the polynomial of degree 4 p(x) has a root of multiplicity 2 at x=4 and roots of multiplicity 1 at x=0 and x=-2. It goes through the point (5,21)

Question

the polynomial of degree 4 p(x) has a root of multiplicity 2 at x=4 and roots of multiplicity 1 at x=0 and x=-2. It goes through the point (5,21)

Answer 1

Since p(x) has a root of multiplicity 2 at x=4, we know that (x-4) is a factor of p(x) twice. Similarly, since p(x) has roots of multiplicity 1 at x=0 and x=-2, we know that (x-0) and (x+2) are factors of p(x).

Therefore, we can write p(x) in factored form as:

p(x) = a(x-4)^2(x-0)(x+2)

where a is a constant of proportionality that we need to determine. We can use the fact that p(x) goes through the point (5,21) to solve for a.

Substituting x=5 and p(x)=21 into the equation above, we get:

21 = a(5-4)^2(5-0)(5+2)

Simplifying, we get:

21 = 63a

Solving for a, we get:

a = 21/63 = 1/3

Therefore, the polynomial p(x) is:

p(x) = (1/3)(x-4)^2(x)(x+2)

We can expand this polynomial to get:

p(x) = (1/3)(x^4 - 8x^3 + 20x^2 - 16x)

Therefore, the polynomial of degree 4 that has a root of multiplicity 2 at x=4 and roots of multiplicity 1 at x=0 and x=-2, and goes through the point (5,21) is:

p(x) = (1/3)(x-4)^2(x)(x+2) = (1/3)(x^4 - 8x^3 + 20x^2 - 16x)

Answer 2

P(x) = ax(x - 4)²(x + 2)

21 = a(5)(5 - 4)²(5 + 2)

Can you solve for a and answer?