The polynomial of degree 5, P(x), has leading coefficient 1, 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)

Answer 1

(x+2)x^2)(x-2)^2, multiplicity of 2 means exponent of 2. factors are the root with changed sign and an x stuck in front of it

= (x^2)(x+2)(x^2 -4x +4)

= (x^2)(x^3 -2x^2-4x -8)

P(x) = x^5 -2x^4 -4x^3 - 8x^2

multiplicity 2 means graphically the curve is tangent to the x axis at that point where x=2 or 0

Answer 2

Answer 3

All polynomials can be written

a(x-r₁)(x-r₂)···(x-r_n).

1. They told you n=5, so P(x) = a(x-r₁)(x-r₂)(x-r₃)(x-r₄)(x-r₅)
2. They told you a=1.
3. They told you two of the r's are 2, two are 0, and one is -2.

Plug them in and you are done.

Answer 4

Hi! I've put together some notes that you can reference as a guide to solving this type of problem. I'll also change the numbers to better suit your problem, so that you have a clear example! :)

1. The multiplicity= the degree of the particular zero (for example, if there is a multiplicity of 2 at x=2, there are 2 zeros at (2,0), or an exponent of 2).
2. Remember that the zeros of a function can be expressed as (x-the zero's value). If the zero is negative, such as -2 it is expressed as the opposite (x+the zero's value)
3. For polynomials, follow the format a(x-___)... (a is the co-efficient, and may be given, or found, via substitution, with the y-intercept/asymptote if not provided)

With this information in mind, we can write out the problem as follows.

1. a= leading coefficient= 1 (given)
2. zeros/x-intercepts: 2 (multiplicity 2), 0 (multiplicity 2) and -2 (multiplicity 1)

To substitute these given zeros into the formula a(x-__)(x-___)(x-___) => 3 sets of (x-) since there are 3 different zeros

1. Since 2 is a positive number: (x-2)
2. Since 0 is non-negative: (x-0)
3. Since -2 is a negative number: (x+2)

Because we are also given multiplicities, we have to also add the appropriate exponents.

1. Multiplicity of 2= ^2
2. Multiplicity of 1= ^1 (does not need to be written, as any number to the 1st power is just that number)
3. Because 2 and 0 have multiplicities of 2, each of their parts of the equation will be squared: (x-2)² and (x-0)²
4. Because -2 only has a multiplicity of 1, we can leave it as is, as (x+2)¹= (x+2)

Last, put all parts of the equation together, resembling the formula y=a(x-__)

 

Solution: y=1(x-2)²(x-0)²(x+2)

 

Please let me know if there's anything else I can do to help you solve this!!

Note: If you are not given a leading coefficient, you would just substitute the y-point you are given (y-intercept or y-asymptote), with x=0 and y=the given value, so "a" is the only variable left to solve for and re-add into the equation, and returning the other numbers you changed back to x and y, respectively.