You can put this solution on YOUR website!
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 -1
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.
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is -4)
Add -1 and -4 to get -5. Place the sum right underneath -1.
Multiply -1 by -5 and place the product (which is 5) right underneath the third coefficient (which is 1)
Add 5 and 1 to get 6. Place the sum right underneath 5.
Multiply -1 by 6 and place the product (which is -6) right underneath the fourth coefficient (which is 6)
Add -6 and 6 to get 0. Place the sum right underneath -6.
Now lets look at the bottom row of coefficients:
The first 3 coefficients (1,-5,6) form the quotient
Now lets break
Looking at the expression
, we can see that the first coefficient is
, the second coefficient is
, and the last term is
Now multiply the first coefficient
by the last term
Now the question is: what two whole numbers multiply to
(the previous product) and
add to the second coefficient
To find these two numbers, we need to list all
of the factors of
(the previous product).
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to
Now let's add up each pair of factors to see if one pair adds to the middle coefficient
|First Number||Second Number||Sum|
From the table, we can see that the two numbers
(the middle coefficient).
So the two numbers
both multiply to and
Now replace the middle term
. So this shows us that
Replace the second term
Group the terms into two pairs.
Factor out the GCF
from the first group.
from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
Combine like terms. Or factor out the common term
completely factors to