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Question

Find all the zeros of 2x49x3+5x2+3x1, if two of its zeros are 2+3 and 23

A
2, 3, 1
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B
4, 7, -1
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C
6, 5, 8
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D
None of these
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Solution

The correct option is D None of these
Let, p(x)=2x49x3+5x2+3x1
Since, 2+3 and 23 are zeros of p(x)
(x23) and (x2+3) divides p(x) (Factor thm.)
(x23)(x2+3) divides p(x)
(x24x+1) divides p(x)
x24x+1)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2x49x3+5x2+3x1( 2x2x1
2x4+8x3+2x2––––––––––––––
x3+3x2+3x1
+x3+4x2+x–––––––––––––
x2+4x1
+x2+4x+1––––––––––––
0
Now, the other zeros can be obtained from on solving 2x2x1=0
2x22x+x1=0
2x(x1)+1(x1)=0
(2x+1)(x1)=0
x=12,1
Hence, all the zeros are 12,1,2+3,23

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