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. 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
All numbers of the form 6+12x are, for all integer x.
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)
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.
that would be limited to 3 and -3 for values of x
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
If we write the problem as 6x-5=x2, then we can write in the form of ax2+bx+c: -x2+6x-5. Then we can use the quadratic equation, x = (-b ± √(b2 - 4ac))/2a, and put in our own values to get the equation x = 3 ± 2. Therefore, x1= 1 and x2=5.
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
(x + 3)(x - 3)
x2 + 2x - 3 = (x + 3)(x - 1)
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.
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
x2-9 = (x-3)(x+3) when factored
9 - x2 will factor into (3 - x)(3 + x)
x2 + 2x - 3 factors to (x - 1)(x + 3)
if x is divisible by 4 then x/4 = y, where y is an integer. so it follows that y = x/(2*2) and therefore 2y = x /2. since y is an integer, so must 2y. since x/2 yields an integer (2y), x must be even.
2 x 2 x 2 x2 x2 x2 x 3 x 11
X is divisible by Y if there is an integer Z such that X = Y*Z that is, Y will go into X without remainder.