- WELCOME TO A LESSON ON SYNTHETIC DIVISION
USED FOR POLYNOMIAL DIVISION.
THE GOAL OF THIS VIDEO IS TO PERFORM SYNTHETIC DIVISION
TO DIVIDE BY A BINOMIAL IN THE FORM OF X - K.
SO NOTICED THE DIVISOR WILL BE IN THE FORM
WHERE THE DEGREE IS 1 AND THE LEADING COEFFICIENT IS ALSO 1.
FOR AN EXPLANATION OF SYNTHETIC DIVISION
WE'LL DISCUSS SYNTHETIC DIVISION
OF A THIRD DEGREE POLYNOMIAL,
AND SO WE'LL USE SYNTHETIC DIVISION
TO DIVIDE A POLYNOMIAL IN THIS FORM BY X - K.
SO YOU CAN SEE THE FIRST STEP
IS TO LIST THE COEFFICIENTS HERE,
AND LIST THE VALUE OF K HERE.
IT'S IMPORTANT TO NOTE THAT K IS FOUND
BY HAVING THE BINOMIAL IN THE FORM OF X - K.
SO IF THIS WAS X - 3, K WOULD BE 3.
TO PERFORM SYNTHETIC DIVISION
WE START BY BRINGING THE "A" VALUE DOWN.
THEN WE MULTIPLY BY K, AND PUT IT IN THE NEXT COLUMN.
MULTIPLY BY K ON THE DIAGONAL, ADD DOWN, AND CONTINUE.
SO THE PROCESS HERE,
THE VERTICAL PATTERN IN RED, IS TO ADD,
AND THE DIAGONAL PATTERN IN BLUE IS TO MULTIPLY BY K.
AND THE BOTTOM ROW REPRESENTS THE COEFFICIENTS
OF THE QUOTIENT,
WHICH IS 1 DEGREE LESS THAN THE DIVIDEND.
SO SINCE THIS IS DEGREE 3 AND THIS IS DEGREE 1,
OUR QUOTIENT WOULD BE DEGREE 2.
LET'S GO AHEAD AND GIVE IT A TRY.
HERE I HAVE THE NOTES FROM THE PREVIOUS PAGE.
LET'S SEE IF WE CAN SET THIS UP.
THE FIRST STEP IS TO IDENTIFY THE COEFFICIENTS
OF THE DIVIDEND.
SO WE HAVE A 1, -3, 4, AND -7.
NEXT WE NEED TO IDENTIFY THE VALUE OF K
IN THE FORM OF X - K.
SO THE VALUE OF K IN THIS PROBLEM,
BECAUSE WE'RE DIVIDING BY X - 3, WOULD BE 3.
I ALWAYS FIND IT EASY TO JUST TAKE THE OPPOSITE SIGN
IN WHAT WE SEE HERE IN THE DIVISOR.
NOW, THE PROCESS STARTS
BY BRINGING DOWN THE LEADING COEFFICIENT OF 1,
AND THEN ON THE DIAGONAL WE MULTIPLY BY K.
SO 3 x 1 = 3, AND THE VERTICAL PATTERN IS TO ADD.
SO -3 + 3 = 0.
NOW, THIS IS QUITE DIFFERENT THAN LONG DIVISION.
REMEMBER WHEN WE DID LONG DIVISION
WE SUBTRACTED EACH TIME,
BUT FOR SYNTHETIC DIVISION WE ADD.
THE DIAGONAL PATTERN, AS WE MULTIPLY IT BY K, 3 x 0 = 0.
THE VERTICAL PATTERN IS TO ADD.
DIAGONAL PATTERN IS TO MULTIPLY BY K, SO 3 x 4 = 12,
AND THE RESULT HERE WOULD BE 5.
THESE ARE THE COEFFICIENTS AND THE REMAINDER OF OUR QUOTIENT.
SO SINCE WE STARTED WITH A DEGREE 3 POLYNOMIAL
AND DIVIDED BY A DEGREE 1,
OUR QUOTIENT WOULD BE 1X SQUARED + 0X + 4.
THIS IS OUR REMAINDER, SO WE HAVE + 5 OVER THE DIVISOR.
REMEMBER THE DIVISOR IS X - 3.
