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THIS LECTURE IS ABOUT QUOTIENT
OF TWO POLYNOMIALS.
AND THIS, IF YOU REFER TO YOUR BOOK,
IS ON SECTION 5.6.
SO, SO FAR YOU'VE SEEN HOW TO ADD AND SUBTRACT POLYNOMIALS,
HOW TO MULTIPLY THEM,
AND NOW YOU'RE LOOKING AT DIVIDING POLYNOMIALS.
THERE ARE TWO TYPES OF DIVISION THAT YOU NEED TO KNOW.
THE FIRST ONE IS WHEN YOU DIVIDE BY A MONOMIAL.
AND YOU REMEMBER WHAT THAT MEANS.
THAT MEANS YOU HAVE TO DIVIDE BY A SINGLE TERM, OKAY?
SO, IF YOU'RE ASKED TO DIVIDE BY A SINGLE TERM,
THE METHOD WOULD BE YOU ARE TO DIVIDE
EACH TERM OF THE DIVIDEND
BY THE MONOMIAL
AND SIMPLIFY.
SO, TAKE FOR EXAMPLE,
YOU HAVE THE POLYNOMIAL 6Y TO THE POWER OF 10
BY THIS 3Y TO THE POWER OF 6
PLUS 4Y CUBED PLUS 8Y.
AND WE WISH TO DIVIDE THE ENTIRE THING
BY, SAY, 3Y CUBED.
SO, WHAT THE METHOD, IT REMINDS YOU TO DO
IS TO DIVIDE EVERY TERM IN THE DIVIDEND
BY THE MONOMIAL THAT'S GIVEN TO YOU IN THE DENOMINATOR.
SO EACH TERM IS BEING WRITTEN OUT
AS INDIVIDUAL FRACTIONS.
NOW, BEING INDIVIDUAL FRACTIONS,
YOU CAN NOW REFER TO WHAT WE LEARNED IN EXPONENTS
AND THAT ALLOWS US TO REDUCE THEM.
NOW, REMEMBER THE WAY TO DO IT,
NUMBERS, WE'RE GOING TO NUMBERS.
IN THIS CASE, 6 GOES INTO-- 3 GOES INTO 6 TWICE.
Y10 AND Y3 MEANS THAT THERE ARE 10 Ys IN THE NUMERATOR
AND 3 Ys IN THE DENOMINATOR.
SO WHEN YOU REDUCE THAT, IT BECOMES 2Y TO THE POWER OF 7.
SO DO THE SAME FOR THE NEXT ONE.
3 GOES INTO 3, Y CUBED GOES INTO Y TO THE POWER OF 6.
AND YOU HAVE Y CUBED LEFT.
BUT DON'T FORGET THERE'S A MINUS HERE.
SO YOUR ANSWER IS GOING TO BE -Y CUBED.
IN THE NEXT FRACTION YOU NOTICE THE NUMBERS CANNOT BE REDUCED,
BUT THE LETTERS CAN.
SO, REMEMBER, NUMBERS AND LETTERS ARE NOT STUCK TOGETHER.
SO, Y CUBED GOES INTO Y CUBED EXACTLY,
SO THAT BECOMES FOUR THIRDS.
AND IN THE LAST CASE, 3 DOES NOT GO INTO 8.
BUT Y AND Y CUBED CAN BE REDUCED
TO BECOME Y SQUARED IN THE DENOMINATOR.
SO THAT BECOMES 8 OVER 3Y SQUARED.
NOW, BEAR IN MIND THAT POLYNOMIALS
DOES NOT NECESSARILY COME WITH ONLY ONE VARIABLE.
SO THERE COULD BE A MIXTURE OF VARIABLES.
IN THIS CASE, FOR EXAMPLE,
35X SQUARED, Y CUBED
PLUS 7XY MINUS 63X CUBED
Y TO THE FOURTH.
AND THE DIVISOR MAY NOT BE POSITIVE.
IN THIS CASE, IT COULD EVEN BE A -7XY.
SO LONG AS IT'S A SINGLE TERM, OKAY? YOU CAN USE THIS METHOD.
AND THE METHOD IS TELLING YOU TO BREAK
EVERY PIECE IN THE NUMERATOR
AND DIVIDE INDIVIDUALLY BY WHAT'S IN THE DENOMINATOR.
