How do I solve this equation?

Question

w3 + 5w2 – w = 5

Answer 1

Solve the "w" variable in the equation.

First: set the equation to equal "0" - subtract 5 from both sides (when you move a term to the opposite side, always use the opposite sign).

w^3 + 5w^2 - w - 5 = 5 - 5
w^3 + 5w^2 - w - 5 = 0

Sec: with 4 terms, group "like" terms & factor both sets of parenthesis.

(w^3 - w) + (5w^2 - 5) = 0
w(w^2-1) + 5(w^2-1) = 0
(w^2-1)(w+5) = 0

Third: solve the "w" variables - set both parenthesis to "0"

a. w^2 - 1 = 0

* Add 1 with both sides.

w^2 - 1 + 1 = 0 +1
w^2 = 1

*Eliminate the exponent - find the square root of both sides.

V`(w^2) = +/- V`1
w = +/- 1
w = 1, -1

b. w+5 = 0

*Subtract 5 from both sides.

w+5 -5 = 0 - 5
w = 0 - 5
w = - 5

Solutions: 1, - 1, -5

Answer 2

Ok, so you have w^3 + 5w^2 - w = 5

First you need to set it equal to 0 by moving the 5 over to the other side of the equation, giving you:

w^3 + 5w^2 - w - 5 = 0

This can be simplified basically by a guess and check type of thing.

Let's choose a factor, say (x + 5) and see if it fits, we will have to use synthetic division:

-5 | 1 5 -1 -5
1 -5 0 5
1 0 -1 0

This turns out to say this:
w^2 - 1

So now we have two factors, (w + 5) and (w^2 -1)

(w + 5)(w^2 - 1) = 0

w^2 - 1 is a difference of squares so it simplifies even further:

(w + 5)(w + 1)(w - 1) = 0

So the solutions are as follows:

w = -5
w = -1
w = 1

And there you go.

Answer 3

w3+5w2-w-5=0 You can factorize it as
(x-1)( x2+6x+5) and the second factor is (x+5)(x+1) So the roots are
x=1 x=-1 and x=-5

Answer 4

w= -5 w= -1 w= 1; since it is w^3 you know there has to be three answers

Answer 5

abc method