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Q: If x is an integer divisible by 3 then is x2 divisible by 3?

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x2+x-6 = (x-2)(x+3) when factored

x2 + x - 6 = 0 x2 + 3x - 2x - 6 = 0 x(x+3) -2(x+3) = 0 so (x+3)(x-2) = 0 so x+3 = 0 or x-2 = 0 so that x = -3 or x = 2

x2 + 8x = -15 x2 + 8x + 15 = 0 since 15 = 5 x 3 and 5 + 3 = 8, then (x + 5)(x + 3) = 0 x + 5 = 0 or x + 3 = 0 x = -5 or x = -3

x2 - 6x = 16 ∴ x2 - 6x + 9 = 25 ∴ (x - 3)2 = 25 ∴ x - 3 = 25 ∴ x = 28

x2 + 5x + 6 = (x + 2)(x + 3)

Related questions

Suppose x = sqrt(3*a) where a is an integer that is not divisible by 3. then x2 = 3*a which is divisible by 3. but x is not even rational and so is not an integer and is certainly not divisible by 3.

Yes

5

Yes. If x is not divisible by 3 then x leaves a remainder of 1 or 2 when it is divided by 3. That is, x is of the form 3y+z where z = 1 or 2. Then x2 = (3y+z)2 = 9y2 + 6yz + z2 = 3(3y2 + 2yz) + z2 The first part of this expression is clearly a multiple of 3, but z2 is not. Whether z = 1 or 2, z2 leaves a remainder of 1 when divided by 3.

Call the unknown integer x. Then, from the problem statement, x - 3/x = 26/3, or:x2 - 3 = 26x/3; or x2 - (26/3)x - 3 = 0x = 9

x2=3x+4 x2-3x-4=0 (x-4)(x+1)=0 x=4 or x=-1 Since you specified a positive integer, the number in question is 4.

All numbers of the form 6+12x are, for all integer x.

yes * * * * * Absolutely not! If it is divisible by 9 it must be divisible by 3. Here, if you want it, is a proof: x is divisible by 9 implies that x = 9*y for some integer y. Now 9 = 3*3, so writing 3*3 instead of 9 gives x = (3*3)*y so, by the associative property of multiplication, x = 3*(3*y) and then, by the closure of multiplication of integers, 3*y is also an integer. Say, z, for example. That is to say, x = 3*z which is equivalent to saying that x is divisible by 3.

For an integer x to be evenly divisible by another integer y, x/y must also be an integer. Here are some examples: 4/2=2. 15/3=5. 18/6=3. From these examples, you can see that 4 is divisible by 2, 15 is divisible by 3, and 18 is divisible by 6. When two numbers are not evenly divisible by each other, you end up with a remainder, or fractional/decimal answer. Here is an example: 25/2 =12.5

x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)

that would be limited to 3 and -3 for values of x

(x + 3)(x - 3)

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