Question

A polynomial of degree 3 has a double zero at 2. When x = 1 it takes the value 6 and when x = 3 it

Answer 1

Answer:

p(x)=(x-2)^2(x+5)

Step-by-step explanation:

p(2)=0 then p(x) =(x-2)^2(x-x1), x1=?

p(1)=6

p(3)=8

p(1)=(1-2)^2(1-x1)

6= (-1)^2 (1-x1)

6=1(1-x1)

6=1-x1

x1=-5;

then p(x) =(x-2)^2(x-(-5))=(x-2)^2(x+5)

Really: f(3)=(3-2)^2(3+5)=1*8=8