AND WHEN YOU DO THAT,
THEY BECOME INDIVIDUAL FRACTIONS,
AND YOU CAN NOW USE RULES OF EXPONENTS TO REDUCE THEM.
SO, IF YOU LOOK AT THIS ONE,
THE FIRST FRACTION, 7 GOES INTO 35 FIVE TIMES.
X INTO X SQUARED X, Y INTO Y CUBED IS Y SQUARED.
NOW, THE ANSWER IS GOING TO BE -5XY SQUARED.
SO, -5XY SQUARED.
IN THE SECOND ONE YOU NOTICE 7 CANCELS, X CANCELS,
EVEN Y CANCELS.
WHEN EVERYTHING IS CANCELED, IT MEANS 1. OKAY.
AND PLUS AND MINUS MEANS MINUS.
SO THE ANSWER IS GOING TO BE -1.
SO, REMEMBER, WHEN EVERYTHING REDUCES, IT BECOMES 1.
IT'S JUST LIKE HOW WE HAVE 5 OVER 5 IS 1, 10 OVER 10 IS 1.
LIKE THIS. 10 OVER 10 IS 1.
5 OVER 5 IS 1.
SO WHEN THE TWO NUMBERS MATCH, IT MEANS 1, OKAY?
IT DOESN'T MEAN ZERO.
NOW, NEXT ONE RIGHT HERE, 7 GOES INTO 63 9 TIMES.
X INTO X CUBED IS X SQUARED.
Y INTO Y FOUR IS Y CUBED.
BUT THE TWO MINUSES MAKES IT A PLUS, OKAY?
SO IT'S GOING TO BE 9X SQUARED Y CUBED.
SO, THIS IS THE FIRST TYPE OF DIVISION.
LET'S LOOK AT THE SECOND TYPE OF DIVISION.
AND THE SECOND TYPE OF DIVISION
IS WHERE YOU ARE DIVIDING BY ANOTHER POLYNOMIAL.
SO, WE'RE NO LONGER TALKING ABOUT SINGLE TERMS.
WE COULD BE TALKING ABOUT TWO OR THREE TERMS.
AND THE METHOD WE ARE GOING TO BE USING
IS CALLED LONG DIVISION.
SO, CONSIDER AN ARITHMETIC EXAMPLE.
FOR EXAMPLE,
CONSIDER YOU HAVE TO DIVIDE,
WHAT'S 205 DIVIDED BY 3?
SO, WHAT IS THE TECHNIQUE INVOLVED?
OKAY, YOU NOTICE THAT FIRST AND FOREMOST
YOU ARE GOING TO PUT THE 205 INSIDE THE DIVISION SYMBOL
AND 3 ON THE OUTSIDE. OKAY.
SO, THE DIVIDEND GOES INSIDE AND THE OUTSIDE IS THE DIVISOR.
SO, THIS IS THE DIVISOR ON THE OUTSIDE.
AND THE ONE INSIDE IS THE DIVIDEND.
ALL RIGHT.
AND THE PROCEDURE IS YOU'RE GOING TO USE THE 3
AND YOU'RE GOING TO GO INTO 20, FOR EXAMPLE.
AND YOU'RE GOING TO ASK YOURSELF,
HOW MUCH CAN 3 GO INTO 20?
WHAT'S THE MAXIMUM?
SO THE STEP IS GOING TO BE, YOU'RE GOING TO FIRST DECIDE.
OKAY, SO 3 GOES INTO 20, SAY, 6 TIMES.
AND THEN YOU'RE GOING TO PUT 18.
AND 18 IS OBTAINED BY MULTIPLYING.
SO THAT'S STEP NUMBER TWO.
SO AFTER YOU DECIDE, YOU NEED TO MULTIPLY.
OKAY, AND THAT'S HOW YOU GET YOUR 18.
AND THE NEXT THING YOU WANT TO DO
IS YOU'RE GOING TO SUBTRACT 18 FROM 20.
AND THAT MAKES IT 2.
AND THEN THE FOURTH STEP
IS WHERE YOU CARRY DOWN THE NEXT TERM.
IN THIS CASE THE NEXT DIGIT.
AND SO, THAT'S 5.
AND THEN THE STORY REPEATS.
YOU'RE GOING TO DECIDE AGAIN.
OKAY, THAT MAKES IT MAXIMUM IS 8 TIMES.
THEN YOU'RE GOING TO MULTIPLY, SO 8 TIMES 3 IS 24.
