How to factor this out completely?

Question

how to I even begin to factor this equation?? Any help would be appreciated.


X^5 - X^3 + 4X^2 - 4


Thanks.

Also, how would I factor this equation too?

16X^3 + 54

thanks

Answer 1

X^5 - X^3 + 4X^2 - 4

Try grouping the terms.

(X^5 - X^3) + (4X^2 - 4)
X^3 (X^2 - 1) + 4(X^2 -1)
(X^3 + 4)(X^2 -1)

(X^2 -1) is a difference of two squares and can still be factored. But you finish it.

16X^3 + 54

Take note that if we factor out 2 we get:

2(8X^3 + 27)

(8X^3 + 27) can be factored further since it is a sum of two cubes. Again you do the rest.

Given you enough clues I believe. :)

Answer 2

x^5 -x³ + 4x² - 4 = x³(x² - 1) + 4(x² - 1)
= (x² - 1)(x³ + 4)
=(x - 1)(x + 1)(x³ + 4)

now a³ + b³ = (a + b)(a² - ab + b²)

then it's equal to
(x - 1)(x + 1)(x + 4^(1/3))(x² - x*4^(1/3) + 4^(2/3))


the second is 2(x³ + 3³) = 2(x + 3)(x² - 3x + 9)

Answer 3

X^5 - X^3 + 4X^2 - 4

x^2 (x^3-x+4) -4

(x *squareroot x^3-x+4* +2) (x *squareroot x^3-x+4* -2)

hope that helps

Answer 4

X^5 - X^3 + 4X^2 -4 = 0
X^3 ( X^2 - 1) + 4 ( X^2 - 1) = 0
X^3 (X - 1)(X+1) + 4 ( X-1) (X+1)

( X^2-1 ) ( X^3 + 4 ) = 0