Respuesta :
ANSWER
The solutions are:
• x = 0
,• x = -6
,• x = -8
EXPLANATION
First we can see that there's and x in every term of the polynomial, so one of the solutions to the equation is when x = 0, because x is a common factor:
[tex]\begin{gathered} x^3+14x^2+48x=0 \\ x(x^2+14x+48)=0 \end{gathered}[/tex]The other two solutions - we know that there are three because the degree of the polynomial is 3 - are the zeros of the polynomial in parenthesis. We can solve it with the formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In our polynomial, a = 1, b = 14 and c = 48:
[tex]\begin{gathered} x=\frac{-14\pm\sqrt[]{14^2-4\cdot48}}{2} \\ x=\frac{-14\pm\sqrt[]{196-192}}{2} \\ x=\frac{-14\pm\sqrt[]{4}}{2} \\ x=\frac{-14\pm2}{2} \\ x_1=\frac{-14-2}{2}=\frac{-16}{2}=-8 \\ x_2=\frac{-14+2}{2}=\frac{-12}{2}=-6 \end{gathered}[/tex]Therefore the solutions to the equation are x = 0, x = -8 and x = -6
