Question 973355
Factoring your first polynomial and your H(x) function are beyond the typical Intermediate
Algebra level.  Make use of Rational Roots Theorem.


{{{x^3+4x^2+x-6}}} can be tested for possible roots,  plus and minus 1,2,3,6.  This can be done with
synthetic division.  You could expect at most, three roots.
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{{{-2}}} is a root, giving remainder 0, and coefficients of quotient, 1,2,-3.
{{{-3}}} is a root, giving remainder 0, and coefficients of quotient, 1,-1.
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This gives you a factorization {{{highlight((x+2)(x+3)(x-1))}}}
The meaning of those last coefficients of 1,-1 is {{{1*x-1}}} , or as shown, {{{x-1}}}.



{{{H(x)=10x^4+0*x^3+2x^2-x+6}}}  may have any roots (need to test each) of again, plus and minus 1,2,3,6.
Testing for roots plus and minus 1,2,3 seem to give nonzero remainders.  Maybe continue testing
for -6 and +6....
Neither give 0 remainder, so no rational roots.
H(x) <i>may</i> have irrational roots, but checking for them is harder.  Other tests are possible
but are more lengthy.
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Using a graphing tool, like putting y=10x^4+2x^2-x+6  into the text field of Google search engine
shows a fairly symmetric looking U shaped graph, and the minimum is above the x-axis everywhere,
so H(x) has no real roots.  This also means no real irrational roots.
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Note that there ARE more roots to check.  numerators are plus or minus 1,2,3,6 and denominators
are plus and minus 1,2,5,10.  I did not try the possible roots other than those for whole numbers.
You should expect up to FOUR complex roots.


You can try Rational Roots Theorem the same way for {{{h(x)=12x^5+2x^2-x+1}}}.
Roots to test for, using synthetic division, are plus and minus 1/2, 1/3, 1/4, 1/6, maybe 1/12.
The negatives of those are not roots, but still you should expect up to five complex roots, since h(x) is degree 5.
(More work is necessary to find them).
(Graphing tool indicates ONE negative real root <i>approximately</i> {{{-0.7518}}} ).