AND THEN YOU'RE GOING TO SUBTRACT.
SO THAT MAKES IT 1.
AND THEN YOU'RE GOING TO CARRY DOWN THE NEXT TERM.
SO IN THE EVENT WHEN THERE'S NO NEXT TERM,
THAT'S WHEN YOU STOP.
SO, IF THERE IS NO NEXT TERM,
THE DIVISION IS DONE.
OKAY. SO, HOW DOES THIS APPLY TO ARITHMETIC?
BUT LET'S RIGHT DOWN THE ANSWER FIRST.
SO, AFTER WE DO THE DIVISION, HOW DO WE WRITE THE ANSWER?
THE ANSWER IS GOING TO BE THAT 205 DIVIDED BY 3 IS REALLY 68,
WHICH IS YOUR QUOTIENT.
OKAY, THAT'S YOUR QUOTIENT.
AND THE REMAINDER IN THIS CASE IS 1.
SO IT'S PLUS 1 OVER 3.
SO THE WAY WE WRITE THE ANSWER IS GOING TO BE
THE DIVIDEND DIVIDED BY THE DIVISOR
IS EQUAL TO
THE QUOTIENT PLUS THE REMAINDER
OVER THE DIVISOR.
THAT'S HOW WE WRITE THE ANSWER.
OKAY.
SO, HOW DOES THIS APPLY TO ARITHMETIC?
LET'S LOOK AT THE FIRST EXAMPLE.
SAY YOU'RE ASKED TO DIVIDE 3X CUBED MINUS 10X + 6.
AND YOU WISH TO DIVIDE THIS BY X + 2.
SO SIMILARLY, YOU'RE GOING TO PUT THE DIVIDEND ON THE INSIDE
AND THE DIVISOR ON THE OUTSIDE.
THE ONLY THING IS THERE IS A FEW NOTES FOR YOU TO REMEMBER.
FIRST THING IS YOUR DIVIDEND
MUST BE WRITTEN
IN DESCENDING ORDER
OF EXPONENTS.
AND YOU'RE SUPPOSED TO USE ZEROES AS PLACEHOLDERS
FOR ANY MISSING TERMS.
SO, IF YOU LOOK AT THE DIVIDEND THAT WE HAVE HERE,
THE HIGHEST EXPONENT IS CUBED.
SO WE START WITH THAT.
AND AFTER CUBED COMES THE SQUARED.
BUT YOU NOTICE THERE IS MISSING X SQUARED TERM.
SO THAT HAPPENS.
YOU HAVE TO PUT THE ZERO FOR THE X SQUARED TERM,
WHATEVER IS MISSING.
SO YOU START WITH A CUBED TERM.
THERE'S A MISSING X SQUARED TERM.
AND THEN I'M GOING TO THE X TERM,
WHICH IN THIS CASE IS 10X.
AND AFTER THE X TERM COMES THE CONSTANT,
WHICH IN THIS CASE IS 6. OKAY?
SO, NOW I CAN START.
SO, REMEMBER THE STEPS.
THE STEPS IS YOU'RE GOING TO FIRST DECIDE
AND THEN YOU MULTIPLY AND THEN YOU SUBTRACT
AND YOU CARRY DOWN THE NEXT TERM.
SO, WHAT DO WE MEAN BY DECIDE?
OKAY? SO, THIS IS WHERE YOU HAVE TO USE THE LEADING TERM
AND THE LEADING TERM TO DETERMINE.
SO, TO DECIDE WHAT YOU NEED TO DO
IS YOU DECIDE BY USING THE LEADING TERMS TO DIVIDE.
SO I'M TAKING THE 3X CUBED TO DIVIDE IT BY THE X.
SO 3X CUBED DIVIDED BY X MAKES IT 3X SQUARED.
SO THAT'S WHERE I PUT THE DECISION UP THERE,
3X SQUARED.
THEN WHAT I NEED TO DO IS TO MULTIPLY.
AND WHAT MULTIPLY IS SAYING
IS YOU NEED TO MULTIPLY THE RESULT
WITH THE DIVISOR.
SO THAT MEANS I'M TAKING THE 3X SQUARED
AND I'M MULTIPLYING WITH THE ENTIRE THING,
WHICH IS X + 2,
AND THAT MAKES IT 3X CUBED PLUS 6X SQUARED.
AND THEN THE NEXT STEP I NEED TO DO IS TO SUBTRACT.
