Question

factor za^2 - 2a - 2z^a + 4z

Answer 1

factor out za from the first and third terms, factor out 2 from second and fourth, and there should be something more that you can factor :P

Answer 2

In the third term, z is raised to a. How can I factor out za from the third term? Thanks for the help. :)

Answer 3

hmmm wondering if we can do a substitution

Answer 4

if it was factorable, we could equate it to 0

za^2 - 2a - 2z^a + 4z = 0

a(za - 2) - 2z (z^(a-1) -2) = 0

a(za - 2) = 2z (z^(a-1) -2)

a(za - 2)/z = 2(z^(a-1) -2)

not real sure that it can be facotred

Answer 5

completing the square in "a" maybe

Answer 6

Thanks. :)

Answer 7

\[za^2 - 2a - 2z^a + 4z \]
\[z(a^2 - 2a +1) -z- 2z^a + 4z \]
\[z(a-1)^2 - 2z^a +3z \]
assuming that z not equal to 0
\[z(a-1)^2 - 2z^a +3z =0\]
\[(a-1)^2 - 2z^{a-1} +3 =0\]
\[(a-1)^2= 2z^{a-1} -3\]
\[a-1= \pm~\sqrt{2z^{a-1} -3}\]
\[a= 1\pm~\sqrt{2z^{a-1} -3}\]
but thats not factoring it; its just trying to define the roots