Question

X^3+7x^2+13x+4=0 Show all work

Answer 1

   
$\displaystyle\\ x^3+7x^2+13x+4=0\\\\ x^3+\underbrace{4x^2+3x^2}_{7x^2}+\underbrace{12x+x}_{13x}+4=0\\\\ (x^3+4x^2) + (3x^2 + 12x)+(x+4)=0\\\\ x^2(x+4)+3x(x+4)+(x+4)=0\\\\ (x+4)(x^2 + 3x +1)=0\\\\ x+4 = 0~~~\Longrightarrow~~~\boxed{x_1 = -4}\\\\ x^2 + 3x +1=0\\\\ x_{23}= \frac{-b\pm \sqrt{b^2-4ac} }{2a}= \frac{-3\pm \sqrt{3^2-4\cdot 1 \cdot 1} }{2\cdot 1}= \frac{-3\pm \sqrt{5} }{2}\\\\ \boxed{x_2 = \frac{-3-\sqrt{5} }{2}}\\\\ \boxed{x_3 = \frac{-3+\sqrt{5} }{2}}$