Find the midpoint of (1,-7),(1, 11)
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Hello!
**Process pictured below**
When dividing, find how many times the first time in the divisor (2x + 3) fits into the first time of the dividend (2x³ + 5x² - 3x - 5). In this step, it fits x² times.
Multiply x² by the terms in the divisor and subtract from the dividend. Bring down the next term in the dividend to continue the process.
Repeat this step until you reach the last number. In this case, there was a remainder of 4. In order to write the remainder, you must express it over the divisor which makes it 4 / 2x + 3.
When 
, you have
Now assume this is true for
, i.e.
and under this hypothesis show that it's also true for
. You have
In other words, there exists
such that
Rewriting, you have
and this is equivalent to
modulo 9, as desired.
Now assume this is true for
and under this hypothesis show that it's also true for
In other words, there exists
Rewriting, you have
and this is equivalent to