Question

P(x)= 3x^3-5x^2-14x-4

Answer 1

   
$\displaystyle\\ P(x)=3x^3-5x^2-14x-4\\\\ D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\ \text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\ 3x^3-5x^2-14x-4=0$

$\displaystyle\\ \text{Verification}\\\\ 3x^3-5x^2-14x-4=\\\\ =3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\ =-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\ =-\frac{6}{9}+\frac{14}{3}-4=\\\\ =-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\ =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\ \Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\ \Longrightarrow~~~P(x)~\vdots~(3x+1)$


$\displaystyle\\ 3x^3-5x^2-14x-4=0\\ ~~~~~-5x^2 = x^2 - 6x^2\\ ~~~~~-14x =-2x-12x \\ 3x^3+x^2 - 6x^2-2x-12x-4=0\\ x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\ (3x+1)(x^2-2x -4)=0\\\\ \text{Solve: } x^2-2x -4=0\\\\ x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\ x_1 =1+\sqrt{5}\\ x_2 =1-\sqrt{5}\\ \Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$