If p(-3) = p(-1) = p(2)= 0 and p(0) =6 find the cubic polynomial
p = (x+3)(x+1)(x-2)(x-a) which is not cubic
at x = 0, p = 6
6 = 3*1*-2(-a)
6 = 6 a
a = 1
so
p = (x+3)(x+1)(x-2)(x-1)
multiply that out
To find the cubic polynomial given these conditions, we can use the fact that a cubic polynomial can be written in the form:
p(x) = a(x - r1)(x - r2)(x - r3)
Where a is a constant and r1, r2, and r3 are the roots of the polynomial.
In this case, we are given that p(-3) = p(-1) = p(2) = 0, which means that -3, -1, and 2 are the roots of the polynomial. We are also given p(0) = 6, which allows us to determine the constant a.
Substituting the roots into the polynomial, we have:
p(x) = a(x - (-3))(x - (-1))(x - 2)
= a(x + 3)(x + 1)(x - 2)
To find the value of a, we use the fact that p(0) = 6:
6 = a(0 + 3)(0 + 1)(0 - 2)
= -6a
Dividing both sides by -6, we get:
a = -1
So the cubic polynomial can be written as:
p(x) = -1(x + 3)(x + 1)(x - 2)