Question 1023621
1. A polynomial with a root of multiplicity 2 at x=3 has {{{(x-3)^2}}} as a factor.
A polynomial with roots of multiplicity 1 at x=0 and x=−2
has {{{(x-0)=x}}} and {{{(x-(-2))=(x+2)}}} as factors.
The factors listed make
{{{P(x)=a*x*(x+2)*(x-3)^2}}} , with {{{a<>0}}} a polynomial f degree 4.
If the graph of such polynomial goes through point (5,14), then {{{P(5)=14}}} ,and {{{a*5*(5+2)*(5-3)^2=14}}} , so
{{{a*5*7*2^2=14}}}-->{{{a*5*7*4=14}}}-->{{{a=14/(5*7*4)}}}-->{{{a=0.1}}} .
So, {{{highlight(P(x)=0.1*x*(x+2)*(x-3)^2)}}} is a formula for such a polynomial.
 
2. Similarly, the polynomial must be of the form
{{{P(x)=a*(x+1)*(x-5)^2}}} , with {{{a<>0}}} ,
ana if the y-intercept is {{{-12.5}}} ,
{{{a}}} must be such that {{{P(0)=12.5}}} .
So, {{{a*(0+1)*(0-5)^2=12.5}}}-->{{{a*1*25=12.5}}}-->{{{a*25=12.5}}}-->{{{a=12.5/25}}}-->{{{a=0.5}}},
and {{{highlight(P(x)=0.5*(x+1)*(x-5)^2)}}} is a formula for such a polynomial.