Question 1192629


The zeros are​ {{{x[1]=-3}}}, {{{x[2]=-1}}}, and {{{x[3]=4}}}
 
{{{p(-2)=-18}}}


{{{p(x)=a(x-x[1])(x-x[2])(x-x[3])}}}.....substitute zeros

{{{p(x)=a(x-(-3))(x-(-1))(x-4)}}}

{{{p(x)=a(x+3)(x+1)(x-4)}}}

{{{p(x)=a(x+3)(x+1)(x-4)}}}.....to find {{{a}}} use given {{{p(-2)=-18}}}

{{{-18=a(-2+3)(-2+1)(-2-4)}}}

{{{-18=a(1)(-1)(-6)}}}

{{{-18=6a}}}

{{{a=-18/6}}}

{{{a=-3}}}

substitute in {{{p(x)=a(x+3)(x+1)(x-4)}}}

{{{p(x)=-3(x+3)(x+1)(x-4)}}}.........expand

{{{p(x)=-3x^3 + 39x + 36}}}


{{{ graph( 600, 600, -10, 10, -30, 30, -3x^3+39x+36) }}}