AND IN SUBTRACTING,
REMEMBER TO CHANGE SIGNS
OF THE SECOND POLYNOMIAL.
SO, 3X CUBED PLUS 6X SQUARED IS PUT HERE,
BUT I'M TRYING TO SUBTRACT THEM.
SO, THE PLUS 3X CUBED BECOMES MINUS.
AND SAME HERE,
THE PLUS 6X SQUARED BECOMES MINUS.
SO IN THE CHANGE OF SIGNS,
YOU NOTICE THE FIRST TERMS NOW CANCEL.
AND ZERO MINUS 6 IS -6X SQUARED.
AND THE FOURTH STEP IS WHERE YOU CARRY DOWN THE NEXT TERM.
SO, WHEN I CARRY DOWN THE NEXT TERM, IT BECOMES -10X.
SO, MY NEW POLYNOMIAL IS NOW -6X SQUARED MINUS 10X.
AND NOW THE STORY REPEATS.
THIS -6X SQUARED BECOMES MY NEW LEADING TERM.
SO I HAVE TO TAKE MY LEADING TERM
TO DIVIDE BY THE LEADING TERM,
WHICH IS X,
AND THAT MAKES IT -6X.
SO, PUT YOUR DECISION IN THE QUOTIENT RIGHT AT THE TOP.
AND THEN WHAT YOU NEED TO DO IS MULTIPLY BY THE DIVISOR,
WHICH IS X + 2.
THAT MAKES IT -6X SQUARED MINUS 12X.
AND THAT'S WHERE YOU PUT IT, UNDERNEATH.
AND THIS LINE IS TO REMIND YOU YOU'RE SUBTRACTING.
AND WHEN YOU SUBTRACT,
DON'T FORGET TO CHANGE THE SIGNS.
SO, MINUS BECOMES PLUS IN THIS CASE.
AND SO THE FIRST TERM CANCELS.
THEY'RE THE SWITCH OF SIGNS.
AND I BRING DOWN THE NEXT, AND I'LL SUBTRACT THEM.
-10 WITH 12 IS 2X.
AND THEN I BRING DOWN THE NEXT TERM.
SO, NOW THE 2X BECOMES THE NEW LEADING TERM.
AND I'M GOING TO TAKE 2X TO DIVIDE BY X,
WHICH MAKES IT 2. OKAY.
SO I PUT THE DIVISION UP THERE, 2.
AND THIS 2 IS GOING TO BE USED TO MULTIPLY BY THE DIVISOR,
WHICH IS X + 2,
AND THAT MAKES IT 2X + 4.
SO, I PUT THE 2X + 4 HERE.
BUT I'M REMINDING MYSELF THAT I'M SUBTRACTING,
SO I HAVE TO CHANGE THE SIGN OF THE SECOND POLYNOMIAL.
AND, SO, PLUS BECOMES MINUS,
PLUS BECOMES MINUS.
THEY CANCEL.
AND 6 - 4 IS 2.
WHEN I'M OUT OF TERMS TO CARRY DOWN,
THAT'S WHEN I STOP. OKAY.
AND THE WAY TO WRITE THE ANSWER IS THE DIVIDEND.
SO, THE ANSWER TO THIS QUESTION
IS WHEN YOU TAKE 3X CUBED MINUS 10X + 6
AND YOU WISH TO DIVIDE THIS BY X + 2,
YOUR ANSWER IS GOING TO BE 3X SQUARED MINUS 6X + 2,
PLUS THE REMAINDER, WHICH IS 2.
THEN YOU DIVIDE BY X + 2.
THAT BECOMES THE ANSWER.
NOW, JUST A REMINDER, THIS IS VERY CRUCIAL.
YOU MUST PUT THE SIGN HERE.
SO THERE MUST BE A SIGN
BETWEEN THE QUOTIENT AND THE REMAINDER
OVER DIVISOR.
REMEMBER, THERE MUST BE A SIGN
AND USUALLY THE SIGN BETWEEN THEM IS A PLUS.
LET'S LOOK AT ANOTHER EXAMPLE.
AND THIS ONE I'M USING THE ONE IN YOUR BOOK.
IT'S NUMBER 51 ON PAGE 250.
SAY, YOU ARE ASKED TO FIND THE QUOTIENT
TO 3X SQUARED PLUS 13X
PLUS 16X TO THE POWER OF 4 PLUS 3 DIVIDED BY 4X + 3.
