



 
Hi Tehmas, In this problem it's not the intercepts that are important, it's the xcoordinates of the places where the graphs cross. Let me illustrate with a simpler question.
I first solve $x^2  2 = x$ to determine where the graphs cross. Simplifying and solving I found that the graphs cross at $x = 1$ and $x = 2.$ I now sketch a graph. In my case $f(x) = x^2  2$ (a parabola) and $g(x) = x$ (a line) are simple enough that I can graph them easily. Asking where $x^2  2 > x$ is asking for the region or regions where the parabola lies above the line. It is clear from the graph that there are two such regions $\{x\; \; x < 1\}$ and $\{x \;\; x > 2\}.$ Penny Tehmas wrote back
If you understand the example then I think the most difficult and labourious part of the solution is finding where the graphs intersect, that is solving \[2x^4+x^337x = 26x^2+12\] or \[2x^4+x^3  26 x^2 37x 12 = 0.\] This problem is designed so that some of the intersection points have integer xcoordinates so that you can use the Factor Theorem to factor the quartic polynomial. Once you have the intersection points solving the inequality is straightforward. Suppose that $x = a$ and $x = b$ are the xcoordinates of two consecutive points where the graphs intersect. Choose some number $c$ between $a$ and $b$ and evaluate $f(x) = 2x^4+x^337x$ and $g(x) = 26x^2+12$ at $x = c.$ If $f(c) > g(c)$ then $f(x) > g(x)$ for the entire interval $a < x < b$ since the graphs don't meet again until $x = b.$ Likewise if $g(c) > f(c)$ then $g(x) > f(x)$ for all x satisfying $a < x < b.$ This is a example where I would use some software either to help with the calculations or to check my work. I used Wolfram Alpha and typed
into the input field. Penny  


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