Question

130z-z^3+3z^2

Answer 1

what are you trying to do?

Answer 2

factor

Answer 3

factor out a z and then it will be a quadratic. you can either factor or use the quadratic formula

Answer 4

ok how do i do that

Answer 5

rewrite in descending order:
\[130z-z ^{3}+3z ^{2}= -z ^{3}+3z ^{2} +130z= z \left( -z ^{2}+3z +130 \right)\]

Answer 6

you can factor out a -1 also

Answer 7

ok i figured it out thanks

Answer 8

well i got z(130-z+z^2) is that right

Answer 9

it should be z(130 - z^2 + 3z)

Answer 10

but (130 - z^2 +3z) will factor too

Answer 11

ok

Answer 12

(130 - z^2 + 3z) = (130 +3z - z^2)