)\cdot \left(q^7+5\right)&=&0 \\&&\phantom{xxx}\blue{\text{left hand side factorized}}\\ q^7+1=0 &\lor& q^7+5=0 \\&&\phantom{xxx}\blue{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ q=5^{{{1}\over{7}}} &\lor& q=1 \\&&\phantom{xxx}\blue{\text{constant terms to the right hand side and taken the root}}\\ \end{array} \] 
Step 2  We sketch the graphs #y=q^{14}+6\cdot q^7+10\cdot q+5# (blue) and #y=10\cdot q# (green dashed). 
Step 3  We can read the solutions to the inequality from the graph. \[q\gt 5^{{{1}\over{7}}}\land q\lt 1\] 
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