Prove that ... is always an integer number

[skrapmejl112]

Help! I Don’t know what to do...

Prove that (x^3-x)(x^2-4)/10 is always an integer number if x is a natural number.

 

[Romsek]

what happens when you factor the numerator?

 

[Dr.Peterson]

Help! I Don’t know what to do...

Prove that (x^3-x)(x^2-4)/10 is always an integer number if x is a natural number.

You'd like to show that the numerator is always a multiple of 10. It may help to also factor the denominator. Then you'll have to think.

Any ideas now?

 

[pka]

Prove that (x^3-x)(x^2-4)/10 is always an integer number if x is a natural number.

Can you show that \(\forall n\in\mathbb{N}\) the number \((n^3-n)(n^2-4)\) an even multiple of five?

 

[skrapmejl112]

what happens when you factor the numerator?

(x^3-x)(x^2-4)/10 = x(x^4 -5x^2 +4) which can be written x^3(x^2-5)+4x. I have literally NO idea what to do.

 

[Romsek]

[MATH](x^3-x)(x^2-4) =\\ x(x^2-1)(x^2-4) = \\ x(x-1)(x+1)(x-2)(x+2) = \\ (x-2)(x-1)x(x+1)(x+2)[/MATH]
Somewhere in those 5 consecutive factors is going to be an even number and a multiple of 5. That gets you a factor of 10 in the numerator
that cancels the 10 in the denominator.

 

[Dr.Peterson]

(x^3-x)(x^2-4)/10 = x(x^4 -5x^2 +4) which can be written x^3(x^2-5)+4x. I have literally NO idea what to do.

That isn't factoring! Factoring means writing the entire expression as a product. It's already factored into two parts; you need to continue, by factoring each of those.

 

[HallsofIvy]

You'd like to show that the numerator is always a multiple of 10. It may help to also factor the denominator. Then you'll have to think.

Any ideas now?

THINK?? Surely you can just plug numbers into a formula. Isn't that how mathematics works?

 

[Steven G]

When you are given a product and asked to factor, at your level, you should never multiply out and then factor. Be happy that this was partially factored for you and then try to further factor each factor.

 

[Steven G]

Whatever x equals, x-2 will be 2 less than x, x-1 will be 1 less than x, x+1 will be 1 more than 1 and x+2 will be 2 more than x.
Suppose x is a natural number (1,2,3,4,....). Then you have 5 consecutive numbers.

Think about the 5 times table. Two consecutive multiples of 5 always differ by 5. You can not write down five consecutive natural numbers and not have a multiple of 5. Consider 7, 8, 9, 10, 11, 12 or 2, 3, 4, 5, 6 or 19, 20, 21, 22, 23. They all have a multiple of 5 in their list.

Same situation with a multiple of 2. Given any two consecutive natural numbers one of those numbers will be a multiple of 2. Surely given 5 consecutive natural numbers one of them (actually 2 or 3 of them) will be a multiple of 2.