YOU NOTICE THAT IF YOU LOOK AT THE DIVISOR,
YOU NOTICE THAT DIVISOR IS NOT A MONOMIAL,
SO THAT TELLS YOU YOU HAVE TO USE LONG DIVISION.
AND REMEMBER WHAT THAT MEANS, THERE'S A FOUR-STEP PROCEDURE.
YOU HAVE TO FIRST DECIDE, THEN YOU HAVE TO MULTIPLY,
AND THEN YOU HAVE TO SUBTRACT, AND THEN YOU HAVE TO CARRY.
SO, KEEP THAT IN MIND.
BUT THERE'S TWO BIG NOTES.
YOU HAVE TO MAKE SURE THEY ARE PUT IN ORDER.
SO THERE'S TWO BIG NOTES.
FIRST ONE IS IT HAS TO BE DESCENDING ORDER.
AND THE OTHER ONE IS YOU HAVE TO USE ZEROES IF NECESSARY.
SO IT'S NOT ALWAYS THAT YOU HAVE TO USE ZEROES.
YOU ONLY USE IF THERE'S ANY MISSING TERMS.
SO, IN OUR CASE,
THE HIGHEST EXPONENT IS 16X TO THE FOURTH.
THE HIGHEST EXPONENT IS 4.
AND AFTER THE FOURTH EXPONENT COMES THE THIRD,
WHICH IS NOT THERE.
SO, THAT'S WHY I HAVE TO PUT ZERO X CUBED.
AND AFTER THE CUBE COMES THE SQUARES,
WHICH IS THERE, WHICH IS 3X SQUARED.
SO, I'M JUST GOING DOWN THE LINE.
16X4 IS RIGHT THERE.
AND THEN THIS MISSING X CUBED, THAT'S WHY I PUT A ZERO.
AND THEN THERE'S THE X SQUARED TERM,
WHICH IN THIS CASE IS 3X SQUARED.
AND AFTER THE SQUARE TERMS COMES THE X TERM,
WHICH IS RIGHT THERE.
SO IT'S 13X.
AND AFTER THE X TERM COMES THE CONSTANT,
WHICH IS 3.
AND THE DIVISOR IS 4X + 3. OKAY.
SO, LET'S START THE PROCESS.
SO WE FIRST HAVE TO DECIDE.
AND THE DECISION IS MADE BY TAKING THE LEADING TERM
DIVIDE BY THE LEADING TERM.
OKAY. SO, I'M TAKING THE 16X4 AND I'M DIVIDING BY 4X,
SO THAT GIVES ME 4X CUBED.
AND THAT'S WHERE I PUT IT, 4X CUBED.
AND THEN I'M GOING TO TAKE THE 4X CUBED
TO MULTIPLY BY THE DIVISOR, WHICH IN THIS CASE IS 4X + 3.
AND THAT COMES OUT TO BE 16X TO THE FOURTH PLUS 12X CUBED.
AND THEN I PUT IT OVER AT THE BOTTOM.
16X4 PLUS 12X CUBED.
SORRY, THAT WAS A MISTAKE.
SO, NOW I'M GOING TO SUBTRACT,
OKAY? SO, SUBTRACT IS WHERE I CHANGE THE SIGNS
SO THE PLUS BECOMES MINUS AND THE PLUS BECOMES MINUS.
SO, THIS CANCELS AND ZERO - 12 MAKES IT -12X CUBED.
OKAY? THEN I CARRY THE NEXT TERM,
WHICH IN THIS CASE IS PLUS 3X SQUARED.
SO I'M GOING TO TAKE THE -12X CUBED,
WHICH IS NOW THE NEW LEADING TERM
TO DIVIDE BY THE 4X.
SO, I'M TAKING THE LEADING DIVIDE BY THE LEADING, OKAY?
AND THAT COMES UP TO BE -3X SQUARED.
AND THAT'S WHY I PUT IT UP THERE, -3X SQUARED.
SO, THIS -3X SQUARED IS BEING MULTIPLIED TO THE 4X + 3,
WHICH MAKES IT -12X CUBED MINUS 9X SQUARED.
SO, THAT'S WHERE I PUT IT UNDERNEATH
-12X CUBED MINUS 9X SQUARED.
SO, REMEMBER THE STEPS.
