If #A_1x^3+A_2x^2+A_3x+A_4=0# Can be simplified Down to #a_1x^2+a_2x+a_3=0#, what are the Values of #a_1,a_2,a_3#?

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Answer 1 · 2018-03-07T14:51:22

See the explanation below

We need

#(x^n)'=nx^(n-1)#

#A_1x^3+A_2x^2+A_3x+A_4=0#

Calculate the derivative

#(A_1x^3+A_2x^2+A_3x+A_4)'=0#

#3A_1x^2+2A_2x+A_3=0#

Comparing this to

#a_1x^2+a_2x+a_3=0#

Therefore,

#a_1=3A_1#

#a_2=2A_2#

#a_3=A_3#