SOLUTION: Divide: 5x^3-5x+1 by x-3
I tried factoring out the 5 and ended up with 5(x^2-x-3)+x as the answer
Am I anywhere close?????
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Question 117090: Divide: 5x^3-5x+1 by x-3
I tried factoring out the 5 and ended up with 5(x^2-x-3)+x as the answer
Am I anywhere close?????
Answer by jim_thompson5910(35256) (Show Source): 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 3
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.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 5)
Multiply 3 by 5 and place the product (which is 15) right underneath the second coefficient (which is 0)
Add 15 and 0 to get 15. Place the sum right underneath 15.
Multiply 3 by 15 and place the product (which is 45) right underneath the third coefficient (which is -5)
Add 45 and -5 to get 40. Place the sum right underneath 45.
Multiply 3 by 40 and place the product (which is 120) right underneath the fourth coefficient (which is 1)
Add 120 and 1 to get 121. Place the sum right underneath 120.
Since the last column adds to 121, we have a remainder of 121. This means is not a factor of
Now lets look at the bottom row of coefficients:
The first 3 coefficients (5,15,40) form the quotient
and the last coefficient 121, is the remainder, which is placed over like this
Putting this altogether, we get:
So
which looks like this in remainder form:
remainder 121
You can use this online polynomial division calculator to check your work
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