- WELCOME TO ANOTHER EXAMPLE ON PARTIAL FRACTION DECOMPOSITION
OF A GIVEN RATIONAL EXPRESSION.
IN THIS VIDEO WE'LL SEE THE EXAMPLE OF WHEN
WE HAVE REPEATED LINEAR FACTORS IN THE DENOMINATOR.
BUT BEFORE WE SET THIS UP, WE FIRST WANT TO VERIFY
THE DEGREE OF THE NUMERATOR
IS LESS THAN THE DEGREE OF THE DENOMINATOR.
NOTICE WE HAVE DEGREE 2 NUMERATOR,
A DEGREE 3 DENOMINATOR,
WHICH MEANS WE CAN SET UP OUR PARTIAL FRACTIONS.
IF THE DEGREE OF THE NUMERATOR
WAS GREATER OR EQUAL TO THE DEGREE OF THE DENOMINATOR,
WE'D HAVE TO PERFORM LONG DIVISION FIRST.
SO WE'LL START BY FACTORING THE DENOMINATOR COMPLETELY.
LET'S GO AHEAD AND DO THIS ON THE SIDE.
WE WANT TO FACTOR X CUBED + 2X SQUARED + X.
FIRST STEP IN FACTORING IS TO FACTOR OUT
THE GREATEST COMMON FACTOR, WHICH IN THIS CASE IS X.
SO WE'LL HAVE X x THE QUANTITY X SQUARED + 2X + 1.
NOTICE THIS TRINOMIAL FACTORS AGAIN.
IT FACTORS INTO TWO BINOMIALS.
THE FIRST POSITIONS WILL BE X AND X.
THE FACTORS OF 1 THAT ADD TO 2 ARE 1 AND 1,
SO WE HAVE 1 AND 1.
NOTICE HOW WE HAVE TWO FACTORS OF X + 1,
SO WE CAN WRITE THIS AS X x THE QUANTITY X + 1 SQUARED.
SO LET'S REWRITE THIS AS 5X SQUARED + 20X + 6
DIVIDED BY X x THE QUANTITY X + 1 SQUARED.
SO BECAUSE WE HAVE THREE LINEAR FACTORS IN OUR DENOMINATOR,
EVEN THOUGH ONE'S REPEATED,
WE'LL HAVE THREE FRACTIONS
IN OUR PARTIAL FRACTION DECOMPOSITION
WHERE THE FIRST FRACTION WE'LL HAVE THE DENOMINATOR OF X,
OUR FIRST LINEAR FACTOR.
AND BECAUSE WE HAVE TWO FACTORS OF X + 1
THIS WILL GIVE US TWO MORE FRACTIONS
WHERE THE FIRST FRACTION WILL HAVE A DENOMINATOR OF X + 1
OR ONE FACTOR OF X + 1.
BUT THE SECOND FRACTION WILL HAVE TWO FACTORS OF X + 1,
SO WE'LL HAVE X + 1 SQUARED.
AND NOW BECAUSE WE HAVE LINEAR FACTORS IN THE DENOMINATOR,
OUR NUMERATORS WILL BE CONSTANTS,
BUT FOR RIGHT NOW THEY'RE UNKNOWNS.
SO WE'LL USE THE VARIABLE A HERE, THE VARIABLE B HERE,
AND THE VARIABLE C.
AND WE WANT TO FIND THE VALUES OF A, B, AND C.
TO DO THIS, WE'LL FIRST CLEAR THE FRACTIONS
BY MULTIPLYING BOTH SIDES OF THE EQUATION BY THE LCD,
WHICH WOULD BE X x THE QUANTITY X + 1 SQUARED.
SO THAT'S OUR NEXT STEP.
NOTICE ON THE LEFT SIDE, X/X SIMPLIFIES TO 1
AND SO DOES THE QUANTITY X + 1 SQUARED
OR THE QUANTITY X + 1 SQUARED.
SO WE'RE LEFT WITH 5X SQUARED + 20X + 6 ON THE LEFT.
AND NOW ON THE RIGHT NOTICE THIS FIRST PRODUCT
WE WOULD HAVE AN X/X THAT SIMPLIFIES TO 1.
SO WE'D HAVE "A" x THE QUANTITY X + 1 SQUARED.
FOR THE SECOND PRODUCT,
NOTICE THAT ONE FACTOR OF X + 1 SIMPLIFIES OUT,
SO WE'D HAVE + BX, AND THEN ONE FACTOR OF X + 1.
AND THEN FILLING FOR THE THIRD PRODUCT,
NOTICE HOW WE HAVE X + 1 SQUARED/X + 1 SQUARED,
SO THE LAST TERM IS JUST + CX.
THIS IS CALLED THE GENERAL EQUATION.
WE WANT TO USE THIS TO FIND THE VALUES OF A, B, AND C
THAT WILL GIVE US OUR PARTIAL FRACTION DECOMPOSITION.
