How do you factor "0 = x^4 - 6x^2 + 5"? Please show all steps.
Here's my work, but I just went around in circles.....
0 = x^4 - 6x^2 + 5
-5 = x^4 - 6x^2
-5 = x^2 (x^2 - 6)
-5 / x^2 = x^2 - 6
(-5 / x^2) + 6 = x^2
(-5 + 6x^2) / (x^2) = x^2
-5 + 6x^2 = x^2 * x^2
-5 + 6x^2 = x^4
0 - x^4 - 6x^2 + 5
NO, you have not factored anything at all
x^4 - 6x^2 + 5 = 0
(x^2 - 1)(x^2 - 5) = 0
(x-1)(x+1)x^2 - 5) = 0 if you factor over the rationals
(x-1)(x+1)x-√5)(x+√5) = 0 if you factor over the reals
start by treating the problem like it is a quadratic. Let u = x^2
u^2 - 6u + 5
Factors as: (u-5)(u-1)
(x^2 -5)(x^2 -1) =0
Whenever there is = 0, Normally, you have to solve for x and not stop at the factoring step.
Set each factor = to zero.
(x^2-5)=0 or (x^2 - 1) = 0
x^2 = 5 or x^2 = 1
x = square root of 5 divided by 2
x = minus the square root of 5 divided by 2
x = 1
x = -1
Factoring a polynomial involves finding its roots or zeros, which are the values of the variable that make the equation equal to zero. In this case, we have the equation 0 = x^4 - 6x^2 + 5.
To factor this equation, let's start by rearranging the terms:
x^4 - 6x^2 + 5 = 0
Now, we can try factoring this equation by recognizing that it resembles a quadratic equation in terms of x^2. Let's substitute a new variable, let's say y, to represent x^2:
y^2 - 6y + 5 = 0
Next, let's factor this quadratic equation:
(y - 5)(y - 1) = 0
Now, we substitute back the original variable:
(x^2 - 5)(x^2 - 1) = 0
Finally, let's factor each binomial:
(x - √5)(x + √5)(x - 1)(x + 1) = 0
So, the factored form of the equation 0 = x^4 - 6x^2 + 5 is:
(x - √5)(x + √5)(x - 1)(x + 1) = 0
Therefore, the roots or zeros of the polynomial are: x = √5, x = -√5, x = 1, and x = -1.