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4x^3-12x^2-36x+27/(2x-3)
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Updated on: 09 Aug 2024, 09:29
Updated on: 09 Aug 2024, 09:29
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Expert Reply
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Question Stats:
47% (02:42) correct
52% (01:48) wrong based on 19 sessions
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\(\frac{4x^3-12x^2-36x+27}{(2x-3)}=0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Solution for x |
0 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
4x^3-12x^2-36x+27/(2x-3)
[#permalink]
09 Aug 2024, 09:34
09 Aug 2024, 09:34
Expert Reply
I fixed it because the book did not equate it to zero which it should be
Now
\(\frac{4x^3-12x^2-36x+27}{(2x-3)}=0\)
\(4x^3-12x^2-36x+27=0\)
The best approach is to use values. Algebra is a bit cumbersome
If you use 0 to see if the answer is C then we do have, substituting
\(27 >0\) So C is not possible
If we use 2
we have
\(32-48-72+27=-61<0\)
You have already two conflict answers
D is right away
Now
\(\frac{4x^3-12x^2-36x+27}{(2x-3)}=0\)
\(4x^3-12x^2-36x+27=0\)
The best approach is to use values. Algebra is a bit cumbersome
If you use 0 to see if the answer is C then we do have, substituting
\(27 >0\) So C is not possible
If we use 2
we have
\(32-48-72+27=-61<0\)
You have already two conflict answers
D is right away
