Quick Math Questions

Write a polynomial function, P, in factored form by using the given information:

1. P is of degree 2; P(0) = -2; zeros: -1 and 4

2. P is of degree 4; P(0) = -4; zeros: 1/2(multiplicity of 2) and 2 + i

Answer please! First correct answer that shows all work (preferably with an explanation =D) gets best answer.

Thanks!

1) Since P has zeros -1 and 4 and is of degree 2 then you know it has (x+1)(x-4) as a factor, hence it is of the form:

P(x) = a(x+1)(x-4)

Now just use your P(0)=-2 to work out a:

P(0) = a * 1 * (-4) = -4a = -2
a = 1/2

so P(x) = (1/2)(x+1)(x-4)


2) There's a key assumption here that's missing, otherwise there would be infinitely many solutions - that is that the coefficients of your polynomial are all real. If you know that, then you can assume that since you have 2 + i as a root, you also have its conjugate 2 - i as a root. This is a theorem that has a name but I forget what the name is. So since you know 1/2 is a root with multiplicity 2 and 2 + i and 2 - i are roots you know that P(x) has a factor of (x-1/2)(x-1/2)(x-(2-i))(x-(2+i)). Hence P is of the form:

P(x) = a(x-1/2)(x-1/2)(x-(2-i))(x-(2+i))

Let's first combine (x-(2-i))(x-(2+i)) to get rid of the complex numbers:

(x-(2-i))(x-(2+i))
= x^2 - x(2+i) - x(2-i) + (2-i)(2+i)
= x^2 - 4x + (4 - i^2)
= x^2 - 4x + 5

So P(x) = a(x - 1/2)(x - 1/2)(x^2 - 4x + 5)

Now we know that P(0) = 4 so we use that to determine a:

P(0) = a * (-1/2) * (-1/2) * 5 = (5/4)a = -4
(5/4)a = -4
a = -16/5

So P(x) = -(16/5)(x - 1/2)(x - 1/2)(x^2 - 4x + 5)

1) P(x)=ax^2+bx+c

c=-2
0=a-b+c
so
a-b=2

0=16a+4b+c
so
16a+4b=2

Combine
4a-4b=8
16a+4b=2
so
20a=10
a=1/2

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

2)

Not sure what you mean by multiplicity of 2....are the zeros at 1/2 (= one half) and (2+i)?