Help!!...I really need help with this question badly and by tonight?

Question

Let P(x)= X^4+6X^2+9

A. Find All Zero of P(x)

B. Factor P(x) Completely



PLease Help...thank so much

Answer 1

well, lets see.
9 = 3*3, and 3+3 = 6

if we can use the form ay^2 + by + c, we can use the quadratic equation.

Now, lets let y = x^2, and we also have your equation. It's looking good here.

hmm. since a = 1, b = 6 and c = 9, it looks like
(y+3)^2, which leads to (x^2 + 3)^2.

Now for roots. Obviously, if x^2 + 3 = 0, then x^2 = -3, so x = jsqrt(3) or (imaginary number) or -jsqrt(3) . so you have two roots, positive and negative jsqrt(3).

Good luck. Nifty problem.

Answer 2

P(x)= X^4+6X^2+9
=(x^2 + 3)^2
P(x) = 0 => (x^2 + 3)^2 = 0
=> x^2 + 3=0 => x^2 = -3 => x=+-sqrt(-3)
=> x= - i sqrt(3) (i is a complex number and that means -1)
and
x= i sqrt(3)

Answer 3

(x^2+3)(x^2+3) = 0

this has 4 zeros which are imaginary and equal
+-sqrt(3) i. 2 negative and 2 positive sqrt (3)i.

part B.
the above answer can not be factored any more.

Answer 4

X^4+6X^2+9 can be written as (x^2 + 3) * (x^2 + 3).
and (x^2 + 3) can be written as (x+ sqrt(3)) * (x- sqrt(3)).

Answer 5

P(x) = x^4 + 6 x^2 + 9
. . . . = (x^2+3)(x^2+3) = 0
x = + - sqr(3) i

Answer 6

+ or - radical negative 3