Asked by Lavern
please help having trouble with this
(x+7)(x-12)(x+5)>0
the solution set is or is it all real numbers.
(x+7)(x-12)(x+5)>0
the solution set is or is it all real numbers.
Answers
Answered by
Bosnian
All parts of the uneqiations must be large of 0
Less part is ( x - 12 )
Solution x > 12
Less part is ( x - 12 )
Solution x > 12
Answered by
Lavern
multiply and simplify by factoring assume all expression under radicals represent non negative numbers
¡¼∛y¡½^7 ∛(¡¼81y¡½^8 )
¡¼∛y¡½^7 ∛(¡¼81y¡½^8 )
Answered by
Reiny
knowing the general shape of a cubic, and knowing that the intercepts are
x = -7, 12 , and -5
we can see that
(x+7)(x-12)(x+5)>0
has solution -7 < x < -5 OR x > 12
x = -7, 12 , and -5
we can see that
(x+7)(x-12)(x+5)>0
has solution -7 < x < -5 OR x > 12
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