Question

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). I got x^5-5x^4-2x^4+3x^2, and I got it wrong. :(

Answer 1

Answer:

Step-by-step explanation:

(x-3)(x-3)(x²)(x+2)

(x²-6x+9)(x³+2x²)

x^5+2x^4-6x^4-12x^3+9x^3+18x^2

x^5 - 4x^4 - 3x^3 +18x^2

Answer 2

Using the Factor Theorem, it is found that the polynomial is given by:

$P(x) = x^5 - 4x^4 - 3x^3 + 18x^2$

The Factor Theorem states that if a polynomial has roots $x_1, x_2, ..., x_n$, it can be written as:

$P(x) = a(x - x_1)(x - x_2)...(x - x_n)$

In which a is the leading coefficient.

In this problem:

• Leading coefficient of 1, thus $a = 1$.
• Roots with multiplicity 2 at x = 3 and x = 0, thus $x_1 = x_2 = 3, x_3 = x_4 = 0$
• Root with multiplicity 1 at x = -2, thus $x_5 = -2$.

Then, the polynomial is:

$P(x) = x^2(x - 3)^2(x + 2)$

$P(x) = x^2(x^2 - 6x + 9)(x + 2)$

$P(x) = x^2(x^3 - 4x^2 - 3x + 18)$

$P(x) = x^5 - 4x^4 - 3x^3 + 18x^2$

A similar problem is given at https://brainly.com/question/24380382