Answer:
1. roots may be +/-1,2,3,6 and fractions
A root must be minus since if they were positive, the function would be >0.
-1 works by inspection, so (x+1) is a factor.
-1/1==6==11==6
1===5===6==0
the factors are (x-1)(x^2+5x+6)=(x+1)(x+2)(x+3)
Roots are -1, -2,-3
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2.x^3-7x+6, Try+/-1,2,3,6
start with 1, and1-7+6=0; (x-1) is a factor
can divide
1/1=0==-7===6
1---1-- -6===0
The factors are (x+1)(x^2+x-6)=(x-1)(x+3)(x-2)
roots are 1,2,-3
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3.x^3+x^2-12x
factor out an x
x(x^2+x-12)=x(x+4)(x-3)
roots are 0, -4, 3
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4.9x^3-7x+2
roots can be +/-1, and +/-2 along with (1/3) (2/3),(1/9), (2/9)
-1
(x+1) is a factor.
-1/9===0== -7===2
9----- -9===2====0
(x+1)(9x^2-9x+2)
factor the last with +/- 1, 2.
second factors by noting x^2-9x+18 factors Into (x-6)(x-3)
divide the constant by 9 and put the denominator in front
(3x-2)(3x-1)
so the roots are -1, (2/3), and (1/3)
-
5x³+4x²-31x+6, note: I will assume 31 x.
Roots include (1/5)(2/5)(3/5)(6/5) along with
+/- 1,2,3,6, and 1, -1 do not work by inspection
2/5===4=== -31===6
5------14---- -3===0
so (x-2)(5x^2+14x-3)
or (x-2)(5x-1)(x+3) are the factors
roots are 2, (1/5), -3
hope this helps