For what value of m is x3–2mx2+16 is divisible by x + 2?
A
0
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B
1
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C
2
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D
3
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Solution
The correct option is B 1 Let p(x)=x3−2mx2+16
Since, p(x) is divisible by (x+2), the remainder = 0 ∴p(−2)=0 [ by remainder theorem] ⇒(−2)3–2m(−2)2+16=0 ⇒−8–8m+16=0 ⇒8=8m ∴m=1.
Hence, the value of m is 1.