- WELCOME TO AN EXAMPLE ON HOW TO FIND
THE PARTIAL FRACTION DECOMPOSITION
OF A RATIONAL EXPRESSION THAT CONTAINS REPEATED QUADRATIC
FACTORS IN THE DENOMINATOR.
FIRST, NOTICE THE DEGREE OF THE NUMERATOR
IS LESS THAN THE DEGREE OF THE DENOMINATOR,
WHICH MUST BE THE CASE BEFORE IT PERFORM
PARTIAL FRACTION DECOMPOSITION.
IF IT WASN'T, WE'D HAVE TO PERFORM LONG DIVISION FIRST.
NOTICE THE DEGREE OF THE NUMERATOR IS 3,
AND THE DEGREE OF THE DENOMINATOR,
IF WE MULTIPLY THIS OUT WOULD BE 4.
ALSO NOTICE THE DENOMINATOR IS ALREADY FACTORED
SINCE X SQUARED + 2 DOES NOT FACTOR AGAIN.
SO STARTING WITH THE ORIGINAL RATIONAL EXPRESSION,
BECAUSE WE HAVE TWO FACTORS IN OUR DENOMINATOR,
WHICH ARE REPEATED, WE'LL HAVE TWO PARTIAL FRACTIONS
HERE ON THE RIGHT.
AND, AGAIN, BECAUSE THIS IS A REPEATED FACTOR,
OUR FIRST FRACTION WILL HAVE ONE FACTOR OF X SQUARED + 2,
AND THE SECOND FRACTION WILL HAVE TWO FACTORS
OF X SQUARED + 2.
BECAUSE THESE ARE QUADRATIC FACTORS,
THE NUMERATORS COULD BE LINEAR FACTORS,
WHICH MEANS FOR THIS FIRST FRACTION
WE'LL HAVE THE NUMERATOR OF AX + B.
AND FOR THE SECOND FRACTION,
BECAUSE THEY'RE ALSO QUADRATIC FACTORS,
WE COULD POSSIBLY HAVE ANOTHER LINEAR FACTOR IN THE NUMERATOR,
SO WE'LL HAVE CX + D.
NOTICE HOW WE ENDED THIS EXPRESSION WITH A VARIABLE B,
SO WE START THIS EXPRESSION WITH C.
NOW, OUR MAIN GOAL HERE IS TO FIND THE VALUES
OF A, B, C, AND D, WHICH WILL GIVE US
OUR PARTIAL FRACTION DECOMPOSITION.
SO FOR THE NEXT STEP WE'RE GOING TO CLEAR THE FRACTIONS
FROM THIS EQUATION BY MULTIPLYING BOTH SIDES
OF THE EQUATION BY THE LEAST COMMON DENOMINATOR,
WHICH WOULD BE THE QUANTITY X SQUARED + 2 TO THE SECOND POWER.
NOTICE THE LEFT SIDE SIMPLIFIES NICELY,
X SQUARED + 2 SQUARED/X SQUARED + 2 SQUARED
THAT SIMPLIFIES TO 1, LEAVING US WITH 18X CUBED + 13X
ON THE LEFT SIDE.
AND THEN ON THE RIGHT SIDE,
NOTICE FOR THIS PRODUCT ONE FACTOR OF X SQUARED + 2
SIMPLIFIES OUT, WHICH WOULD LEAVE US WITH THE QUANTITY
AX + B x THE QUANTITY X SQUARED + 2.
AND THEN FOR THE SECOND PRODUCT,
BOTH FACTORS OF X SQUARED + 2 SIMPLIFY OUT OR SIMPLIFY TO 1,
LEAVING US WITH CX + D.
THIS IS CALLED THE BASIC EQUATION,
WHICH WE'LL USE TO FIND THE VALUES OF A, B, C, AND D.
AND THERE ARE TWO MAIN METHODS FOR DOING THIS.
ONE METHOD IS CALLED EQUATING COEFFICIENTS,
AND THE SECOND METHOD,
WHICH WE USED IN MOST OF THE PREVIOUS VIDEOS,
WE SELECT CONVENIENT VALUES OF X,
WHICH MEANS VALUES OF X
THAT WOULD SIMPLIFY OUT SEVERAL OF THE TERMS
ALLOWING US TO FIND A, B, C, AND D.
BUT FOR THIS EXAMPLE WE'RE GOING TO USE
EQUATING THE COEFFICIENTS,
WHICH MEANS WE'RE GOING TO START
BY MULTIPLYING OUT THE RIGHT SIDE,
AND THEN GROUPING THE LIKE TERMS IN TERMS OF X NEXT TO EACH OTHER
TO FACTOR THEM.
LET'S GO AHEAD AND START WITH THAT.
WE HAVE 8X TO THE THIRD + 13X.
NOW, HERE WE'RE GOING TO HAVE
FOUR PRODUCTS, ONE, TWO, THREE, FOUR.
