Question

Factor the polynomial completely and find al zeros

P(x)=x^6-729

Please help!

Answer 1

\[P(x)= x^{6}-729\]

Answer 2

x^6 - 729
= (x^3)^2 - 27^2
= (x^3 - 27)(x^3 + 27)

Answer 3

now you can factor each parentheses using the cube identities

a^3 + b^3 = (a + b)(a^2 - ab + b^2) and
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Answer 4

putting a^3 = x^3 and b^3 = 27
where a= x and b = 3

Answer 5

to find the zeros
let (x^3 - 27)( x^3 + 27) = 0
and solve for x

Answer 6

\[(x+3)\times (x^{2}-3x+9)\times (x-3)\times (x ^{2}+3x+9)\]

Answer 7

is that right?.. for the zeros.. should i just use the the quadratic equations to get the remained zeros?

Answer 8

yes that's it

Answer 9

AAAHH thank you!

Answer 10

hmm maybe its better to find the zeroes using the last expression
x = 3 and -3 are obvious roots
to find the other 4 roots you need to solve
x^2 - 3x + 9 = 0 and x^2 + 3x + 9 = 0

Answer 11

could i solve them with the quadratic equation?

Answer 12

they will all be complex roots

Answer 13

quadratic formula yes

Answer 14

cool! thank you.. you were really helpful!

Answer 15

yw