Solve this Cubic Equation 5a^3 - 3a^2 - 2 = 0 ???

Question

5a^3 - 3a^2 - 2 = 0

Answer 1

1 real positive root... 2 non-real roots

The only possible rational real roots are +/- 1, 2, 1/5, 2/5

a = 1 is one solution

factor out (a-1) you get
5a^2 + 2a + 2 = 0

The rest is as easy as the quadratic formula:
a = [ -b +/- sqrt(b^2 - 4'a'c) ] / 2'a'

Because the coefficients on the quadratic are real, you know that the complex roots come as a conjugate pair.

The roots come out as
5a^2 + 2a + 2
a = [ -(2) +/- sqrt((2)^2 - 4(5)(2)) ] / 2(5)
a = [ -2 +/- sqrt(4 - 40 ] / 10
a = [ -2 +/- sqrt(-36) ] / 10
a = [ -2 +/- 6i ] / 10
a = -1/5 +/- 3/5i

Answer 2

Let's try a-1

5*1^3 - 3*1^2 - 2 = 5-3-2=0 Bingo!

Let's divide 5a^3 - 3a^2-2 by (a - 1)

5a^3 - 3a^2 - 2 - 5a^2(a-1) = 5a^3 - 3a^2 - 2 - 5a^3 + 5a^2 =
= 2a^2 - 2

2a^2 - 2 = 2(a^2 - 1) = 2(a + 1)(a - 1)

5a^3 - 3a^2 - 2 = 5a^2(a - 1) + 2(a + 1)(a - 1) =
= (5a^2 + 2a + 2)(a - 1)

Let's solve 5a^2 + 2a + 2 = 0 in order to find other roots:

a2,3 = [-2 +- sqrt(2^2 - 4*2*5)]/10 = [-2 +- sqrt(-36)]/10 =
= [-2 +- 6i]/10 = -0.2 +- 0.6i

Answer 3

5a^3 - 3a^2 - 2 = 0

f(a) = 5a^3 - 3a^2 - 2
Use the Factor Theorem
Try f(1)
f(1) = 5 - 3 - 2 = 0

Therefore (a-1) is a factor

Divide by (a-1) to get
(a-1)(5a^2 + 2a + 2) = 0
a=1 is one root

Use the quadratic formula to get the other two roots
a = [-2±sqrt(4-40)]/10
= [-2±6i]/10
= {(-1+3i)/5, (-1-3i)/5}

a = {1, (-1+3i)/5, (-1-3i)/5}

Answer 4

5a³ - 3a² - 2 = 0
delta = -3² - 4*5*-2
delta = 9 + 40
so = 3 +/- 7]) : 2*5
a' = 3 -7 = -4 : 10 = -0,4
a" = [3 + 7] : 10 = 1
\/1 = 1 or -1
Solution:[aa belongs to R| a' = -0.4, a" = 1 and a"' = 1
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