Answer :
We want to solve the cubic equation
[tex]$$
6x^3 + 19x^2 - 19x + 4 = 0.
$$[/tex]
A good strategy is to factor the polynomial. In this case, we can rewrite the polynomial in its factored form:
[tex]$$
(x+4)(2x-1)(3x-1)=0.
$$[/tex]
Now, we set each factor equal to zero:
1. For the factor [tex]$x+4=0$[/tex]:
[tex]$$
x+4=0 \quad \Rightarrow \quad x=-4.
$$[/tex]
2. For the factor [tex]$2x-1=0$[/tex]:
[tex]$$
2x-1=0 \quad \Rightarrow \quad 2x=1 \quad \Rightarrow \quad x=\frac{1}{2}.
$$[/tex]
3. For the factor [tex]$3x-1=0$[/tex]:
[tex]$$
3x-1=0 \quad \Rightarrow \quad 3x=1 \quad \Rightarrow \quad x=\frac{1}{3}.
$$[/tex]
Thus, the solutions to the equation are
[tex]$$
x = -4, \quad x = \frac{1}{2}, \quad x = \frac{1}{3}.
$$[/tex]
This completes the step-by-step solution of the cubic equation.
[tex]$$
6x^3 + 19x^2 - 19x + 4 = 0.
$$[/tex]
A good strategy is to factor the polynomial. In this case, we can rewrite the polynomial in its factored form:
[tex]$$
(x+4)(2x-1)(3x-1)=0.
$$[/tex]
Now, we set each factor equal to zero:
1. For the factor [tex]$x+4=0$[/tex]:
[tex]$$
x+4=0 \quad \Rightarrow \quad x=-4.
$$[/tex]
2. For the factor [tex]$2x-1=0$[/tex]:
[tex]$$
2x-1=0 \quad \Rightarrow \quad 2x=1 \quad \Rightarrow \quad x=\frac{1}{2}.
$$[/tex]
3. For the factor [tex]$3x-1=0$[/tex]:
[tex]$$
3x-1=0 \quad \Rightarrow \quad 3x=1 \quad \Rightarrow \quad x=\frac{1}{3}.
$$[/tex]
Thus, the solutions to the equation are
[tex]$$
x = -4, \quad x = \frac{1}{2}, \quad x = \frac{1}{3}.
$$[/tex]
This completes the step-by-step solution of the cubic equation.
