Solve (4D+-8D³-7D² +11D+6)y=0.

Answer 1

set each factor = 0

y=0

or 4D+8D^3-7D^2 + 11D + 6 = 0

or 8D^3 -7D^2 +15D + 6 =0 has potentially maximum 3 real solutions, 2 or at least 1 real solution

It's a curve that graphically crosses the x axis 1 or 3 times, or crosses once and just touches the x axis for 2 real solutions

D = 0 cubic=+6

D = -1 cubic=-14

D=-1/2 cubic = -10.75

D=-1/4 cubic = +3 11/16

D= -1/3 cubic = -2/27

one real solution is a negative proper fraction, between -1/3 and -1/4

if there are 2 or 1 more real solution, they are also between - 0.33 and -0.25 that's where the cubic changes sign from - to +

the graph has 2 turning points

take the derivative of the cubic to get

15 +24D^2 -14D -=0

D = 14/48 + or - (1/48)sqr(14^2 +4(15)(14)) = about 0.96 or - -.38

D = -0.38 or +0.96

there is 1 real solution between -0.25 and -0.33,

there are two other imaginary solutions for D

but also one solution for y, y=0

when D= -0.322, the cubic =nearly zero