3y^4 - 2y^2 - 5 Factor completely
To factor completely, we can start by factoring out the common factor, if any, then try to factor the resulting expression further.
Given expression:
3y^4 - 2y^2 - 5
There is no common factor that can be factored out.
So, now we need to try factoring the remaining expression further.
Let's denote z = y^2.
Now we have:
3z^2 - 2z - 5
We can use the quadratic formula to find the roots of this quadratic equation.
The quadratic formula is given by:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
For the quadratic equation 3z^2 - 2z - 5, the coefficients are:
a = 3, b = -2, c = -5
Using the quadratic formula, we have:
z = (-(-2) ± sqrt((-2)^2 - 4(3)(-5))) / (2(3))
= (2 ± sqrt(4 + 60)) / 6
= (2 ± sqrt(64)) / 6
= (2 ± 8) / 6
So, the two possible values for z are:
z = (2 + 8) / 6 = 10 / 6 = 5/3
z = (2 - 8) / 6 = -6 / 6 = -1
Now substitute back z = y^2:
For z = 5/3:
y^2 = 5/3
y = ±sqrt(5/3)
For z = -1:
y^2 = -1
No real solutions exist for y in this case.
Thus, the factored form of the given expression is:
3y^4 - 2y^2 - 5 = (y - sqrt(5/3))(y + sqrt(5/3))