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Bunuel
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Bunuel
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. \(x^3 – x\)

Lowest is x hence x, x+2, x+4

On multiplication we will get \(x^3 + 6x^2 + 8x\)

Hence A
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Bunuel
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. \(x^3 – x\)

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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Bunuel
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. \(x^3 – x\)

Check using the least even integers - 2 , 4 & 6 , product is 48

(A) 8 + 6*4 + 8*2 = 48

We don't need to go any further, just plugin in three numbers and checking option (A) confirms the answer.....
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three consecutive even integers: x, x+2, x+4

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

IMO A
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Bunuel
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. \(x^3 – x\)

The smallest even integer is x, the next consecutive even integer is (x + 2), and the third consecutive even integer is (x + 4). We can express the product of our 3 integers as:

x(x + 2)(x + 4) = x(x^2 + 6x + 8) = x^3 + 6x^2 + 8x

Answer: A
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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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Three consecutive EVEN Integers are 2n, 2n+2 & 2n+4
As per the given info the smallest even integer is x .
Therefor 2n=x, 2n+2=x+2 & 2n+4=x+4

Product of x, x+2 & x+4
= x*(x+2)*(x+4)
= x*(x^2+6x+8)
= x^3+6x^2+8x

Ans A

Rgds
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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
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Bunuel
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. \(x^3 – x\)

Say 3 consecutive even integers are\(x, x+2, x+4\)

Multiply them,

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

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

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

(A)
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Bunuel
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. \(x^3 – x\)

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