How do you factor 2x^-21x+40 ?
I really need help with this! I have a test coming up next week and knowing how to solve this would really help! Also after solving this I need help setting it equal to zero and solving.
Update:I need help with this by the end of today PLEASE!
Comments
I'll try to make it as simple as I can
2 x 40 =80
Find 2 numbers which when you multiply gives you 80 and when you add (or subtract) gives you -21
the 2 numbers are : -16 and -5 because (-16x-5)=80 and (-16-5) =-21
**There is an error in your question:you forgot the power of 2'**
So, 2x^2 -21x+40 =2x^2 -16x-5x+40
=>2x(x-8) -5(x-8)
=>(2x-5) (x-8)<--
Now ,if the equation is set to 0
=>(2x-5) (x-8) =0
either 2x-5=0 or x-8=0
either 2x=5 or x=8
either x=5/2 or x=8 <--