Question 152793


Since {{{-5}}}, {{{4}}}, and {{{6}}} are given zeros this means that:



{{{x=-5}}}, {{{x=4}}}, and {{{x=6}}}



Get all terms to the left side in each case



{{{x+5=0}}}, {{{x-4=0}}}, and {{{x-6=0}}}




{{{(x+5)(x-4)(x-6)=0}}} Now use the zero product property in reverse to join the factors.



{{{(x+5)(x^2-10x+24)=0}}} FOIL



{{{x(x^2-10x+24)+5(x^2-10x+24)=0}}} Distribute



{{{x^3-10x^2+24x+5x^2-50x+120=0}}} Distribute again



{{{x^3-5x^2-26x+120=0}}} Combine like terms



-------------------------------------------

Answer:


So the polynomial with roots of  {{{-5}}}, {{{4}}}, and {{{6}}} is


{{{y=x^3-5x^2-26x+120}}}




Notice how if we graph {{{y=x^3-5x^2-26x+120}}}, we can visually verify our answer


{{{ graph( 500, 500, -10, 10, -50, 150, x^3-5*x^2-26*x+120 ) }}} Graph of {{{y=x^3-5*x^2-26*x+120}}} with roots of {{{x=-5}}}, {{{x=4}}}, and {{{x=6}}}