How do you factor #f(x)=2x^3-10x^2-8x+40#?

Question

↳Redirected from "How does the rate of reaction change with concentration?"

3214 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-14T14:01:26

#(x-2)# #(2x-10)# #(x+2)#

First, find a factor of #f(x)#

try #f(1)#

#f(1)# = 24 therefore is not a factor

next try #f(2)#

#f(2)# = 0 therefore #(x-2)# is a factor of #f(x) = 2x^3 - 10x^2 - 8x - 40#

Now, divide #f(x) = 2x^3 - 10x^2 - 8x - 40# by #(x-2)# using long division.

enter image source here

= #(x-2)# #(2x^2 -6x -20)#

Then simplify this so

= #(x-2)# #(2x-10)# #(x+2)#