can any1 factor x^4 +6x^2+25 by adding and subtracting x^2?

Question

X^2 must be added nd subtracted to factor the problem

Answer 1

Well if you add 4x² and subtract 4x² you will have:

x^4 + 10x² + 25 - 4x²

Now you can factor this as:
(x² + 5)(x² + 5) - 4x²

Notice that is a perfect square now:
(x² + 5)² - 4x²

This is a difference of squares:
(x² + 5)² - (2x)²

Remember a² - b² = (a + b)(a - b):
(x² + 5 + 2x)(x² + 5 - 2x)

Or equivalently:
(x² + 2x + 5)(x² - 2x + 5)

If this were equal to zero, then you could use the quadratic formula on each part to come up with the roots:
x = 1 + 2i
x = 1 - 2i
x = -1 + 2i
x = -1 - 2i

Answer 2

To factor this with just x^2, it wont give a perfect square.
However, it can be done by adding and subtracting 4x^2 which is,
x^4 + 6x^2 + 4x^2 + 25 - 4x^2
= x^4 + 10x^2 + 25 -4x^2
= (x^2 + 5)^2 -4x^2
= (x^2 + 5)^2 - (2x)^2
= (x^2 + 2x + 5)(x^2 - 2x + 5)
Further factorization is possible to get non integral or non real roots.

Answer 3

Pretty sure this has complex roots. Let u=x^2 then using the QF, you get
3+/- sqrt [36-100]/2 or u=3+/- i4 Given that, are you sure the problem is as shown?

Answer 4

(x^2-2x+5)(x^2+2x+5)