How can I solve these inequalities???
1)
x^3-5x^2+2x+8≥0
2)
2(x^3-2x^2+3)
please help!!
1)
x^3-5x^2+2x+8≥0
2)
2(x^3-2x^2+3)
please help!!
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1) If you plug in x=(-1) in x^3-5x^2+2x+8, we get (-1)-5-2+8 which is 0
This shows x+1 is a factor of x^3-5x^2+2x-8.
Dividing we get the quotient to be x^2-6x+8 which can be factored as
(x-4)(x-2)
Thus given inequality is (x+1)(x-4)(x-2)<=0
Now product of 3 numbers is <=0 if 2 of them are same sign and the third is opposite sign
Note that if x+1<=0, then x<=(-1) and then x<2 and x<4 so that all 3 quantities are <=0, and their product is <=0
But if x+1>0, then 2<=x<=4 in order for the product of the 3 to be <=0
So the inequality is true when x<=(-1) or 2<=x<=4,
i.e., when x is in (-infinity,(-1)] union [2,4]
2) If you subtract the right hand quantity and simplify you get
x^3-4x^2+x+6 <0
x^3-4x^2++6 can be factored as (x+1)(x^2-3x+3) just as in (1) above
x^2-3x+3 is (x-3/2)^2+(3/4) and is therefore always >=0
So for the product to be <0, x+1 must be <0, i.e., x<(-1)
So when x is in (-infinity,(-1))
This shows x+1 is a factor of x^3-5x^2+2x-8.
Dividing we get the quotient to be x^2-6x+8 which can be factored as
(x-4)(x-2)
Thus given inequality is (x+1)(x-4)(x-2)<=0
Now product of 3 numbers is <=0 if 2 of them are same sign and the third is opposite sign
Note that if x+1<=0, then x<=(-1) and then x<2 and x<4 so that all 3 quantities are <=0, and their product is <=0
But if x+1>0, then 2<=x<=4 in order for the product of the 3 to be <=0
So the inequality is true when x<=(-1) or 2<=x<=4,
i.e., when x is in (-infinity,(-1)] union [2,4]
2) If you subtract the right hand quantity and simplify you get
x^3-4x^2+x+6 <0
x^3-4x^2++6 can be factored as (x+1)(x^2-3x+3) just as in (1) above
x^2-3x+3 is (x-3/2)^2+(3/4) and is therefore always >=0
So for the product to be <0, x+1 must be <0, i.e., x<(-1)
So when x is in (-infinity,(-1))