What are the real or imaginary solutions of the polynomial equation. 64x^3+27=0
Please help!!!! medal and fann!!!!

Question

What are the real or imaginary solutions of the polynomial equation. 64x^3+27=0
Please help!!!! medal and fann!!!!

anonymous · Answer

do you know how to factor the sum of two cubes?

anonymous · Answer

no i cant say i do.

anonymous · Answer

i skipped 3 units so idk how to do any of this?

anonymous · Answer

ok i will show you 
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

anonymous · Answer

in this case you have 
$$64x^3+27=(4x)^3+3^3$$ put $a=4x,b=3$ and plug them in

anonymous · Answer

so 4x^3+3^3=(4x+3)(4x^2-4x*3+3^2)
right?

anonymous · Answer

is that what you meant by plug them in?
cause that was a formula right?

anonymous · Answer

yeah

anonymous · Answer

oops no

anonymous · Answer

but close

anonymous · Answer

$$4x^3+3^3=(4x+3)(4x^2-4x*3+3^2)$$ is what you wrote but you need to square the 4 as well, should be 
$$4x^3+3^3=(4x+3)(16x^2-12x+9)$$

anonymous · Answer

oh ok that makes more sense

anonymous · Answer

what do i do from there?

geerky42 · Answer

$$0 = (4x)^3+3^3=(4x+3)(16x^2-12x+9)$$

So either $4x+3 = 0$   OR   $0 = 16x^2-12x+9$

So from here, you can easily solve first equation, then for second equation, just use quadratic formula.