HELP! Factorise: 4x^4 - 5x² + 1?

Question

I am so stuck, I understand factorising and everything but i just dont no what to do when there are no common factors, what are the tricks to sorting out brackets?

Please dont miss any steps when explaining, that what my lecturer does and that why i dont get it!

Thanks :)

Answer 1

In this case, because there are no odd powers of x, it is actually a quadratic in the variable x^2. Until you get the hang of these, you'll likely find it helpful to substitute y = x^2:
4x^4 - 5x^2 + 1, let y = x^2
= 4y^2 - 5y + 1
= (4y - 1) (y - 1)
= (4x^2 - 1) (x^2 - 1)
= (2x-1) (2x+1) (x-1) (x+1)

Answer 2

let F = 4x^4 - 5x^2 + 1

look for zeros (ie which x values give F = 0 )

1 and -1 are zeros

so (x-1) and (x+1) are factors [ = (x^2-1) ]

so F = (x-1) (x+1) (4x^2-1)

factorizing the last bracket gives

F = (x-1) (x+1) (2x-1) (2x+1)

Answer 3

4x^4 - 5x^2+1

here is the trick: change - 5x^2 into - 4x^2 - x^2

4x^4 - 4x^2 - x^2 +1

factoring 4x^2 from the first two terms and -1 from the last two, we get:

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

factoring (x^2 - 1), we get

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

now both factors are a difference of two squares
factoring we get

( 2x + 1) (2x - 1)(x +1)(x - 1)

Answer 4

4x^4-5x^2+1
This can be written as 4x^4-4x^2-x^2+1
This can be written as 4x^2(x^2-1) -1(x^2-1)
The factorization is (4x^2-1)(x^2-1)

Answer 5

Could you perhaps try completing the square, but with a function to the 4th power?

SO something like (ax^2+bx)?

Answer 6

take x^2=t
so equation becomes
=4t^2-5t+1=0
=4t^2-4t-1t+1=0
=4t(t-1)-1(t-1)=0
=(t-1)(4t-1)=0
substitute back the value of t ....................

hope iam clear...............