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x - 2 = 0 ,    x + 2 = 0  ,   x + 1 = 0 ,  x -1 = 0   x  = 2 ,      x  =  - 2  ,  x  = - 1    ,   x = 1So we shall check these by putting in 5x^3+8x^2−7x−6   one by one.

Putting x = - 2 ,  we get 5(-2)^3+8(-2)^2−7(-2)−6= 5 (-8) + 8(4) +14 - 6= -40 + 32 + 8 = 0So  x + 2  is a factor of given equation.

Putting x = 2 ,  we get 5(2)^3+8(2)^2−7(2)−6= 5 (8) + 8(4) -14 - 6= 40 + 32 -20= 52So  x - 2  is a not a factor of given equation.

Putting x = - 1 ,  we get 5(-1)^3+8(-1)^2−7(-1)−6= 5 (-1) + 8(1) +7 - 6= -5 + 8 + 1 = 4So  x + 1  is not a factor of given equation.

Putting x = 1 ,  we get 5(1)^3+8(1)^2−7(1)−6= 5 (1) + 8(1) -7 - 6= 5 + 8 -13 = 0So  x -1  is a factor of given equation.


Answer x-1 and x+2

Using the Factor Theorem, it is found that these following binomial expressions are factors of [tex]2x^3 + 5x^2 - x - 6[/tex].

[tex]x - 1[/tex]

[tex]x + 2[/tex]

The Factor Theorem states that if [tex]x_1, x_2, ..., x_n[/tex] are roots of [tex]f(x)[/tex], then [tex](x - x_1), (x - x_2), ..., (x - x_n)[/tex] are binomial factors of [tex]f(x)[/tex].

In this problem:

[tex]f(x) = 2x^3 + 5x^2 - x - 6[/tex]

[tex]x = 1[/tex] is a root, since [tex]f(1) = 0[/tex], thus a binomial factor is given by [tex]x - 1[/tex], and:

[tex](ax^2 + bx + c)(x - 1) = 2x^3 + 5x^2 - x - 6[/tex]

[tex]ax^3 +(b - a)x^2 + (c - b)x - c = 2x^3 + 5x^2 - x - 6[/tex]

Thus [tex]a = 2, c = 6, b - a = 5 \rightarrow b = 7[/tex].

The other roots are the solutions of:

[tex]2x^2 + 7x + 6[/tex]

Then:

[tex]\Delta = 7^{2} - 4(2)(6) = 1[/tex]

[tex]x_{1} = \frac{-7 + \sqrt{1}}{2(2)} = -\frac{3}{2}[/tex]

[tex]x_{2} = \frac{-7 - \sqrt{1}}{2(2)} = -2[/tex]  

Then, [tex]x + 2[/tex] and [tex]x + \frac{3}{2}[/tex] are also binomial factors.

[tex]x + \frac{3}{2}[/tex] is not an option, only [tex]x - 1[/tex] and [tex]x + 2[/tex]

A similar problem is given at https://brainly.com/question/24380382

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