Question

3. Find a polynomial, P(x), of degree 3 with zeros of 4,1 and −1, if P(0) = 8.

Answer 1

$$P(x)=a(x-4)(x-1)(x+1)$$

If $P(0)=8$

$$P(0)=a(0-4)(0-1)(0+1)=8=>a=2$$

$$P(x)=2(x-4)(x-1)(x+1)$$

$$=(2x-8)(x^2-1)$$

$$=2x^3-8x^2-2x+8$$