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We can know the answer to this by letting p(x) be equal to zero and by substituting the values of x to the function.

[tex]p(x)= x^{3} -2 x^{2} -5x+6=0[/tex]

To know what to substitute, we equate all factors to zero since even if only one of these factors is zero, the product of all factors would immediately be zero. 

1. [tex]x+1=0; x=-1[/tex]
2. [tex]x+2=0; x=-2[/tex]
3. [tex]x-1=0; x=1[/tex]
4. [tex]x-3=0; x=3[/tex]

We then try to substitute each value of x to the original polynomial and see if it will equal to zero.

1. [tex](-1)^{3} -2(-1)^{2} -5(-1)+6=8\neq0[/tex]
2. [tex](-2)^{3} -2(-2)^{2} -5(-2)+6=0[/tex]
3. [tex](1)^{3} -2(1)^{2} -5(1)+6=0[/tex]
4. [tex](3)^{3} -2(3)^{2} -5(3)+6=0[/tex]

As we can see, all but number 1 equated to zero therefore numbers 2, 3, and 4 are all linear factors of p(x).

ANSWER: Numbers 2, 3, and 4.
To be a linear factor, the value of x must make the value of the polinomial equal to zero:
1. x+1=0→x+1-1=0-1→x=-1
p(-1)=(-1)^3-2(-1)^2-5(-1)+6
p(-1)=-1-2(1)+5+6
p(-1)=-1-2+5+6
p(-1)=8 different to zero, then x+1 is not a linear factor of p(x)

2. x+2=0→x+2-2=0-2→x=-2
p(-2)=(-2)^3-2(-2)^2-5(-2)+6
p(-2)=-8-2(4)+10+6
p(-2)=-8-8+10+6
p(-2)=0, then x+2 is a linear factor of p(x)

3. x-1=0→x-1+1=0+1→x=1
p(1)=(1)^3-2(1)^2-5(1)+6
p(1)=1-2(1)-5+6
p(1)=1-2-5+6
p(1)=0, then x-1 is a linear factor of p(x)

4. x-3=0→x-3+3=0+3→x=3
p(3)=(3)^3-2(3)^2-5(3)+6
p(3)=27-2(9)-15+6
p(3)=27-18-15+6
p(3)=0, then x-3 is a linear factor of p(x)

Answer: The factor of p(x) are options 2, 3, and 4

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