If x is the least of three consecutive even integers, what is the product of these three integers in terms of x?

A. \(x^3 + 6x^2 + 8x\)

B. \(x^3 – 3x^2 + 2x\)

C. \(x^3 – 3x^2 - 2x\)

D. \(x^3 – x^2\)

E. \(x^3 – x\)
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The three consecutive even integers would be \(= x , (x + 2), (x + 4)\)

Product of these 3 integers would be;

\(x(x+2)(x+4)\)

\(x(x^2+ 2x + 4x + 8)\)

\(x(x^2 + 6x + 8)\)

\(x^3 + 6x^2 + 8x\) . Answer (A) ...

Let x be the least even integer of the consecutive integers.
So, the other even integers will be (x+2) and ( x+4)

The product of the integers = x(x+2)(x+4) = (x^2+2x) (x+4) = (x^3 + 6x^2 + 8x)



Answer A
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three consecutive even integers: x, x+2, x+4

product: x (x^2 + 6x + 8)

IMO A

Based on given information we have \(x, x+2, x+4\) and we need to find the product of these numbers.

\((x)(x+2)(x+4)\)

\((x)(x^2+4x+2x+8)\)

\((x)(x^2+6x+8)\)

\(x^3 + 6x^2 + 8x\)

Hence, Answer is A
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Speed tip
you don't have to multiply x(x+2)(x+3), all options have a negative sign except A

note the abundance of minus signs in B-E
if 2*4*6=48,
only A>8