Answer:
Answer:
2x • (x2 - 2xy + 5x - 10y)
Step-by-step explanation:
Step 1 :
Equation at the end of step 1 :
(((2•(x3))+(10•(x2)))-(22x2•y))-20xy
Step 2 :
Equation at the end of step 2 :
(((2 • (x3)) + (2•5x2)) - 22x2y) - 20xy
Step 3 :
Equation at the end of step 3 :
((2x3 + (2•5x2)) - 22x2y) - 20xy
Step 4 :
Step 5 :
Pulling out like terms :
5.1 Pull out like factors :
2x3 - 4x2y + 10x2 - 20xy =
2x • (x2 - 2xy + 5x - 10y)
Final result :
2x • (x2 - 2xy + 5x - 10y)
Step-by-step explanation:
Answer:
Thanks I guess
Step-by-step explanation:
If a polynomial with real coefficient has a complex zero, the conjugate of that number will be a zero as well.
So, if 4-6i is a zero, 4+6i will be a zero as well
If -2+11i is a zero, -2-11i will be a zero as well
So, the zeroes are
Minus 6 both sides
4>x/3
times3 both sides
12>x
Im pretty sure tge answers -1.68426
