HOW DO I FACTOR THIS: x^3 - 8y^3?

Question

Also, how do I solve for x: x^2 + 5x + 4 = 0. If the solutions are complex numbers, then how do I give the solutions in the usual a + bi form!?!

Answer 1

[04]
x^3-8y^3
=(x)^3-(2y)^3
=(x-2y){(x)^2+x*2y+(2y)^2}
=(x-2y)(x^2+2xy+4y^2)

x^2+5x+4=0
x^2+x+4x+4=0
x(x+1)+4(x+1)=0
(x+1)(x+4)=0
x= -1 or -4

Answer 2

x^3 - 8y^3
(x - 2y)(x^2 + 2yx + y^2)

x^2 + 5x + 4 = 0
(x + 4)(x + 1) = 0
x = -1; x = -4

Answer 3

x^3-8y^3

(x-2y)(x^2+2xy+4y^2)

x^2+5x+4=0
(x+1)(x+4)=0
x=-1 or -4

Answer 4

Hi,

Lets solve this equation

x^2+5x+4=0

The best and fast way to slove it is to use quadratic formula.

a=1 b=5 c=4

x=-b+/-sq.root of b^2-4ac/2a

x=-5+/-sq.root of 25-4*1*4/2

x=-5+/-sq.root of 25-16/2

x=-5+/-sq.root of 9/2

x1=-5+3/2

x1=-2/2

x1=-1

x2=-5-3/2

x2=-8/2

x2=-4

So the two answers are -1 and -4

All the best,
Mika

Answer 5

x^3 -- 8y^3
= x^2 -- (2y)^3
= (x -- 2y)^3 + 6xy(x - 2y)
= (x -- 2y){(x -- 2y)^2 + 6xy}
= (x -- 2y){x^2 -- 4xy + 4y^2 + 6xy}
= (x -- 2y)(x^2 + 2xy + 4y^2)

x^2 + 5x + 4 = 0
Or x^2 + 4x + x + 4 = 0
Or x(x + 4) + 1(x + 4) = 0
Or (x + 4)(x + 1) = 0
giving x = -- 4, -- 1
x = -- 4 + 0i, -- 1 + 0i