यदि दो वास्तविक अचरों `a` और `b` के लिए व्यंजक `ax^(3)+3x^(2)-8x+b,(x+2)` और `(x-2)` से पूर्णतः विभाजित है तो

Options

A. `a=2, b=12`
B. `a=12, b=2`
C. `a=2,b=-12`
D. `a=-2, b=12`

Answer

C

Explanation

`ax^(3)+3x^(2)-8x+b` is divisible
by `(x+2)` and `(x-2)`
`:. (x=2)` and `(x-2)` are factors
`:.x+2=0impliesx=-2`
`x-2-0impliesx=2`
Put `x=-2`
`:.a(-2)^(3)+3(-2)^(2)-8(-2)+b=0`
`=-8a+12+16+b=0`
`-8a+b+28=0`
`-8a+b=-28`.........i
and
put `x=2`
`implies a(2)^(3)+3(2)^(2)-8xx2+b=0`
`implies 8a+12-16+b=0`
`8a+b-4=0`

`8a+b=4`..........ii
From equation i & ii
`:. -8a+b=-28`
`ul(8a+b=4)`
`2b=-24`
`b=-12`
`a=2`