SOLUTION: I still need help:
please help me solve the following synthetic division i am struggling with
Use synthetic division and the remainder theroem to find the indicated function valu
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-> SOLUTION: I still need help:
please help me solve the following synthetic division i am struggling with
Use synthetic division and the remainder theroem to find the indicated function valu
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Question 134154: I still need help:
please help me solve the following synthetic division i am struggling with
Use synthetic division and the remainder theroem to find the indicated function value.
f(x)=6x to the 4th power +11x³+6x²-6x+53; f(4) Found 2 solutions by checkley71, jim_thompson5910:Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! F(X)=6X^4+11X^3+6X^2-6X+53
F(4)=6*4^4+11*4^3+6^4^2-6*4+53
F(4)=6*256+11*64+6*16-24+53
F(4)=1536+704+96-24+53
F(X)=2365 ANSWER.
First 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.
4
|
6
11
6
-6
53
|
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)
4
|
6
11
6
-6
53
|
6
Multiply 4 by 6 and place the product (which is 24) right underneath the second coefficient (which is 11)
4
|
6
11
6
-6
53
|
24
6
Add 24 and 11 to get 35. Place the sum right underneath 24.
4
|
6
11
6
-6
53
|
24
6
35
Multiply 4 by 35 and place the product (which is 140) right underneath the third coefficient (which is 6)
4
|
6
11
6
-6
53
|
24
140
6
35
Add 140 and 6 to get 146. Place the sum right underneath 140.
4
|
6
11
6
-6
53
|
24
140
6
35
146
Multiply 4 by 146 and place the product (which is 584) right underneath the fourth coefficient (which is -6)
4
|
6
11
6
-6
53
|
24
140
584
6
35
146
Add 584 and -6 to get 578. Place the sum right underneath 584.
4
|
6
11
6
-6
53
|
24
140
584
6
35
146
578
Multiply 4 by 578 and place the product (which is 2312) right underneath the fifth coefficient (which is 53)
4
|
6
11
6
-6
53
|
24
140
584
2312
6
35
146
578
Add 2312 and 53 to get 2365. Place the sum right underneath 2312.
4
|
6
11
6
-6
53
|
24
140
584
2312
6
35
146
578
2365
Since the last column adds to 2365, we have a remainder of 2365.
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