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 } (-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*\Big(-(1)/(3)\Big)^3-5*\Big(-(1)/(3)\Big)^2-14*\Big(-(1)/(3)\Big)-4=\\\\ =-(1)/(9)-(5)/(9)+(14)/(3)-4=\\\\ =-(6)/(9)+(14)/(3)-4=\\\\ =-(6)/(9)+(42)/(9)- (4* 9)/(9)=\\\\ =-(6)/(9)+(42)/(9)- (36)/(9)= (42-6-36)/(9)=(42-42)/(9)=(0)/(9)=0\\\\ \Longrightarrow~~~P(x)~\vdots~\Big(x+ (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)= (-b\pm √(b^2-4ac))/(2a)=\\\\=(2\pm √(4+16))/(2)=(2\pm √(20))/(2)=(2\pm 2√(5))/(2)=1\pm√(5)\\\\ x_1 =1+√(5)\\ x_2 =1-√(5)\\ \Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-√(5))(x-1+√(5))}`