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What is the highest common facter of \( 2x^{3}\)+\( x^{2}\)-\( x\)-\( 2\) and \( 3x^{3}\)-\( 2x^{2}\)+\( x\)-\( 2\)?
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A) \( x-1\) |
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B) \( x+1\) |
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C) \( 2x+1\) |
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D) \( 2x-1\) |
Correct Answer:
| A) \( x-1\) |
Description for Correct answer:
H.C.F. of \( \Large 2x^{3} + x^{2} -x -2 \)
and \( \Large 3x^{3} - 2x^{2} + x - 2\)
From option (A)
For x = 1
\( \Large 2x^{3} + x^{2} -x -2 = 2 + 1 -1 -2 = 0 \)
and \( \Large 3x^{3} - 2x^{2} + x - 2 = 3 - 2 + 1 - 2 = 0 \)
Hence (x - 1) is the factor of both equation
From option (B)
For x = 1
\( \Large = 2 + 1 + 1 - 2 =-2 \ne 0 \)
Hence (x + 1) is not the factor of given equation
From option (C)
\( \Large x = -\frac{1}{2} \)
\( \Large 2x^{3} + x^{2} -x -2 = 2 \left( - \frac{1}{2}\right)^{3} + \left( - \frac{1}{2}\right)^{2} - \left( - \frac{1}{2}\right)^{-2} \)
\( \Large = -\frac{2}{8} + \frac{1}{4} + \frac{1}{2} - 2\)
\( \Large \frac{-2 + 2 + 4 -16}{8} \ne 0 \)
\( \Large \left( x + \frac{1}{2}\right) \) is not the factor of given equation.
From option(D)
\( \Large x = \frac{1}{2} \)
\( \Large 2x^{3} + x^{2} -x -2 = 2 \left( \frac{1}{2}\right)^{3} + \left( \frac{1}{2}\right)^{2} - \left( \frac{1}{2}\right)^{-2} \)
\( \Large = -\frac{2}{8} + \frac{1}{4} - \frac{1}{2} - 2\)
\( \Large \frac{2 + 2 - 4 -16}{8} = -2 \ne 0 \)
\( \Large \left( x - \frac{1}{2}\right) \) is not the factor of given equation.
Option (A) is only factor, hence it in H.C.F.
Option (A)is correct.
H.C.F. of \( \Large 2x^{3} + x^{2} -x -2 \)
and \( \Large 3x^{3} - 2x^{2} + x - 2\)
From option (A)
For x = 1
\( \Large 2x^{3} + x^{2} -x -2 = 2 + 1 -1 -2 = 0 \)
and \( \Large 3x^{3} - 2x^{2} + x - 2 = 3 - 2 + 1 - 2 = 0 \)
Hence (x - 1) is the factor of both equation
From option (B)
For x = 1
\( \Large = 2 + 1 + 1 - 2 =-2 \ne 0 \)
Hence (x + 1) is not the factor of given equation
From option (C)
\( \Large x = -\frac{1}{2} \)
\( \Large 2x^{3} + x^{2} -x -2 = 2 \left( - \frac{1}{2}\right)^{3} + \left( - \frac{1}{2}\right)^{2} - \left( - \frac{1}{2}\right)^{-2} \)
\( \Large = -\frac{2}{8} + \frac{1}{4} + \frac{1}{2} - 2\)
\( \Large \frac{-2 + 2 + 4 -16}{8} \ne 0 \)
\( \Large \left( x + \frac{1}{2}\right) \) is not the factor of given equation.
From option(D)
\( \Large x = \frac{1}{2} \)
\( \Large 2x^{3} + x^{2} -x -2 = 2 \left( \frac{1}{2}\right)^{3} + \left( \frac{1}{2}\right)^{2} - \left( \frac{1}{2}\right)^{-2} \)
\( \Large = -\frac{2}{8} + \frac{1}{4} - \frac{1}{2} - 2\)
\( \Large \frac{2 + 2 - 4 -16}{8} = -2 \ne 0 \)
\( \Large \left( x - \frac{1}{2}\right) \) is not the factor of given equation.
Option (A) is only factor, hence it in H.C.F.
Option (A)is correct.
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