can someone explain this to me algebraically???

[wet canyon]

And maybe visually on a number line?

[edgy ermine]

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[wet canyon]

I get that they plugged in the numbers. But I dont understand the results.

[dire vessel]

they evaluated the function and checked when they got 0

and obviously reduced mod 6 since we're in Z/6Z

$1^2+2(1)+4=7\equiv 1\mod 6$

[south dove]

Omegabet_

[wet canyon]

So for 2 its 12 mod 6? And theres no remainder thats why its 0?

[dire vessel]

yes, 6|12

you just do the computation in Z, then project back down to Z/6Z

since that's how the operations are defined

$[a]+[b]:=[a+b]$ for any equivalence classes of mod 6 (likewise for the product)

[south dove]

Omegabet_

[wet canyon]

Thank you so much!

[gaunt moat]

@wet canyon has given 1 rep to @dire vessel

[silver plank]

for modular arithmetic, a circle might be more suitable than a number line

[brave citrus]

I don't really agree. For these problems, it's much easier to visualize the polynomial over Z/6Z by considering its image over Z and marking every point in 6Z with a dot or something

But this isn't really important.

Sorta like this

[dire vessel]

+close