Question

I need help doing these kinds of problems. Thanks.

Answer 1

ok

Answer 2

/i thought it goes into e^x(x) but am unsure if that is correct or how to find the x's

Answer 3

\[xe^x+e^x=0\]
\[e^x(x+1)=0\] and since \(e^x>0\) always, set \(x+1=0\) and solve for \(x\)

Answer 4

second one is similar
factor out the common factor of \(e^{9x}\) then ignore it and set the other factor equal to zero and solve

Answer 5

the second one I factor down to \[\large x^3e^9x(9x+4) ?\]

Answer 6

x^2e^9x(9x+4)

Answer 7

\[x^3e^{9x}(9x+4)\] is the right factorisation

Answer 8

just check your sign for the zeroes

Answer 9

so x^3 = 0 and e^9x disapears right? then 9x+4 is -4/9 so my x's would be 0, -4/9?

@unklerhaukus

Answer 10

Yes that is right,

(the reason the e^{9x} 'disappears' is because e^{9x} is never zero