YOU HAVE TO FIRST DECIDE THAT YOU NEED TO MULTIPLY,
WHICH I JUST DID.
AND NOW I NEED TO SUBTRACT.
SO IN SUBTRACTING,
MAKE SURE YOU CHANGE THE SIGN OF THE SECOND POLYNOMIAL.
THAT'S WHY THE MINUSES BECAME THE PLUSES.
AND NOW THE SIGNS--
BECAUSE OF THE CHANGE OF SIGNS,
YOU NOTICE THE FIRST TERM CANCELLED.
AND 3 AND 9, THEY BECOME 12X SQUARED.
THEN THE FOURTH STEP IS WHERE I CARRY THE NEXT TERM DOWN.
SO, THAT'S WHERE I BRING DOWN THE 13X.
OKAY.
SO, WHAT HAPPENS IS YOU TAKE THE NEW LEADING TERM
AND DIVIDE BY THE NEW LEADING TERM.
SO IT'S GOING TO BE 12X SQUARED DIVIDED BY 4X,
WHICH MAKES IT 3X.
SO THAT'S WHERE YOU PUT IT UP HERE, 3X.
AND THEN YOU'RE GOING TO MULTIPLY.
THAT'S THE NEXT STEP.
3X MULTIPLIED TO 4X + 3,
WHICH MAKES IT A 12X SQUARED PLUS 9X.
AND THAT'S WHERE I PUT IT UNDERNEATH 12X SQUARED PLUS 9X.
SO WHEN I SUBTRACT, I'M CHANGING THE SIGN,
SO THE PLUS BECOMES MINUS.
AND THE PLUS BECOMES MINUS, THAT CANCELS.
13 MINUS 9 IS 4X,
AND I BRING DOWN THE NEXT TERM, WHICH IS 3.
SO, THE STORY REPEATS.
I'M TAKING THE LEADING WITH THE LEADING.
SO, 4X OVER 4X IS 1.
AND SO I PUT THE 1 THERE.
AND THE 1 IS MULTIPLIED BY 4X + 3,
WHICH IS 4X + 3.
SO, WHEN YOU PUT THIS UNDERNEATH AND YOU CHANGE THE SIGN,
IT ACTUALLY COMPLETELY CANCELS OUT,
LEAVING YOU A REMAINDER OF ZERO.
SO, LET'S MAKE SURE YOU WRITE THE CORRECT ANSWER.
SO, HOW YOU WOULD WRITE THE ANSWER IS THAT
3X SQUARED PLUS 13X PLUS 16X TO THE FOURTH PLUS 3
DIVIDED BY 4X + 3 IS REALLY 4X CUBED MINUS 3X SQUARED
PLUS 3X + 1.
THERE'S MY QUOTIENT RIGHT THERE.
THE QUOTIENT THAT'S UP HERE.
AND YOU HAVE TO PUT THE PLUS.
DON'T FORGET.
AND THE REMAINDER IS ZERO DIVIDED BY THE DIVISOR,
WHICH IS 4X + 3.
BUT ZERO DIVIDED BY 4X + 3 IS STILL GOING TO BE ZERO.
SO, THIS IS REALLY ZERO.
SO YOU COULD LEAVE THAT OUT
AND SAY THE ANSWER IS 4X CUBED MINUS 3X SQUARED + 3X + 1.
THAT'S HOW YOU WRITE THE ANSWER.
SO, YOU REMEMBER HOW WE WRITE THE ANSWER
IS YOU ARE GIVEN THE DIVIDEND DIVIDED BY A DIVISOR.
AND THE WAY WE WRITE THE ANSWER
IS THAT THE QUOTIENT PLUS REMAINDER OVER THE DIVISOR
IS THE RESULT.
OKAY, UNLESS THE REMAINDER IS A ZERO.
SO, THIS IS ONE WAY TO DO IT.
ANOTHER WAY IS YOU MULTIPLY THROUGHOUT BY THE DIVISOR.
AND YOU NOTICE THAT
IF YOU MULTIPLY BY THE DIVISOR THROUGHOUT,
WHAT YOU END UP WITH IS
THAT THE DIVIDEND IS GIVEN BY THE QUOTIENT
MULTIPLIED BY THE DIVISOR.
SO YOU MULTIPLY THESE TOGETHER AND YOU ADD TO THE REMAINDER.
SO THIS IS YET ANOTHER WAY TO CHECK.