NOTICE THE FIRST PRODUCT WOULD BE AX CUBED,
AND THEN WE'LL HAVE 2AX, AND THEN WE'LL HAVE BX SQUARED,
AND THEN + 2B.
AND THEN WE STILL HAVE + CX + D.
NOW, FOR THIS NEXT STEP,
WE'LL GROUP THE LIKE TERMS IN TERMS OF X NEXT TO EACH OTHER,
WHICH WE'LL THEN FACTOR.
SO WE HAVE 8X CUBED + 13X.
THERE'S ONLY ONE X CUBE TERM, IT'S AX CUBE.
THERE'S ONLY ONE X SQUARED TERM THAT'S + BX SQUARED.
NOTICE HOW THERE ARE TWO X TERMS.
WE HAVE 2AX AND WE ALSO HAVE CX, SO WE HAVE + 2AX + CX.
AND THERE ARE TWO TERMS THAT DON'T HAVE X's, 2B AND D.
SO + 2B + D, AND NOW WE'LL FACTOR THE X
FROM THESE TWO TERMS,
AND THEN MATCH UP THE COEFFICIENTS
ON THE LEFT SIDE OF THE EQUATION WITH THE LIKE TERMS.
SO TO HELP US DO THIS I'M GOING TO WRITE 8X CUBED + 13X
AS 8X CUBED + 0X SQUARED, WHICH IS STILL 0, + 13X.
AND OUR CONSTANT WOULD BE 0, SO WE'LL + 0 HERE.
SO ON THE RIGHT SIDE WE'LL HAVE AX CUBED + BX SQUARED.
AND NOW FOR THESE TWO TERMS I'LL FACTOR OUT THE X,
SO WE'LL HAVE + THE QUANTITY 2A + C x X PLUS
AND I'M GOING TO GO AHEAD AND JUST PUT THESE TWO TERMS
IN PARENTHESIS (2B + D).
SO NOW THE METHOD OF EQUATING THE COEFFICIENTS MEANS
IF THESE TWO SIDES OF THE EQUATION
ARE EQUAL TO EACH OTHER THEN THE COEFFICIENTS
OF THE LIKE TERMS MUST BE EQUAL.
WHICH MEANS IF WE HAVE 8X CUBED HERE, 8 MUST BE = TO A.
THE COEFFICIENT OF THE X SQUARED TERM IS 0,
WHICH MEANS 0 MUST BE EQUAL TO B.
THE COEFFICIENT OF THE X TERM IS 13,
WHICH MEANS 13 MUST BE EQUAL TO 2A + C.
AND THEN FINALLY ON THE LEFT SIDE THE CONSTANT IS 0,
WHICH MEANS 0 MUST EQUAL TO 2B + D.
SO KNOWING THIS INFORMATION
WE CAN NOW SET UP A SYSTEM OF EQUATIONS
JUST TO FIND A, B, C, AND D.
SO, AGAIN, LET'S START WITH A = 8.
NEXT WE KNOW THAT B MUST = 0.
WE ALSO KNOW THAT 2A + C MUST = 13.
AND 2B + D MUST = 0.
SO, AGAIN, WE ALREADY KNOW A = 8 AND B = 0,
SO USING THE THIRD EQUATION 2A + C = 13,
WE CAN FIND C SINCE WE KNOW A = 8.
YOU WOULD HAVE 2 x 8 + C = 13.
WELL, 2 X 8 OF COURSE IS 16.
SUBTRACTING 16 ON BOTH SIDES WE HAVE C = -3.
AND NOW FOR THE LAST VARIABLE D,
IF B = 0 WE'D HAVE 2 x 0 + D = 0, SO WE ALSO KNOW D = 0.
NOW THAT WE KNOW THE VALUES OF A, B, C, AND D,
WE KNOW OUR PARTIAL FRACTION DECOMPOSITION.
WE'LL JUST SUBSTITUTE THESE VALUES
INTO OUR TWO FRACTIONS HERE.
LET'S GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SINCE WE KNOW "A" IS 8 AND B IS 0,
FOR THIS FIRST FRACTION WE'D HAVE 8X + 0 OR JUST 8X.
SO WE'D HAVE 8X DIVIDED BY THE QUANTITY X SQUARED + 2.
AND THEN FOR THE SECOND FRACTION
WHERE WE HAVE A NUMERATOR OF CX + D, NOTICE C IS -3 AND D IS 0,
SO THIS WOULD BE -3X.
BUT INSTEAD OF WRITING 3X OVER OUR DENOMINATOR,
WE CAN GO AHEAD AND JUST WRITE -3X DIVIDED BY OUR DENOMINATOR.
AND THIS WOULD BE OUR PARTIAL FRACTION DECOMPOSITION
OF THE GIVEN RATIONAL EXPRESSION,
WHICH MEANS IF YOU WERE TO SUBTRACT THESE TWO FRACTIONS
THE DIFFERENCE WOULD BE THE ORIGINAL RATIONAL EXPRESSION.
OKAY, I HOPE YOU FOUND THIS EXPLANATION HELPFUL.