SOLUTION: Use synthetic division to divide the polynomial 2x^3 – 7x + 10 by x + 3, and write the quotient polynomial and the remainder.

Algebra ->  Rational-functions -> SOLUTION: Use synthetic division to divide the polynomial 2x^3 – 7x + 10 by x + 3, and write the quotient polynomial and the remainder.       Log On


   



Question 148297: Use synthetic division to divide the polynomial 2x^3 – 7x + 10 by x + 3, and write the quotient polynomial and the remainder.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Let's simplify this expression using synthetic division


Start with the given expression %282x%5E3+-+7x+%2B+10%29%2F%28x%2B3%29

First lets find our test zero:

x%2B3=0 Set the denominator x%2B3 equal to zero

x=-3 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 2x%5E3 to -7x%5E1 there is a zero coefficient for x%5E2. This is simply because 2x%5E3+-+7x+%2B+10 really looks like 2x%5E3%2B0x%5E2%2B-7x%5E1%2B10x%5E0
-3|20-710
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
-3|20-710
|
2

Multiply -3 by 2 and place the product (which is -6) right underneath the second coefficient (which is 0)
-3|20-710
|-6
2

Add -6 and 0 to get -6. Place the sum right underneath -6.
-3|20-710
|-6
2-6

Multiply -3 by -6 and place the product (which is 18) right underneath the third coefficient (which is -7)
-3|20-710
|-618
2-6

Add 18 and -7 to get 11. Place the sum right underneath 18.
-3|20-710
|-618
2-611

Multiply -3 by 11 and place the product (which is -33) right underneath the fourth coefficient (which is 10)
-3|20-710
|-618-33
2-611

Add -33 and 10 to get -23. Place the sum right underneath -33.
-3|20-710
|-618-33
2-611-23

Since the last column adds to -23, we have a remainder of -23. This means x%2B3 is not a factor of 2x%5E3+-+7x+%2B+10
Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,-6,11) form the quotient

2x%5E2+-+6x+%2B+11

and the last coefficient -23, is the remainder, which is placed over x%2B3 like this

-23%2F%28x%2B3%29



Putting this altogether, we get:

2x%5E2+-+6x+%2B+11%2B-23%2F%28x%2B3%29

So %282x%5E3+-+7x+%2B+10%29%2F%28x%2B3%29=2x%5E2+-+6x+%2B+11%2B-23%2F%28x%2B3%29

which looks like this in remainder form:
%282x%5E3+-+7x+%2B+10%29%2F%28x%2B3%29=2x%5E2+-+6x+%2B+11 remainder -23