find the real number solutions of the equation.
w^4+w^3-w=1
could you please show step by step..thank you
is it only 1 b/c if u substitute 1 into w, it'll equal to 1
w^4+w^3-w=1
w^4+w^3-w-1=0
(w-1)(w^3+2w^2+2w+1)=0
(w-1)(w+1)(w^2+w+1)=0
so real solutions are 1 and -1
complex solutions are 1 -1 j j^2 with j=exp(2Pi/3)
To find the real number solutions of the equation w^4 + w^3 - w = 1, we can start by rewriting the equation as a polynomial equation equal to zero:
w^4 + w^3 - w - 1 = 0
Now, let's break this down step by step to solve for w:
Step 1: Rearrange the equation:
w^4 + w^3 - w - 1 = 0
Step 2: Factor out a common term (if possible):
Since there are no common terms to factor out, we proceed to the next step.
Step 3: Look for possible rational roots:
The Rational Root Theorem helps us identify possible rational roots by listing all the factors of the constant term, which is -1 in this case. Factors of -1 are ±1.
Step 4: Test the possible rational roots:
We can test the two possible rational roots, w = ±1, by substituting them into the equation to see if they satisfy it.
If we substitute w = 1 into the equation, we get:
(1)^4 + (1)^3 - (1) - 1 = 0
1 + 1 - 1 - 1 = 0
0 = 0
Since the left-hand side equals the right-hand side, w = 1 is a solution.
If we substitute w = -1 into the equation, we get:
(-1)^4 + (-1)^3 - (-1) - 1 = 0
1 - 1 + 1 - 1 = 0
0 = 0
Again, the left-hand side equals the right-hand side, so w = -1 is also a solution.
Therefore, the real number solutions of the equation w^4 + w^3 - w = 1 are w = 1 and w = -1.