NOW, TO CLEAN THIS UP,
OF COURSE, WE WOULD NOT WANT TO LEAVE THE 0 TERM
IN OUR FINAL ANSWER,
OR THE LEADING COEFFICIENT OF 1,
SO WE WOULD WRITE THIS AS X SQUARED + 4 + 5/X - 3.
SO THIS IS A VERY NICE WAY TO DIVIDE BY A BINOMIAL,
AS LONG AS IT FITS THE FORM
REQUIRED FOR SYNTHETIC DIVISION.
YOU CAN SEE IT'S MUCH FASTER THAN FORMING LONG DIVISION.
LET'S TRY ANOTHER.
FIRST WE NEED OUR COEFFICIENTS IN DESCENDING ORDER.
SO WE'RE GOING TO HAVE A 2, 6.
NOW, THERE'S NO X TERM,
SO WE HAVE TO HAVE A 0 COEFFICIENT FOR X,
AND THEN 29.
SO WE HAVE 2, 6, 0, AND 29.
IF WE LEAVE OFF THE 0,
THE PROCESS WILL NOT WORK CORRECTLY.
NEXT WE NEED THE VALUE OF K.
IT SAYS X + 4, SO I KNOW THE VALUE OF K WILL BE -4.
SO WE START OFF BY BRINGING THE 2 DOWN,
AND WE MULTIPLY BY K ON THE DIAGONAL.
THIS WOULD BE -8.
ADD. IT WOULD BE -2.
AGAIN MULTIPLY BY K, AND THIS WOULD BE 8 NOW.
ADD, MULTIPLY BY K.
-4 x 8 = -32.
THE SUM HERE WOULD BE -3.
THIS IS ALL WE NEED TO FIND THE QUOTIENT.
SO IF WE TAKE A DEGREE 3 AND DIVIDE BY A DEGREE 1,
THE QUOTIENT WILL BE A DEGREE 2.
SO OUR QUOTIENT WILL BE
2X SQUARED - 2X + 8 - THE REMAINDER/THE DIVISOR,
WHICH IS X + 4.
HERE'S OUR FINAL QUOTIENT.
LET'S GO AHEAD AND TRY ANOTHER ONE.
FIRST LIST THE COEFFICIENTS.
NOTICE WE WILL NOT HAVE ANY 0 TERMS HERE.
SINCE THE DIVISOR IS X + 5, K WILL BE - 5 OR -5.
START BY BRINGING THE LEADING COEFFICIENT DOWN.
MULTIPLYING BY K, THAT'D BE -10.
ADDING, MULTIPLYING BY K, 20.
ADDING, MULTIPLYING BY K,
AND HERE WE SEE WE HAVE A REMAINDER OF 0.
SO OUR QUOTIENT WILL BE 2X SQUARED - 4X + 3.
AGAIN WE HAVE DEGREE 3 DIVIDED BY DEGREE 1,
WHICH MEANS OUR QUOTIENT WILL BE DEGREE 2.
SINCE THIS DOES DIVIDE EVENLY INTO THIS POLYNOMIAL,
WE COULD SAY THAT X + 5 IS A FACTOR OF THIS POLYNOMIAL.
LET'S TAKE A LOOK AT A COUPLE MORE.
NOW WE'RE TAKING A DEGREE 5 AND DIVIDING BY A DEGREE 1.
THIS QUOTIENT WILL BE A DEGREE 4 POLYNOMIAL.
LET'S SET IT UP.
FIRST WE NEED OUR COEFFICIENTS.
LEADING COEFFICIENT IS A 1Y TO THE 5th.
WE'D HAVE 0Y TO THE 4th, 0Y TO THE 3rd, 0Y TO THE 2nd,
0Y, AND -32.
SO LET'S CHECK THAT AGAIN.
Y TO THE 5th IS GIVEN, 0Y TO THE 4th, 0Y TO THE 3rd,
0Y SQUARED, 0Y, AND OUR CONSTANT.
OUR VALUE OF K, BECAUSE WE SEE Y - 2, K WILL BE 2.
START BY BRINGING THE LEADING COEFFICIENT DOWN,
MULTIPLY BY K, ADDING, MULTIPLY BY K, ADD,
MULTIPLY BY K, ADD, AND MULTIPLY BY K, AND ADD,
AND LASTLY, MULTIPLY BY K AND ADD.
