Question 1138584


a polynomnomial of degree {{{4 }}} with
 
{{{1}}} as a zero of multiplicity {{{2}}}->a zero appears twice

so you have {{{x[1]=1}}} and {{{x[2]=1}}}

 and {{{-3}}} and {{{5}}} as zeros of multiplicity {{{1}}}->each appears ones

so you have {{{x[3]=-3}}} and {{{x[4]=5}}}

using zero product rule, we have

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])(x-x[4])}}}...substitute given values

{{{f(x)=(x-1)(x-1)(x-(-3))(x-5)}}}

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

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

{{{f(x)=(x^2-2x+1)(x^2-2x-15)}}}

{{{f(x)=x^4 - 4x^3 - 10x^2 + 28x - 15}}}