How to solve the roots of following cubic equation $a^3-6a^2+9a-4=0$

Question

How to solve the roots of following cubic equation

$$a^3-6a^2+9a-4=0$$

I am solving roots of characteristics equation

so I use casio 991ms calculator so it gave me roots $-0.355301,3.177650,3.1776506$

please help me with this

Answer 1

HINT

We have

$$a^3-6a^2+9a-4=a^3-2a^2+a-4a^2+8a-4=a(a-1)^2-4(a-1)^2$$

Answer 2

Get the critical values of the equation by testing for a number that produces the result 0. Then use that equation to divide the polynomial, you'll be left with a quadratic equation. I think you can continue from there

Answer 3

you did something wrong, not sure calculator issue or a typo.

For cubic equations, if it's a setup question better to try a simple root first by inspection.

Hint: try $a=1$.

Once you have one value, you can reduce to second order and find the other two roots.

Answer 4

The numbers $1,4$ are roots of your equation. Can you find the rest? In case you want to have a view of the graph you can use a nice software like "desmos".

Answer 5

$a^3-6a^2+9a-4=0$

$a(a^2+6a+9)=4$

$a(a-3)^2=4\times 1^2$

$a_1=4$

$\frac{a^3-6a^2+9a-4}{a-4}= a^2-2a+1$

$a^2-2a+1=0$

$(a-1)^2=0$

$a_2=a_3=1$