AND, IN FACT,
THIS IS THE DIVISION ALGORITHM.
SO, HOW CAN WE TEST
IF YOU UNDERSTAND THE DIVISION ALGORITHM?
YOU CAN LOOK AT NUMBER 68 ON PAGE 250.
SO, THIS IS ONE WAY WE CAN ASK YOU THE QUESTION.
LET ME SHOW YOU.
THE QUESTION READS,
"WHAT POLYNOMIAL WHEN DIVIDED BY X CUBED PLUS 3,
USE THE QUOTIENT 2X - 1, AND 3X + 4 IS THE REMAINDER?"
SO, YOU NOTICE IN THIS QUESTION
WE'RE NOT REALLY ASKING YOU TO DIVIDE
'CAUSE WE ARE TELLING YOU THE ANSWER ALREADY.
QUOTIENT IS 2X - 1 AND 3X + 4 IS THE REMAINDER.
THIS BEING THE DIVISOR.
SO, WHAT YOU ARE REALLY SUPPOSED TO USE
IS THE DIVISION ALGORITHM.
AND WHAT POLYNOMIAL IS REALLY ASKING YOU, WHAT'S THE DIVIDEND?
SO, FOR YOU TO KNOW WHAT IS THE POLYNOMIAL,
ALL YOU HAVE TO DO IS TAKE THE QUOTIENT
AND MULTIPLY TO THE DIVISOR
AND THEN YOU ADD TO THE REMAINDER.
SO, WHEN YOU DO THAT,
YOU WILL GET THE RESULT THAT YOU NEED.
THE QUOTIENT IS GIVEN HERE, WHICH IS 2X - 1.
THE DIVISOR IS ALSO GIVEN,
WHICH IS X CUBED PLUS 3 AND THE REMAINDER IS 3X + 4.
SO, YOU CAN MULTIPLY THIS OUT.
SO, 2X WITH X CUBED IS 2X TO THE FOURTH.
2X WITH 3 IS 6X - 1 WITH X CUBED
IS -X CUBED MINUS 1 WITH 3 IS -3.
AND THEN YOU HAVE TO ADD THE 3X + 4.
AND YOU MIGHT WANT TO REARRANGE THEM
IN DESCENDING ORDER OF EXPONENTS.
SO IT'S GOING TO BE 2X4 MINUS X CUBED.
THAT TAKES CARE OF THESE TWO TERMS.
THE 6X AND 3X COMBINES TO MAKE IT 9X.
AND THE MINUS 3 AND THE PLUS 4 MAKES IT PLUS 1.
SO, THIS IS ACTUALLY NOT A NEW CONCEPT.
IF YOU THINK ABOUT THE EARLIER EXAMPLE,
THE ARITHMETIC EXAMPLE THAT WE DID EARLIER,
WE DID 205.
SO, LET'S RECAP, OKAY?
205, WHEN I DIVIDE IT BY 3,
THE ANSWER CAME UP TO BE 68 AND 1 OVER 3.
ALTHOUGH IN ARITHMETIC WE WRITE IT AS 68 AND 1/3
BECAUSE THE PLUS THAT'S BETWEEN THE WHOLE NUMBER
AND THE FRACTION
IS UNDERSTOOD, OKAY?
BUT HOW CAN WE CHECK THAT WE ARE CORRECT?
SO, ONE WAY TO CHECK
IS YOU TAKE THE 68 AND YOU MULTIPLY BY THE 3.
SO, WHAT WE ARE USING IS WE'RE USING THE QUOTIENT
AND WE'RE MULTIPLYING BY THE DIVISOR, OKAY?
AND THIS CAME UP TO BE 204.
AND THEN WHAT WE HAVE TO DO IS WE HAVE TO ADD 1.
AND THIS ADDING OF 1 IS WHERE YOU ADD THE REMAINDER.
SO YOU TAKE THE QUOTIENT, MULTIPLY THE DIVISOR,
BUT YOU ADD TO THE REMAINDER AND THAT COMES OUT TO BE 205.
AND WHAT 205 IS IS THE DIVIDEND.
SO, WHAT YOU ARE DOING
IS EXACTLY WHAT THE DIVISION ALGORITHM SAYS,
WHERE YOU TAKE THE QUOTIENT TIMES THE DIVISOR
AND IF YOU ADD TO THE REMAINDER,
YOUR ANSWER HAS TO BE THE DIVIDEND. �