Answer to Question #279504 in Algebra for Romart

Question

P(x)=6x³+5x²-2x-1

Answer 1

$$P(x)=6x^3+5x^2-2x-1$$

$$=6x^3+6x^2-x^2-x-x-1$$

$$=6x^2(x+1)-x(x+1)-(x+1)$$

$$=(x+1)(6x^2-x-1)$$

Let $6x^2-x-1=0.$

$$D=(-1)^2-4(6)(-1)=25$$

$$x=\dfrac{1\pm\sqrt{25}}{2(6)}=\dfrac{1\pm5}{12}$$

$$x_1=\dfrac{1-5}{12}=-\dfrac{1}{3}, x_2=\dfrac{1+5}{12}=\dfrac{1}{2}$$

Then

$$6x^2-x-1=6(x+\dfrac{1}{3})(x-\dfrac{1}{2})$$

$$=(3x+1)(2x-2)$$

$$P(x)=6x^3+5x^2-2x-1$$

$$=(x+1)(3x+1)(2x-2)$$

Zeros: $-1, -\dfrac{1}{3}, \dfrac{1}{2}$