Respuesta :
We want to find the solutions for the following equation:
[tex]x^3+8x^2+4x-48=0[/tex]Since (x + 4) is a factor, we can rewrite this equation as a product of this factor by a second degree equation.
[tex]\begin{gathered} x^3+8x^2+4x-48=(x+4)(ax^2+bx+c) \\ x^3+8x^2+4x-48=ax^3+bx^2+cx+4ax^2+4bx+4c \\ x^3+8x^2+4x-48=ax^3+(4a+b)x^2+(c+4b)x+4c \end{gathered}[/tex]Comparing the coefficients from the left side with the coefficients from the right side, we get the following equations:
[tex]\begin{gathered} a=1 \\ 4a+b=8 \\ c+4b=4 \\ 4c=-48 \end{gathered}[/tex]Solving this system, we get the following results:
[tex]\begin{gathered} a=1 \\ b=4 \\ c=-12 \end{gathered}[/tex]Rewriting our polynomial, we have
[tex]x^3+8x^2+4x-48=(x+4)(x^2+4x-12)[/tex]We can rewrite our original equation as
[tex](x+4)(x^2+4x-12)=0[/tex]We have a product of two terms, then, if one is zero the product is zero.
To find the roots of this polynomial, we have to solve them individually
[tex]\begin{cases}x+4=0 \\ x^2+4x-12=0\end{cases}[/tex]The solution for the first equation and the first root, is x = -4. Solving the second one we get the two remaining roots. We could solve it using the Bhaskara Formula, but I'm going to factorize it again using the same process.
[tex]\begin{gathered} x^2+4x-12=(x+k)(x+l) \\ x^2+4x-12=x^2+(k+l)x+kl \end{gathered}[/tex][tex]\begin{cases}k+l=4 \\ kl=-12\end{cases}\Rightarrow\begin{cases}k=-2 \\ l=6\end{cases}[/tex]Our second degree polynomial can be rewritten as
[tex]x^2+4x-12=(x-2)(x+6)[/tex]This means our other roots are x = 2 and x = -6.
[tex]\begin{gathered} x_1=-4 \\ x_2=2 \\ x_3=-6 \end{gathered}[/tex]