SOLUTION: Please help me solve this problem: DIVIDE x^3-6 by x-2 Thank you!

Question 806049: Please help me solve this problem:
DIVIDE x^3-6 by x-2
Thank you!
Found 3 solutions by josgarithmetic, math_tutor2020, ikleyn:
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
Include the power of 2 and of 1 in the dividend.

___________|______x^2_______2x___________4___
___________|_________________________________
x-2________|______x^3______0*x^2________0*x_________-6
___________|______x^3______-2x^2
___________|______________________
__________________0_________2x^2________0*x
____________________________2x^2________-4x
____________________________________________
_____________________________0___________4x________-6
_________________________________________4x________-8
________________________________________________________
____________________________________________________+2

Start with asking . The numerator is from the dividend and the denominator is from the divisor. Put the partial quotient in the top line above the x^3 of the dividend. The rest of this step is just like regular long division.

Continue the process for the next term of the dividend.

The result gives remainder of +2.
The resulting division is
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Polynomial long division is a valid method as shown by tutor @josgarithmetic.

Synthetic division is the more efficient route (in my opinion), which is what I'll use below.

The denominator x-2 has the test root x = 2
We write 2 in the upper left corner.

Think of x^3-6 as 1x^3+0x^2+0x+(-6) so we can see the coefficients much more clearly. Those coefficients are: 1, 0, 0, -6 which are laid out across the top row (ignoring the previous "2" mentioned earlier)

2 1 0 0 -6

We'll drop the 1 down like so

2 1 0 0 -6
1

Then multiply the test root (2) with that value we dropped. The result 2*1 = 2 is placed just under the 0 in the next column

2 1 0 0 -6
2
1

Next we add the stuff along the 2nd column: 0+2 = 2

2 1 0 0 -6
2
1 2

Multiply the test root (2) with this new value in the bottom row: 2*2 = 4
Write 4 just under the other 0

2 1 0 0 -6
2 4
1 2

Keep this process going until the table is filled out.
This is what you should get:

2 1 0 0 -6
2 4 8
1 2 4 2

If you are confused with any step of the synthetic division process, then please let me know. I recommend searching out videos of similar examples. Just so you get a feel of what's going on. It's a bit tricky to describe the movement patterns with text only.

The last number in the bottom row is the remainder.
The other numbers in the bottom row are the coefficients of the quotient.
The values 1, 2, 4 are the coefficients to 1x^2+2x+4 aka x^2+2x+4

This is how we end up with
quotient = x^2+2x+4
remainder = 2

The remainder 2 is placed over the denominator (x-2)
Which means we can write

This is an identity which is a true equation for almost all real numbers x as long as

Side note: Because the remainder is not zero, it means (x-2) is not a factor of x^3-6.

Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
Please help me solve this problem:
DIVIDE x^3-6 by x-2
~~~~~~~~~~~~~~~~~~~~~~~~

        I will show you another, nontraditional way to solve the problem,
        which is (I am 100% sure) absolutely unexpected to you   (and to many others).

Start from this innocent transformation x^3 - 6 = (x^3 - 8) + 2. For the parentheses in the right side, use well known decomposition a^3 - b^3 = (a-b)*(a^2 + ab + b^2). Take here a= x, b= 2. You will get then x^3 - 8 = (x-2)*(x^2 + 2x + 4). Therefore, x^3 - 6 = (x^3 - 8) + 2 = (x-2)*(x^2 + 2x + 4) + 2. It means that = x^2 + 2x + 4 with the remainder of 2. ANSWER. = x^2 + 2x + 4 with the remainder of 2.

Solved.

-------------------

It is good solution to show it at the local school Math circle.

The students will be positively shocked (= overwhelmed; = stunned) :   a deafening effect is guaranteed.

Notice that I did not make any division for it - neither polynomial nor synthetic.

I extracted the solution from the air, using the method called " focus-pocus ".

After such an " exercise ", the students will know and understand Math " in two times better " than before it.

And they will have a stimulus to learn it better.

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