Any math pros out there? :) need some help?

Question

fully factor this

16x^4 - 12x^2 + 1



PLEASE SHOW ALL WORK OR DONT BOTHER

best answer gets 10 big points

Answer 1

Add and subtract 8x^2.

16x^4 - 8x^2 + 1 -12x^2 +8x^2 =

(4x^2 -1)^2 - 4x^2 =

(4x^2 -1 -2x) (4x^2 -1 + 2x)

It's clear neither of these factor further.

Mildly interesting problem.

Answer 2

Listing all the factors of 16 and 1 we form all possible "inside" and "outside" products and add or subtract them until we find the center coefficient of -12:


16*1=16 1*1=1, 8*1=8 2*1=2, 4*1=4 4*1=4


Now notice that for each outside and inside product pairs separated by commas there is no way to add or subtract the products to produce -12. The same is true for the other outside and inside products.

This method exhausts all possible factor combinations with INTEGER coefficient so the given trinomial cannot be factored using integer coefficients. If the above is not clear try all possible combinations written in factored form and multiply to see if they produce the trinomial:

(16x^2 - 1)(x^2 - 1) = 16x^4 - 17x^2 + 1 which fails.

Continue with all the other possibilities to verify they also fail.

Answer 3

This is really a quadratic equation in disguise.

Let u = x^2

From inspection, this will not factor nicely into integers. So, to find out factors, we set the equation to 0.
16x^4 - 12x^2 + 1 = 0
16u^2 - 12u + 1 = 0

We solve for u by the quadratic equation:
u = (-(-12) +- Sqrt((-12)^2 - 4(16)(1))) / (2(16))
= (12 +- Sqrt(144 - 48)) / 32
= (12 +- Sqrt(96)) / 32
= (12 +- Sqrt(16)*Sqrt(6)) / 32
= (12 +- 4Sqrt(6)) / 32
= (3 +- Sqrt(6)) / 8

Now, we solve for x in u = x^2 and get x = +-Sqrt(u)
So, we get our solutions:
+-Sqrt((3 +- Sqrt(6)) / 8)

Now, we go back to the original equation. It can be factored into the following form:
(4x^2 - a^2)(4x^2 - b^2)
(2x - a)(2x + a)(2x - b)(2x + b)

Now, we just set that equation above equal to zero and notice that for it to be true, at least one of the components in the product must be 0. Seeing that we have four solutions and four components, this is it.

(2(Sqrt((3 + Sqrt(6)) / 8) - a) = 0; a = 2(Sqrt((3 + Sqrt(6)) / 8))
(2(Sqrt((3 - Sqrt(6)) / 8) - b) = 0; b = 2(Sqrt((3 - Sqrt(6)) / 8))

So, your solution is:
(2x - 2(Sqrt((3 + Sqrt(6)) / 8)))(2x + 2(Sqrt((3 + Sqrt(6)) / 8)))(2x - 2(Sqrt((3 - Sqrt(6)) / 8)))(2x + 2(Sqrt((3 - Sqrt(6)) / 8)))

We see that we can factor 2 out of each component, or 16 out of the equation:
16(x - Sqrt((3 + Sqrt(6))/8)(x + Sqrt((3 + Sqrt(6))/8))(x - Sqrt((3 - Sqrt(6))/8)(x + Sqrt((3 - Sqrt(6))/8))

Not pretty, but that's what you get when you don't have integer factors.

Answer 4

Not sure if this this is as factored as you need it but
16x^4 -12x^2 +1
4x^2(4x^2 - 3) + 1

I'm really rusty on this and wouldnt have answered but noone else was =/

Answer 5

Let 4x^2= y
Then 16x^4=y2
And y2-3y+1 is transformed version.
To factor: y =1/2[3+/- sqrt(5)]
So..... x = +/- 1/4sqrt(2) * y