- WELCOME TO ANOTHER EXAMPLE OF FINDING THE PARTIAL FRACTION
DECOMPOSITION OF A GIVEN RATIONAL EXPRESSION.
NOTICE IN THIS EXAMPLE THE DENOMINATOR
IS ALREADY FACTORED WHERE WE HAVE ONE LINEAR FACTOR
AND ONE QUADRATIC FACTOR,
BUT BEFORE WE START IT'S ALSO IMPORTANT TO NOTICE
THAT THE DEGREE OF THE NUMERATOR
IS LESS THAN THE DEGREE OF THE DENOMINATOR.
NOTICE WE MULTIPLIED OUT THE DENOMINATOR
WE WOULD HAVE DEGREE 3, THE NUMERATOR IS DEGREE 2.
SO WE CAN GO AHEAD AND SET UP OUR PARTIAL FRACTIONS.
WE'LL START WITH THE ORIGINAL RATIONAL EXPRESSION.
NOTICE HOW THIS QUADRATIC HERE DOES NOT FACTOR AGAIN
BECAUSE THERE ARE NO FACTORS OF 1 THAT ADD TO -1.
SO BECAUSE WE HAVE 2 FACTORS IN OUR DENOMINATOR
WE'LL HAVE 2 FRACTIONS IN OUR DECOMPOSITION
WHERE THE FIRST DENOMINATOR WILL BE THE LINEAR FACTOR X - 2.
AND THE SECOND DENOMINATOR
WILL BE THE QUADRATIC FACTOR X SQUARED - X + 1.
FOR THE LINEAR FACTORS
THE NUMERATOR IS ALWAYS GOING TO BE A CONSTANT
SO WE'LL USE THE VARIABLE "A"
SINCE WE DON'T KNOW THAT CONSTANT YET.
AND THEN FOR QUADRATIC FACTORS
THE NUMERATOR COULD BE A LINEAR FACTOR;
MEANING IT COULD BE IN THE FORM OF BX + C.
NOTICE HOW WE STARTED THIS WOULD B BECAUSE WE USED "A" HERE.
AND NOW FOR THE NEXT STEP WE'RE GOING TO CLEAR THE FRACTIONS
FROM THIS EQUATION AND THEN SOLVE FOR A, B, AND C.
SO TO CLEAR THE FRACTIONS WE'RE GOING TO MULTIPLY BOTH SIDES
OF THE EQUATION BY THE LCD,
WHICH WOULD BE THE QUANTITY
X - 2 x THE QUANTITY X SQUARED - X + 1.
NOTICE ON THE LEFT SIDE THIS SIMPLIFIES NICELY,
X - 2/X - 2 AND THE QUADRATIC FACTOR/THE QUADRATIC FACTOR.
SO WE'RE LEFT WITH 5X SQUARED - 8X + 5
AND ON THE RIGHT SIDE, THIS FIRST PRODUCT,
THE LINEAR FACTOR SIMPLIFIES OUT
SO WE HAVE "A" x THE QUADRATIC FACTOR X SQUARED - X + 1.
AND THEN FOR THE SECOND PRODUCT
THE QUADRATIC FACTOR SIMPLIFIES OUT
LEAVING US WITH THE QUANTITY BX + C x THE QUANTITY X - 2.
AND NOW TO FIND THE VALUES OF A, B, AND C
WE CAN USE A METHOD CALLED EQUATING THE COEFFICIENTS,
WHICH IS WHERE WE MULTIPLY THE RIGHT SIDE OUT
THEN GROUP THE LIKE TERMS IN TERMS OF X TOGETHER
WHERE WE'D HAVE X SQUARED TERMS, X TERMS, THE CONSTANTS.
WE WOULD FACTOR THOSE AND THEN SET THE COEFFICIENTS
OF THOSE TERMS EQUAL TO THE COEFFICIENTS
OF THE LIKE TERMS ON THE LEFT.
BUT IN THIS VIDEO WE'RE GOING TO USE A SHORTCUT METHOD,
WHICH IS WHERE WE SELECT SPECIFIC VALUES OF X
MORE SPECIFICALLY THE VALUES OF X
THAT WOULD SIMPLIFY OUT SEVERAL TERMS
IN ORDER TO FIND THE VALUES OF A, B, AND C.
FOR EXAMPLE, NOTICE IF WE SELECT X = 2 THE FACTOR OF X - 2
WOULD BE = 0.
