The expression (3x^3 + 8x^2 + 4x) (x^2 + 3x – 4) has a term with x raised to the power 5. Therefore it has 5 roots. The roots can be found by equating (3x^3 + 8x^2 + 4x) (x^2 + 3x – 4) to 0.

(3x^3 + 8x^2 + 4x)(x^2 + 3x – 4) = 0

Let’s factorize the expression

=> x (3x^2 + 8x + 4) (x^2 + 3x – 4) = 0

=> x (3x^2 + 6x + 2x + 4) (x^2 + 4x – x -4) =0

=> x [3x(x + 2) +2(x+2)] [x(x + 4) – 1(x+4)] =0

=> x (3x+2) (x+2) (x – 1) (x+4) =0

Now equating each of the factors to 0

=> x1= 0

=> 3x +2 = 0, or x2 = -2/3

=> x+2 = 0, x or x3 = -2

=> x- 1 = 0, or x4 = 1

and x+4 = 0 or x5 = -4

**The required roots of (3x^3 + 8x^2 + 4x)(x^2 + 3x – 4) are [ -4 , -2, -2/3 , 0 , 1].**

Let f(x) = (3x^3 + 8x^2 + 4x)(x^2 + 3x – 4).

We need to find the roots of the function f(x).

First we will factor and simplify the function.

We notice that f(x) is a product of two terms.

Let us simplify each term.

==> (3x^3 + 8x^2 + 4x)

We will factor x from all terms.

==> 3x^3 + 8x^2 + 4x = x( 3x^2 + 8x + 4).

Now we will factor between brackets.

==> (3x^3 + 8x^2 + 4x = x( 3x+2)(x+2)

Now we will factor the second term.

(x^2 + 3x - 4) = ( x+4)(x-1).

Now we will rewrite the function f(x).

f(x) = x(3x+2)(x+2)(x+4)(x-1).

Now we have 5 roots for the function.

x1= 0, x2= -2/3, x3=-2, x4= -4 ,and x5= 1

**==> x= { 0, -2/3, -2, -4, 1}**

Given: (3x^3 + 8x^2 + 4x)(x^2 + 3x - 4)

First, do common factor

x(3x^2 + 8x +4)(x^2 + 3x - 4)

Then, factor everything

x(3x + 2)(x + 2)(x + 4)(x - 1)

To get the root, equate each factor to 0.

**x = 0**,

3x + 2 = 0

**x = -2/3**

x + 2 = 0

**x = -2**,

x + 4 = 0

**x = -4**,

x - 1 = 0

**x = 1**.

So the values of x or the roots are -4, -2, -2/3, 0 and 1. Do not forget to check the values of x by substituting to the given equated to 0. If it does not satisfy the equation, the said root is extraneous.

To find the roots of (3x^3 + 8x^2 + 4x)(x^2 + 3x – 4).

The roots of (3x^3 + 8x^2 + 4x)(x^2 + 3x – 4) are got by solving

(3x^3 + 8x^2 + 4x)(x^2 + 3x – 4) = 0.

Or by solving 3x^3 + 8x^2 + 4x = 0 ...(1) and

(x^2 + 3x – 4) = 0.....(2).

we take 3x^3+8x^2+4x = 0

=> x(3x^2+8x+4) = 0.

=> x((3x+2)(x+2) = 0.

=> x= 0, 3x+2= 0 , x+2= 0.

So x = 0, x = -2/3, x= -2.

Now we take the other factor equated to zero at (2):

x^2 + 3x – 4 = 0

=> (x+4)(x-1) = 0

=> x+4= 0, x-1 = 0.

So x= -4 and x = 1.

So the roots are : x = 0, x = -2/3, x= -2 , x= -4 and x = 1.