Help
for synthetic substitution you would set up a synthetic division problem with -3 as the divisor (the number in the top left corner)
okay so just solve for -3
do you remember how to set up/solve a synthetic division problem?
umm no
isn't it 11 -6 +2
|dw:1515692727482:dw|
|dw:1515692771422:dw|
steps for the algorithm: step 1 is to drop the first number in the first column down
|dw:1515692851363:dw|
step 2: multiply in this direction
|dw:1515692882340:dw|
bring the result into the next column
|dw:1515692904257:dw|
take the sum of the second column (-6 + (-33) = -39
|dw:1515692930778:dw|
now that I am looking at it I forgot the placeholder ugh ;_;
since we are missing an x term we substitute a 0 as a placeholder
|dw:1515693001287:dw|
anyway, we will multiply in this direction again
|dw:1515693020453:dw|
and then bring the result into the next column
|dw:1515693064789:dw|
taking the sum , 0 + 117 = 117
|dw:1515693091429:dw|
then multiplying by the divisor again...
|dw:1515693109463:dw|
bring the result into the last column
|dw:1515693143330:dw|
then take the sum of the last column 2 + (-351) = -349 = your answer
I would recommend studying these steps and writing them down b/c this algorithm comes up a lot
when you're ready we can move onto 10
I actually need to take a break b/c I'm starting to get worn out ;;
Lol go ahead
when you are back please let me know
anyway for direct substitution #10 you just need to plug in x = 5 to find m(5)
m(5) = 5 + 5^2 - 1 = 5 + 25 - 1 = 30 - 1 = 29 = your answer got to get to class, have a nice afternoon ^^
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