Question 936620
to identify POTENTIAL roots use THE RATIONAL ZERO THEOREM

If {{{f(x)}}} ={{{a[n]x^n}}}+ .... .+{{{a[1]x}}}+{{{a[0]}}}  has integer coefficients, then every rational zero of {{{f(x)}}} has the following form:

{{{p/q=(factors_of_constant_term_a[0])/(factors_of_leading_coefficient_a)}}}

using that, we have:
1- {{{y(x) = x^3-1}}} 
--> factor {{{1}}} and {{{1}}}:
{{{x }}}= ±{{{1/1}}} which is {{{x}}} = ±{{{1/1}}}

2- {{{h(x) = x^4 -1}}}
--> factor {{{1}}} and {{{1}}}:
{{{x }}}= ±{{{1/1}}} which is {{{x}}} = ±{{{1/1}}}

3- {{{g(x) = x^4 +2x^3-16x^2-2x+15}}}
--> factor {{{1}}}, and {{{15}}}:
{{{1}}} X {{{1}}},{{{3}}},{{{5}}},{{{15}}}
{{{x}}} = ±{{{1/1}}}, ±{{{3/1}}},±{{{5/1}}},±{{{15/1}}}, which is => {{{x}}} = ±{{{1}}},±{{{3}}},±{{{5}}},±{{{15}}}
Test these zeros using synthetic division.
test {{{x=1}}}
{{{g(x) = x^4 +2x^3-16x^2-2x+15}}}


{{{1}}}|--1-----2----..-16------...-2----15
---| ------1----..-1----..-1------17---..-15
---|   1---- 1----.. -17--------15------- 0

so, {{{x=1}}} is the root

same way you check all other potential roots


now find roots:


1- {{{y(x) = x^3-1}}}........factor: set {{{y(x)=0}}} and use {{{a^3 - b^3 = (a -b)(a^2 + ab + b^2)  }}}..(difference of {{{2}}} cubes rule)

{{{0= (x-1)(x^2+x+1)}}}

we already know one root: if {{{0= (x-1)}}}=> {{{x=1}}}

use quadratic formula to find other two roots from {{{0= (x^2+x+1)}}}:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-1 +- sqrt( 1^2-4*1*1 ))/(2*1) }}} 

{{{x = (-1 +- sqrt( 1-4 ))/2 }}} 

{{{x = (-1 +- sqrt( -3 ))/2 }}} 

solution will be complex roots:

{{{x = -1/2 + sqrt( -3 )/2 }}}
and
{{{x = -1/2 - sqrt( -3 )/2 }}}


{{{ graph( 600, 600, -10, 10, -10, 10,x^3-1) }}}


2- {{{h(x) = x^4 -1}}}

{{{0 = x^4 -1^4}}}

{{{0 = (x^2)^2 -(1^2)^2}}}

{{{0 = (x^2 -1^2)(x^2 +1^2)}}}

{{{(x-1)(x+1)(x^2+1)=0 }}}

roots:



{{{(x-1)=0 }}} =>{{{x=1}}}
{{{(x+1)=0 }}}=>{{{x=-1}}}......real roots


{{{(x^2+1)=0 }}}=> {{{x^2=-1 }}}=> {{{x=sqrt(-1) }}}=> {{{x=i }}} or {{{x=-i }}}
{{{x=1}}}.........complex roots

{{{ graph( 600, 600, -10, 10, -10, 10,(x-1)(x+1)(x^2+1)) }}}


3- {{{g(x) = x^4 +2x^3-16x^2-2x+15}}}...write {{{2x^3}}} as {{{5x^3-3x^3}}} and {{{-16x^2}}} as {{{-15x^2-x^2}}}

{{{0 = x^4-x^2+5x^3-5x-3x^3-3x-15x^2-15}}}....group

{{{0 = (x^4-x^2)+(5x^3-5x)-(3x^3-3x)-(15x^2-15)}}}

{{{0 = x^2(x^2-1)+5x(x^2-1)-3x(x^2-1)-15(x^2-1)}}}

{{{0 = (x^2-1)(x^2+5x-3x-15)}}}...group

{{{0 = (x-1)(x+1)((x^2-3x)+(5x-15))}}}

{{{0 = (x-1)(x+1)(x(x-3)+5(x-3))}}}

{{{0 = (x-1)(x+1)(x-3)(x+5)}}}

roots:
{{{0 = (x-1)}}}=> {{{x=1}}}

{{{0 =(x+1)}}}=> {{{x=-1}}}

{{{0 = (x-3)}}}=>{{{x=3}}}

{{{0 = (x+5)}}}=> {{{x=-5}}}........all roots are real roots


{{{ graph( 600, 600, -25, 25, -45, 25, x^4+ 2x^3 -16x^2 -2x +15) }}}



4- {{{f(x) = x^4+ 2x^3 -16x^2 -2x +15 }}} this is same as 3-