The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=2 and x=0, and a root of multiplicity 1 at x=-2. Find a possible formula for P(x).
Well, let me put on my mathematician clown shoes and solve this for you!
Since we know that P(x) has roots of multiplicity 2 at x = 2 and x = 0, we can write two factors as (x - 2)² and x².
Since P(x) has a root of multiplicity 1 at x = -2, we can include another factor of (x + 2).
Now, since the leading coefficient of P(x) is 1, we can multiply all these factors together to get our formula for P(x):
P(x) = (x - 2)² * x² * (x + 2)
So, there you have it! P(x) = (x - 2)² * x² * (x + 2). Voila!
To find a possible formula for P(x), we need to consider the given information about the roots and the leading coefficient.
The roots are given as follows:
Multiplicity 2 at x = 2
Multiplicity 2 at x = 0
Multiplicity 1 at x = -2
The leading coefficient is given as 1.
Based on the multiplicities of the roots, we can write the polynomial P(x) in factored form as:
P(x) = (x - 2)^2 * x^2 * (x + 2)
To find a possible formula, we can expand this expression:
P(x) = (x - 2)(x - 2) * x * x * (x + 2)
= (x^2 - 4x + 4) * x^2 * (x + 2)
= (x^4 - 4x^3 + 4x^2) * (x + 2)
= x^5 + 2x^4 - 4x^3 - 8x^2 + 4x^2 + 8x
= x^5 + 2x^4 - 4x^3 - 4x^2 + 8x
Therefore, a possible formula for P(x) is:
P(x) = x^5 + 2x^4 - 4x^3 - 4x^2 + 8x
To find a possible formula for the polynomial P(x), we need to use the given information about its degree, leading coefficient, and roots.
Let's break down the given information:
1. The degree of the polynomial is 5, which means it will have a total of 5 terms.
2. The leading coefficient is 1, which indicates that the coefficient of the term with the highest power of x is 1. Since the leading coefficient is 1, the highest power term will be x^5.
3. The root at x = 2 has a multiplicity of 2. This means that the factor (x - 2) will be squared in the formula for P(x).
4. The root at x = 0 also has a multiplicity of 2. This means that the factor x will also occur twice in the formula for P(x). So, we need to include (x - 0)^2 = x^2 in the formula.
5. The root at x = -2 has a multiplicity of 1. Therefore, we only need to include the factor (x - (-2)) = (x + 2) once in the formula.
Now, we can put it all together to find a possible formula for P(x):
P(x) = (x - 2)^2 * x^2 * (x + 2) * f(x)
In this formula, f(x) represents the remaining factor(s) that we haven't determined yet. Since we know that the polynomial has degree 5 and the leading coefficient is 1, the remaining factor f(x) must be of degree 5 - 2 - 2 - 1 = 0.
So, f(x) is a constant term, let's call it c.
Therefore, the possible formula for P(x) is:
P(x) = (x - 2)^2 * x^2 * (x + 2) * c
This formula satisfies the conditions given in the problem. Note that c can be any non-zero constant value, which means there are infinitely many possible formulas for P(x) that meet the conditions.
"has roots of multiplicity 2 at x=2 and x=0" ----- (x-2)^2 (x^2)
"a root of multiplicity 1 at x=-2" ---- (x+2)
y = x^2(x+2)(x-2)^2
"has roots of multiplicity 2 at x=2 and x=0" is not clear
does the multiplicity apply to both roots? I assumed thus