if x is a +ve integer, is x³ - 3x² + 2x divisible by 4?If x is a positive integer , is x^3 - 3x^2+2x divisible by 4?
(1) x = 4y + 4, where y is an integer
(2) x = 2z + 2, where z is an integer
in the above (1) is exactly a multiple for 4, so sufficient BUT in (2) if z = 0 , then x = 2, which when substituted in quest will not be divisible by 4. right?
but the answer is D, both alone sufieicient
can anyone guide. do i have a wrong approach towards DS??
(1) x=4y+4, where y is an integer --> since x itself is divisible by 4 then x^3-3x^2+2x is divisible by 4. Sufficient.
(2) x=2z+2, where z is an integer --> x^3-3x^2+2x=x(x^2-3x+2)=(2z+2)(4z^2+8z+4-6z-6+2)=4
(z+1)(2z^2+z) --> hence this expression is divisible by 4. Sufficient.
As for your question: if x=2 then x^3-3x^2+2x=0. Now, zero is divisible by EVERY integer except zero itself, as 0/integer=integer.
For more on this topic check Number Theory chapter of Math Book: math-number-theory-88376.html
Hope it's clear.
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