SO LETS START THERE.
IF X = 2 WE'LL SUBSTITUTE 2 FOR X.
SO HERE WE'D HAVE 2 SQUARED THAT'S 4 x 5
THAT'S 20 - 8 x 2 OR -16 THAT'S 4 + 5 THAT'S 9.
AND NOW ON THE RIGHT SIDE, AGAIN IF X IS 2
WE'D HAVE 2 SQUARED OR 4 - 2 THAT'S 2 + 1 THAT'D BE 3.
SO A x 3 WOULD BE 3A AND THEN OVER HERE IF X = 2
THIS WOULD BE 0.
0 x ANYTHING IS STILL 0
SO NOTICE THAT WE HAVE THE EQUATION 9 = 3A
DIVIDE BOTH SIDES BY 3 SO NOW WE KNOW "A" MUST = 3.
BUT FOR THE NEXT VALUE OF X LET'S GO AND SELECT X = 0
BECAUSE THAT WOULD AT LEAST ELIMINATE THE BX TERM.
SO IF X = 0 WE WOULD HAVE 0 - 0 + 5 THAT'S 5 EQUALS
AGAIN IF X IS 0 THIS WOULD JUST BE 0 - 0 + 1.
A x 1 IS A. AND THEN FOR THIS LAST PRODUCT,
IF X IS 0 WE WOULD JUST HAVE C x -2 OR - 2C.
THIS EQUATION DOES HAVE 2 VARIABLES.
WE ALREADY KNOW A MUST = 3 SO NOW WE'LL SUBSTITUTE 3 FOR A,
THAT WOULD GIVE US 5 = 3 - 2C.
SUBTRACT 3 ON BOTH SIDES, THAT WOULD GIVE US 2 = -2C
DIVIDED BY -2 C = -1.
NOW WE STILL NEED THE VALUE OF B
SO LOOKING AT OUR EQUATION ON THE RIGHT
I CAN'T SEE AN OBVIOUS SUBSTITUTION FOR X
SO LETS GO AHEAD AND USE X = 1.
BUT REMEMBER WE ALSO KNOW THAT A = 3 AND C = -1.
SO IF X = 1 WE'D HAVE 5 - 8 THAT'S -3 + 5.
IT'S GOING TO BE 2 EQUALS AND ON THE RIGHT SIDE
IF X = 1 THIS WOULD BE 1 - 1 + 1 SO THAT'S JUST 1.
REMEMBER A = 3
SO WE'D HAVE 3 x 1 + HERE WE HAVE BX + C.
WE DON'T KNOW B, BUT WE KNOW X IS 1
SO WE JUST HAVE B + C BUT C IS - 1.
SO WE HAVE B - 1 x AGAIN IF X IS 1
THIS WOULD JUST BE 1 - 2 OR -1.
SO NOTICE HOW THIS DOES GIVE US AN EQUATION WITH 1 VARIABLE
WHICH WE CAN SOLVE FOR B.
SO WE HAVE 2 = 3 DISTRIBUTE -1 WE'LL HAVE - B + 1.
IN MONEY-LIKE TERMS WE HAVE 2 = 3 + 1 IS 4
SUBTRACT 4 ON BOTH SIDES SO WE HAVE -B = -2
DIVIDE BOTH SIDES BY -1 AND WE HAVE B = 2.
NOW THAT WE KNOW THE VALUES OF A, B, AND C
WE'LL PERFORM THIS SUBSTITUTION INTO OUR 2 FRACTIONS HERE
FOR OUR PARTIAL FRACTION DECOMPOSITION.
AND WE'LL GO AHEAD AND DO THIS ON THE NEXT SLIDE.
SO NOW THAT WE KNOW THE VALUES OF A, B, AND C;
THE ORIGINAL RATIONAL EXPRESSION IS EQUAL TO "A"
DIVIDED BY THE QUANTITY X - 2
OR 3 DIVIDED BY THE QUANTITY X - 2 + BX + C
DIVIDED BY X SQUARED - X + 1 WHERE B = 2
AND C = -1, WHICH WOULD GIVE US 2X - 1.
THIS WOULD BE THE PARTIAL FRACTION DECOMPOSITION
OF OUR RATIONAL EXPRESSION.
I HOPE YOU FOUND THIS HELPFUL.