If the polynomial f(x)=ax3+bx−c is divisible by the polynomial g(x)=x2+bx+c, then ab=
A
−1c
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B
−1
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C
1
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D
1c
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Solution
The correct option is C1 f(x)=ax3+bx−c =ax3+ox2+bx−c g(x)=x2+bx+c
Dividing f(x) by g(x) ∵f(x) is completely divisible by g(x) ∴r(x)=0 (b−ac+ab2)x+abc−c=0 b−ac+ab2=0 and abc−c=0 ⇒abc=c⇒ab=1