The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=3 and x=0 , and a root of multiplicity 1 at x=− 2, find a possible formula for P(x).
P(x) = x^2(x-3)^2 (x+2) is a possibility
To find a possible formula for the polynomial P(x), we can use its roots and their multiplicities.
We know that P(x) has roots of multiplicity 2 at x = 3 and x = 0, and a root of multiplicity 1 at x = -2.
Since x = 3 is a root of multiplicity 2, it means that (x - 3)² is a factor of P(x). Similarly, since x = 0 is a root of multiplicity 2, it means that x² is a factor of P(x). Lastly, since x = -2 is a root of multiplicity 1, it means that (x + 2) is a factor of P(x).
Therefore, a possible formula for P(x) can be obtained by multiplying these factors together:
P(x) = (x - 3)² * x² * (x + 2)
The leading coefficient is 1, so we don't need to include it explicitly since it is always assumed to be 1 unless stated otherwise.
Thus, a possible formula for P(x) is:
P(x) = (x - 3)² * x² * (x + 2)
To find a possible formula for the polynomial P(x), we need to consider its roots and their multiplicities.
Given that the leading coefficient of P(x) is 1, we can start by expressing the polynomial as P(x) = (x - r₁)(x - r₂)(x - r₃)(x - r₄)(x - r₅), where r₁, r₂, r₃, r₄, and r₅ represent the roots.
Given that x = 3 and x = 0 are roots of multiplicity 2, we can write two factors for each of these roots: (x - 3)² and x².
Given that x = -2 is a root of multiplicity 1, we can include the factor (x - (-2)) = (x + 2).
Applying these factors, we can write the formula for P(x) as: P(x) = (x - 3)²(x - 3)²x²(x + 2)
Now, we can expand and simplify the expression:
P(x) = (x - 3)²(x - 3)²x²(x + 2)
= (x - 3)⁴x²(x + 2)
= (x - 3)⁴x³(x + 2)
So, a possible formula for P(x) is P(x) = (x - 3)⁴x³(x + 2).