Question 117516
{{{a+b+c+d=0}}} Start with the given equation



{{{d=-a-b-c}}} Solve for d


{{{ax^3+bx^2+cx-a-b-c=0}}} Plug in {{{d=-a-b-c}}} into {{{ax^3+bx^2+cx+d=0}}} 



{{{(ax^3-a)+(bx^2-b)+(cx-c)=0}}} Group like terms



{{{a(x^3-1)+b(x^2-1)+c(x-1)=0}}} Factor out the GCF from each individual group



{{{a(x-1)(x^2+x+1)+b(x^2-1)+c(x-1)=0}}} Factor {{{x^3-1}}} by using the difference of cubes formula to get {{{(x-1)(x^2+x+1)}}}



{{{a(x-1)(x^2+x+1)+b(x-1)(x+1)+c(x-1)=0}}} Factor {{{x^2-1}}} by using the difference of squares formula to get {{{(x-1)(x+1)}}}



Notice how we have the common term {{{x-1}}}. We can factor this term out.



{{{(x-1)(a(x^2+x+1)+b(x+1)+c)=0}}} Factor out the GCF {{{x-1}}}



Now if we let {{{H=x-1}}} and {{{K=a(x^2+x+1)+b(x+1)+c}}} we'll get


{{{H*K=0}}}



So by the zero product property we get


{{{H=0}}} or {{{K=0}}}



but since {{{H=x-1}}}, this means


{{{x-1=0}}}


Now solve for x



{{{x=1}}}



So {{{x=1}}} is a zero of {{{ax^3+bx^2+cx+d=0}}} if the coefficients satisfy the equation {{{a+b+c+d=0}}}




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Check:


Unfortunately, there is no formal check for this type of problem. But if we make sure that the coefficients satisfy the equation {{{a+b+c+d=0}}}, then we can graph the equation and see that one root is {{{x=1}}}. For instance, let {{{a=b=1}}} and {{{c=d=-1}}} so we'll get



{{{x^3+x^2-x-1}}} (notice how {{{1+1-1-1=2-2=0}}} which satisfies the equation {{{a+b+c+d=0}}})



Now graph the polynomial to get



{{{ graph( 500, 500, -10, 10, -10, 10, x^3+x^2-x-1) }}}



and we can see that one root is {{{x=1}}}



If we try different values of a,b,c, and d that will satisfy the equation {{{a+b+c+d=0}}}, we'll see that {{{x=1}}} is a root every time. So in a way, our answer has been verified.