Answer:
[tex]x=-1, \quad x=\dfrac{1}{2}, \quad x=-\dfrac{2}{3}[/tex]
Step-by-step explanation:
Given polynomial:
[tex]f(x)=6x^2+7x^2-x-2[/tex]
[tex]\boxed{\begin{minipage}{6 cm}\underline{Factor Theorem}\\\\If $f(x)$ is a polynomial, and $f(a) = 0$,\\ then $(x-a)$ is a factor of $f(x)$.\\\end{minipage}}[/tex]
Since the constant of the given polynomial is -2, the possible factors are ±1 and ±2.
Apply the Factor Theorem to check the possible values:
[tex]f(1)=6(1)^3+7(1)^2-(1)-2=10 \neq 0[/tex]
[tex]f(-1)=6(-1)^3+7(-1)^2-(-1)-2=0[/tex]
[tex]f(2)=6(2)^3+7(2)^2-(2)-2=72 \neq 0[/tex]
[tex]f(-2)=6(-2)^3+7(-2)^2-(-2)-2=-20 \neq 0[/tex]
Therefore, (x + 1) is a factor of the polynomial:
[tex]f(x)=(x+1)(ax^2+bx+c)[/tex]
Compare with the original function to find the coefficients of the quadratic:
[tex]6x^3=ax^3 \implies a=6[/tex]
[tex]7x^2=(b+a)x^2 \implies b=1[/tex]
[tex]-2=c \implies c=-2[/tex]
Therefore:
[tex]f(x)=(x+1)(6x^2+x-2)[/tex]
Factor the quadratic:
[tex]f(x)=(x+1)(6x^2+4x-3x-2)[/tex]
[tex]f(x)=(x+1)[2x(3x+2)-1(3x+2)][/tex]
[tex]f(x)=(x+1)(2x-1)(3x+2)[/tex]
Set the polynomial function to zero:
[tex](x+1)(2x-1)(3x+2)=0[/tex]
Apply the zero product property:
[tex](x+1)=0 \implies x=-1[/tex]
[tex](2x-1)=0 \implies x=\dfrac{1}{2}[/tex]
[tex](3x+2)=0 \implies x=-\dfrac{2}{3}[/tex]
Therefore, the solutions are:
[tex]x=-1, \quad x=\dfrac{1}{2}, \quad x=-\dfrac{2}{3}[/tex]