AND WE HAVE TWO OPTIONS.
WE CAN USE A METHOD CALLED EQUATING COEFFICIENTS,
WHICH MEANS WE WOULD MULTIPLY OUT THE RIGHT SIDE
OF THIS EQUATION, COMBINE LIKE TERMS.
AND THEN COMPARING THE LIKE TERMS
ON THE RIGHT SIDE AND LEFT SIDE OF THE EQUATION,
WE WOULD SET THE COEFFICIENTS EQUAL TO EACH OTHER,
WHICH WOULD GIVE US A SYSTEM OF EQUATIONS
IN TERMS OF A, B, AND C, WHICH WE WOULD THEN SOLVE.
THE OTHER OPTION IS TO SELECT CONVENIENT VALUES OF X
TO HELP US FIND THE VALUES OF A, B, AND C.
SO WE'RE GOING TO USE THE SHORTER METHOD
OF USING SPECIFIC VALUES OF X TO FIND THE VALUES OF A, B, AND C.
VALUES THAT WE'LL USE ARE THE VALUES OF X
THAT WILL MAKE FACTORS INVOLVING X EQUAL TO ZERO.
FOR EXAMPLE, IF WE LET X = 0,
NOTICE HOW ON THE RIGHT SIDE OF THE EQUATION
WE JUST HAVE 0 + 0 + 6 OR 6 = IF X IS 0 HERE
WE'D HAVE 1 SQUARED x A THAT'S JUST A.
IF X IS 0 HERE, THIS PRODUCT WOULD BE 0,
AND SO WOULD THIS PRODUCT HERE.
SO NOW WE KNOW THAT A MUST = 6.
THE NEXT CONVENIENT VALUE OF X WOULD BE X = -1
BECAUSE THAT'LL MAKE X + 1 = 0.
SO LET'S GO AHEAD AND DO THAT NEXT, X = -1.
SO WE'LL SUBSTITUTE -1 IN FOR X, THIS WILL BE 5 + -20
THAT'S -15 + 6, WHICH WOULD BE -9.
ON THE RIGHT SIDE IF X = -1 THIS WOULD BE 0,
AND THIS WOULD BE 0 AS WELL.
SO WE'D HAVE JUST C x -1 OR -C.
WELL, DIVIDING BOTH SIDES BY -1 WE WOULD HAVE C = 9.
NOW, WE'RE STILL LEFT TO FIND THE VALUE OF B,
AND WE'VE ALREADY USED THE CONVENIENT VALUES OF X.
SO NOW WE'LL HAVE TO SELECT ONE MORE VALUE OF X
TO PERFORM SUBSTITUTION.
BUT SINCE WE ALSO KNOW A = 6 AND C = 9,
WE'LL STILL BE ABLE TO FIND THE VALUE OF B.
SO LET'S GO AHEAD AND USE X = 1.
SO IF X = 1 ON THE LEFT SIDE,
WE HAVE 5 + 20 + 6 THAT'S 31 = ON THE RIGHT SIDE,
AGAIN, IF X IS 1 WE HAVE 1 + 1 THAT'S 2 SQUARED THAT'S 4 x A,
BUT A IS 6, 6 x 4 IS 24 + IN THE PARENTHESIS
WE HAVE (1 + 1 OR 2).
SO WE HAVE 2 x 1 x B THAT'S 2B.
AND THEN + C x X, WELL, C IS 9 AND X IS 1,
9 x 1 IS 9.
SO NOW WE CAN SOLVE THIS FOR B, 24 + 9 IS 33.
SUBTRACT 33 ON BOTH SIDES THAT'S -2 = 2B.
DIVIDE BOTH SIDES BY 2.
NOW WE KNOW THAT B = -1.
NOW THAT WE KNOW THE VALUES OF A, B, AND C,
WE HAVE OUR PARTIAL FRACTION DECOMPOSITION WHERE "A" IS 6,
B IS -1, AND C IS 9.
LET'S GO AND WRITE THIS OUT ON THE NEXT SLIDE.
SO WE JUST FOUND THAT 5X SQUARED + 20X + 6
DIVIDED BY THE QUANTITY X TO THE THIRD + 2X SQUARED + X,
WE COULD ALSO LEAVE THIS IN FACTORED FORM.
EITHER WAY THIS IS EQUAL TO A DIVIDED BY X OR 6 DIVIDED BY X.
THEN INSTEAD OF WRITING + -1 DIVIDED BY THE QUANTITY X + 1
WE'LL GO AND JUST WRITE -1 DIVIDED BY THE QUANTITY X + 1.
AND THEN SINCE C IS 9
WE HAVE + 9 DIVIDED BY THE QUANTITY X + 1 SQUARED.
THIS RIGHT SIDE WOULD BE THE PARTIAL FRACTION DECOMPOSITION
OF THE ORIGINAL RATIONAL EXPRESSION.
OKAY. I HOPE YOU FOUND THIS HELPFUL.