Questions on Algebra: Operations with Signed Numbers answered by real tutors!

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Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -2


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.

-2	|	2	-1	3	-1
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)

-2	|	2	-1	3	-1
	|
		2

Multiply -2 by 2 and place the product (which is -4) right underneath the second coefficient (which is -1)

-2	|	2	-1	3	-1
	|		-4
		2

Add -4 and -1 to get -5. Place the sum right underneath -4.

-2	|	2	-1	3	-1
	|		-4
		2	-5

Multiply -2 by -5 and place the product (which is 10) right underneath the third coefficient (which is 3)

-2	|	2	-1	3	-1
	|		-4	10
		2	-5

Add 10 and 3 to get 13. Place the sum right underneath 10.

-2	|	2	-1	3	-1
	|		-4	10
		2	-5	13

Multiply -2 by 13 and place the product (which is -26) right underneath the fourth coefficient (which is -1)

-2	|	2	-1	3	-1
	|		-4	10	-26
		2	-5	13

Add -26 and -1 to get -27. Place the sum right underneath -26.

-2	|	2	-1	3	-1
	|		-4	10	-26
		2	-5	13	-27

Since the last column adds to -27, we have a remainder of -27. This means is not a factor of
Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,-5,13) form the quotient



and the last coefficient -27, is the remainder, which is placed over like this





Putting this altogether, we get:



So

which looks like this in remainder form:
remainder -27