factorize z^4 + 2z^3 - 2z^2 + 8 into linear factors
does not appear to factor, check your spelling
http://www.algebrahelp.com/calculators/expression/factoring/calc.do?expression=x%5E4+-+2x%5E3+-+2x%5E2+%2B+8
no, i entered it correctly. the unit is complex numbers and polynomials
To factorize the expression z^4 + 2z^3 - 2z^2 + 8 into linear factors, we need to look for common factors and then use various factoring techniques.
The first step is to factor out any common factors among all the terms. In this case, we can't factor out any common factors.
Next, we can try to factor by grouping. Let's group the terms and see if any common factors emerge:
Grouping 1: (z^4 + 2z^3) - (2z^2 - 8)
Within the first group, we can factor out z^3 as a common factor: z^3(z + 2). Similarly, within the second group, we can factor out 2 as a common factor: -2(z^2 - 4).
Combining these factors, we get: z^3(z + 2) - 2(z^2 - 4).
Now, we can further factor the terms inside the parentheses:
z^3(z + 2) - 2(z^2 - 4) = z^3(z + 2) - 2(z + 2)(z - 2).
Now, we have a common factor of (z + 2), so we can factor it out:
(z + 2)(z^3 - 2(z - 2)).
At this point, we see that we cannot factorize the remaining terms further. Therefore, the fully factored form of z^4 + 2z^3 - 2z^2 + 8 into linear factors is:
(z + 2)(z^3 - 2(z - 2)).