f(w) = w^4+3w^3-22w^2 + 2 is
f'(w) = 4w^3 +9w^2 -44w right?
I want to set this equal to 0, so can i do this?
4w^3 +9w^2 -44w = 0
w(4w^2 +9w -44) = 0

and then use a quadratic formula?
if i do, what happen to the w at the left?

Question

f(w) = w^4+3w^3-22w^2 + 2 is 
f'(w) = 4w^3 +9w^2 -44w   right?
I want to set this equal to  0, so can i do this?
4w^3 +9w^2 -44w = 0
w(4w^2 +9w -44) = 0

and then use a quadratic formula?
if i do, what happen to the w at the left?

ranga · Answer

If A * B = 0
then either A = 0 or B = 0 or both are 0.

A cubic equations will have 3 roots.
w(4w^2 +9w -44) = 0  implies w = 0 or 4w^2 +9w -44 = 0
solving the quadratic will give you two roots.  The third root is w = 0.

anonymous · Answer

$$-9 \pm \sqrt{9^2 - 4(4)(-44)} $$ is equal to 1.9949 and - 4.2449 right?

ranga · Answer

Don't forget to divide by 2a or 8

anonymous · Answer

(−9±92−4(4)(−44))/2(4)

ranga · Answer

Above 9^2 looks like 92 and square root is missing.  Use your reply before the last one but just divide by 8.

ranga · Answer

w = 1/8(-9 + sqrt(785))  or 1/8(-9 - sqrt(785))   and the third root is w = 0.

anonymous · Answer

Oh yes, when i copy and paste, the square has disappear