The polynomial of degree 4, P ( x ) , has a root of multiplicity 2 at x = 1 and roots of multiplicity 1 at x = 0 and x = − 3 . It goes through the point ( 5 , 64 ) .

Question

The polynomial of degree 4, , has a root of multiplicity 2 at  and roots of multiplicity 1 at  and . It goes through the point .

Find formula in simplest form:

Answer 1

For this problem, we check out our given zeros.

x = 1 (multiplicity of 2)

x = 0 (multiplicity of 1)

x = -3 (multiplicity of 1)

The multiplicities tell us how many times the zeros repeat. In this case, we should have a total of 4 zeros, which also means that our polynomial also has a degree of 4.

We write our polynomial like this:

P(x) = a(x - 1)(x - 1) (x + 0)(x + 3) **x + 0 would just be x and our single variables can be moved to the

front of the polynomial

P(x) = ax(x - 1)(x - 1)(x + 3) *Next we foil but do not distribute the 'a'

P(x) = ax(x² - 2x + 1)(x + 3)

P(x) = a(x³ - 2x² + x)(x + 3)

P(x) = a(x⁴ - 2x³ + x² + 3x³ - 6x² + 3x) **combine your link terms

P(x) = a(x⁴ + x³ - 5x² + 3x)

Once you combine your like terms, use the point that you were given to solve for 'a'

64 = a (5⁴ + 5³ - 5(5²) + 3(5))

64 = a (625 + 125 - 125 + 15)

64 = 640a

a = 1/10

Now that we found our 'a' we plug it back to the equation that had the missing 'a'

P(x) = 1/10 (x⁴ + x³ - 5x² + 3x) *last we distribute the fraction

P(x) = 1/10 x⁴ + 1/10 x³ - 1/2 x² + 3/10 x