Answer :

We start with the polynomial

[tex]$$5x^3 - 19x^2 + 12x.$$[/tex]

Step 1. Factor out the common factor

Notice that each term has a factor of [tex]$x$[/tex]. Factor [tex]$x$[/tex] out:

[tex]$$5x^3 - 19x^2 + 12x = x \Bigl(5x^2 - 19x + 12\Bigr).$$[/tex]

Step 2. Factor the quadratic

Now, we need to factor the quadratic

[tex]$$5x^2 - 19x + 12.$$[/tex]

We look for two numbers that multiply to [tex]$5 \cdot 12 = 60$[/tex] and add to [tex]$-19$[/tex]. The two numbers that satisfy these conditions are [tex]$-15$[/tex] and [tex]$-4$[/tex], because

[tex]$$(-15) \cdot (-4) = 60$$[/tex]
[tex]$$(-15) + (-4) = -19.$$[/tex]

Rewrite the quadratic by decomposing the middle term:

[tex]$$5x^2 - 19x + 12 = 5x^2 - 15x - 4x + 12.$$[/tex]

Now, factor by grouping:

- Group the first two terms and the last two terms:

[tex]$$5x^2 - 15x - 4x + 12 = (5x^2 - 15x) + (-4x + 12).$$[/tex]

- Factor out the common factors in each group:

[tex]$$5x^2 - 15x = 5x(x - 3),$$[/tex]

[tex]$$-4x + 12 = -4(x - 3).$$[/tex]

- Now factor out the common binomial [tex]$(x-3)$[/tex]:

[tex]$$5x(x-3) - 4(x-3) = (x-3) \Bigl(5x-4\Bigr).$$[/tex]

Thus, the quadratic factors as

[tex]$$5x^2 - 19x + 12 = (x-3)(5x-4).$$[/tex]

Step 3. Write the fully factored form

Substituting back into the factored form with the extracted [tex]$x$[/tex], we get:

[tex]$$5x^3 - 19x^2 + 12x = x (x-3)(5x-4).$$[/tex]

Step 4. Find the roots

To find the roots of the polynomial, set each factor equal to zero:

1. From [tex]$x = 0$[/tex], we have:

[tex]$$x = 0.$$[/tex]

2. From [tex]$x-3=0$[/tex], we have:

[tex]$$x = 3.$$[/tex]

3. From [tex]$5x-4=0$[/tex], we have:

[tex]$$5x = 4 \quad \Rightarrow \quad x = \frac{4}{5}.$$[/tex]

Final Answer

The factored form of the polynomial is

[tex]$$x (x-3)(5x-4),$$[/tex]

and the roots are

[tex]$$x = 0, \quad x = 3, \quad x = \frac{4}{5}.$$[/tex]