SOLUTION: Use synthetic division to perform this division 2x^3+4x^2-3x+10/x-3 here are the choices: A. 2x^2+10x+27+-91/x-3 b. 2x^2-8x+29+23/x-3 c. 2x^2+10x+27+91/x-3 d. 2x^2-8x+29+-

Algebra ->  Equations -> SOLUTION: Use synthetic division to perform this division 2x^3+4x^2-3x+10/x-3 here are the choices: A. 2x^2+10x+27+-91/x-3 b. 2x^2-8x+29+23/x-3 c. 2x^2+10x+27+91/x-3 d. 2x^2-8x+29+-      Log On


   



Question 189561: Use synthetic division to perform this division 2x^3+4x^2-3x+10/x-3
here are the choices:
A. 2x^2+10x+27+-91/x-3
b. 2x^2-8x+29+23/x-3
c. 2x^2+10x+27+91/x-3
d. 2x^2-8x+29+-32/x-3
any help would be appreaciated thanks!

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+%2B+4x%5E2+-+3x+%2B+10%29%2F%28x-3%29

First lets find our test zero:

x-3=0 Set the denominator x-3 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.
3|24-310
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
3|24-310
|
2

Multiply 3 by 2 and place the product (which is 6) right underneath the second coefficient (which is 4)
3|24-310
|6
2

Add 6 and 4 to get 10. Place the sum right underneath 6.
3|24-310
|6
210

Multiply 3 by 10 and place the product (which is 30) right underneath the third coefficient (which is -3)
3|24-310
|630
210

Add 30 and -3 to get 27. Place the sum right underneath 30.
3|24-310
|630
21027

Multiply 3 by 27 and place the product (which is 81) right underneath the fourth coefficient (which is 10)
3|24-310
|63081
21027

Add 81 and 10 to get 91. Place the sum right underneath 81.
3|24-310
|63081
2102791

Since the last column adds to 91, we have a remainder of 91. This means x-3 is not a factor of 2x%5E3+%2B+4x%5E2+-+3x+%2B+10
Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,10,27) form the quotient

2x%5E2+%2B+10x+%2B+27

and the last coefficient 91, is the remainder, which is placed over x-3 like this

91%2F%28x-3%29



Putting this altogether, we get:


2x%5E2+%2B+10x+%2B+27%2B91%2F%28x-3%29


So %282x%5E3+%2B+4x%5E2+-+3x+%2B+10%29%2F%28x-3%29=2x%5E2+%2B+10x+%2B+27%2B91%2F%28x-3%29


which looks like this in remainder form:


%282x%5E3+%2B+4x%5E2+-+3x+%2B+10%29%2F%28x-3%29=2x%5E2+%2B+10x+%2B+27 remainder 91


So the answer is C