Question 755192
{{{h(x)=(3x^4-54x^2+96x-45)/(2x^3-x^2-32x+16)=3(x^4-18x^2+32x-15)/(2x^3-x^2-32x+16)}}}
The x-intercepts are the values of x that make h(x)=0.
Those would be the values that make {{{x^4-18x^2+32x-15=0}}},
as long as they do not also make {{{2x^3-x^2-32x+16=0}}}
If there are rational zeros for {{{x^4-18x^2+32x-15}}}, they would be integers that are factors of 15.
The possible zeros are -15, -5, -3, -1, 1, 3, 5, and 15.
The only ones of those that could be a zero of {{{2x^3-x^2-32x+16}}} are -1 and 1, which are factors of 16, but neither is a zero of {{{2x^3-x^2-32x+16}}}.
 
ZEROS OF {{{x^4-18x^2+32x-15}}}:
{{{highlight(x=1)}}} is obviously one of the zeros, since
{{{1^4-18*1^2+32*1-15=1-18+32-15=0}}}
{{{x^4-18x^2+32x-15}}} divides exactly by {{{(x-1)}}} twice, and we find that
{{{x^4-18x^2+32x-15=(x-1)(x^3+x^2-17x+15)=(x-1)(x-1)(x^2+2x-15)}}}
Then, {{{x^2+2x-15}}} is easy to factor as
{{{x^2+2x-15=(x-3)(x+5)}}}
So the full factorization of {{{x^4-18x^2+32x-15}}} is
{{{x^4-18x^2+32x-15=(x-1)^2(x-3)(x+5)}}}
So the x-intercepts of h(x), which are zeros of h(x) and of {{{x^4-18x^2+32x-15}}} are:
{{{highlight(x=1)}}}, {{{highlight(x=3)}}}, and {{{highlight(x=-5)}}}