What are the factors of x to the power 6 plus nine?

Answer 1

x6 + 9

= x6 - (-9) since i2 = -1

= (x3)2 - 9i2 factor the difference of two squares

= (x3 + 3i)(x3 - 3i) since 3 = (31/3)3 and -i = i3 we can write:

= [x3 - (31/3)3i3] [x3 + (31/3)3i3]

= [x3 - (31/3i)3] [x3 + (31/3i)3] factor the sum and the difference of two cubes

= [(x - 31/3i)(x2 + 31/3ix + (31/3)2i2)] [(x + 31/3i)(x2 - 31/3ix + (31/3)2i2)]

= [(x - 31/3i)(x2 + 31/3ix - (31/3)2)][(x + 31/3i)(x2 - 31/3ix - (31/3)2)]

Thus, we have two factors (x - 31/3i) and (x + 31/3i),so let's find four others

Add and subtract x2/4 to both trinomials

[x2 - x2/4 + (x/2)2 + 31/3ix - (31/3)2] [x2 - x2/4 + (x/2)2 - 31/3ix - (32/3)2] combine and factor -1

= {3x2/4 - [((x/2)i))2 - 31/3ix + (31/3)2]}{3x2/4 - [((x/2)i))2 + 31/3ix + (32/3)2]} write the difference of the two squares

= {((3)1/2x/2))2 - [(x/2)i - 31/3]2}{((3)1/2x/2))2 - [(x/2)i + 32/3]2]} factor the difference of two squares

= {[(31/2/2)x - ((1/2)i)x - 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - (((1/2)i)x + 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]}

= {[(31/2/2)x - ((1/2)i)x + 31/3)] [((31/2/2)x + ((1/2)i)x - 31/3)]} {[((31/2/2)x) - ((1/2)i)x - 31/3)] [((31/2/2)x) + ((1/2)i)x + 31/3)]} simplify

= {[((31/2 - i)/2))x + 31/3)] [((31/2 + i)/2))x - 31/3)]} {[((31/2 - i)/2))x - 31/3)] [((31/2+ i)/2))x + 31/3)]}

so we have the 6 linear factors of x2 + 9.

1) (x - 31/3i)

2) (x + 31/3i)

3) [((31/2 - i)/2))x + 31/3)]

4) [((31/2 + i)/2))x - 31/3)]

5) [((31/2 - i)/2))x - 31/3)]

6) [((31/2+ i)/2))x + 31/3)]

Check: Multiply:

[(1)(2)][(3)(5)][(4)(6)]

A) (x - 31/3i)(x + 31/3i) = x +(31/3)2

B) [((31/2 - i)/2))x + 31/3)] [((31/2 - i)/2))x - 31/3)] = [(1 - (31/2)i)/2]x2 - (31/3)2

C) [((31/2 + i)/2))x - 31/3)][((31/2+ i)/2))x + 31/3)] = [(1 + (31/2)i)/2]x2 - (31/3)2

Multiply B) and C) and you'll get x4 - (31/3)2x2 + (31/3)4

Now you have:

[x +(31/3)2][x4 - (31/3)2x2 + (31/3)4] = x6 + 9