SO IT LOOKS LIKE WE HAVE ANOTHER REMAINDER OF 0,
WHICH MEANS Y - 2 IS A FACTOR OF THIS DEGREE 5 POLYNOMIAL.
SO WE HAVE A DEGREE 5 DIVIDED BY A DEGREE 1,
SO THIS WOULD BE A DEGREE 4.
SO 1Y TO THE 4th,
AND THEN DECREASING BY 1 DEGREE EACH TIME,
+ 2Y TO THE 3rd + 4Y SQUARED + 8Y + 16,
AND AGAIN, A REMAINDER OF 0.
SO HERE'S OUR QUOTIENT.
OKAY. LET'S TRY ONE MORE.
NOW, IF YOU TAKE A LOOK AT THIS LAST PROBLEM,
NOTICE HOW IT'S NOT IN THE FORM THAT WE NEED IT TO BE IN.
REMEMBER THAT THE DIVISOR HAS TO BE IN THE FORM OF X - K.
SO NORMALLY WE WOULD JUST PERFORM LONG DIVISION
ON THIS PROBLEM,
BUT THERE IS A WAY TO KIND OF WORK THROUGH THIS.
IF WE DIVIDE EVERYTHING BY 2,
THIS WOULD NOT AFFECT THE QUOTIENT.
SO WE CAN DIVIDE THIS BY 2, AND THIS BY 2,
AND EVERY OTHER TERM BY 2.
AND IF WE DO THAT,
WE CAN STILL WORK THE PROBLEM USING SYNTHETIC DIVISION.
THE NEXT THING YOU'LL NOTICE ON THIS PROBLEM
IS THAT THESE TERMS ARE NOT IN DESCENDING ORDER.
SO LET'S REWRITE THIS IN THE CORRECT ORDER
AT THE SAME TIME.
THIS WOULD BE THE SAME
AS 8X CUBED - 6X SQUARED + 7X - 1 DIVIDED BY X + 1/2.
OKAY, SO LET'S GO AHEAD AND PERFORM SYNTHETIC DIVISION
IN THIS FORM
AND SEE WHAT WE GET.
SO OUR COEFFICIENTS, 8, -6, 7, -1,
AND THE VALUE OF K IT SAYS X + 1/2, SO K WOULD BE -1/2.
OKAY. NOW THE PROCESS WILL BE THE SAME.
WE'LL BRING DOWN THE 8, MULTIPLY IT BY -1/2.
THAT WOULD BE -4.
ADD, MULTIPLY IT BY K.
THAT'D BE 5.
ADD, MULTIPLY IT BY -1/2.
THAT WOULD EQUAL -6, THE REMAINDER OF -7.
NOW, LET'S INTERPRET THESE RESULTS.
DEGREE 3 DIVIDED BY DEGREE 1.
THIS IS THE COEFFICIENT OF THE DEGREE 2 TERM.
SO WE'D HAVE 8X SQUARED - 10X + 12 - 7/OUR DIVISOR,
WHICH WOULD BE X + 1/2.
NOW THIS IS THE CORRECT QUOTIENT,
BUT IT'S IN KIND OF AN ODD FORM,
BECAUSE THE ORIGINAL PROBLEM HAD A DIVISOR OF 2X + 1.
SO WHAT WE COULD DO
IS MULTIPLY BOTH THE TOP AND THE BOTTOM BY 2/2,
WHICH WOULD RESULT IN 14 IN THE NUMERATOR
AND 2X + 1 IN THE DENOMINATOR.
THIS WOULD BE THE CORRECT QUOTIENT
FOR THIS DIVISION PROBLEM.
NOW YOU MIGHT ASK YOURSELF
WHETHER IT'S WORTH DOING SYNTHETIC DIVISION
ON THIS PROBLEM,
OR MAYBE YOU WOULD JUST CONVERT BACK TO LONG DIVISION.
IT REALLY IS YOUR PREFERENCE.
PERSONALLY, I WOULD PROBABLY JUST DO LONG DIVISION,
BUT IF YOU'RE SET ON PERFORMING SYNTHETIC DIVISION,
YOU CAN DO IT THIS WAY.
I HOPE YOU FOUND THIS VIDEO HELPFUL, AND HAVE A GREAT DAY.