- WE WANT TO FIND ALL OF THE ZEROS OR ROOTS
OF THE GIVEN DEGREE THREE POLYNOMIAL FUNCTION.
WE WANT TO START BY FINDING THE RATIONAL ZEROS
OF THE FUNCTION,
AND TO DO THIS WE'RE OFTEN ASKED TO MAKE A LIST
OF ALL THE POSSIBLE RATIONAL ZEROS.
AND ALL THE POSSIBLE RATIONAL ZEROS OF A POLYNOMIAL FUNCTION
ARE FORMED BY THE RATIOS OF THE FACTORS
OF THE CONSTANT TERM,
IN THIS CASE -78,
TO THE FACTORS OF THE LEADING COEFFICIENT
WHICH IN THIS CASE IS JUST +1.
SO LET'S START BY LISTING ALL THE FACTORS OF -78,
ALL THE FACTORS OF 1,
AND THEN WE'LL FIND THE RATIO OF THE TWO LISTS.
FINDING ALL THE FACTORS OF 78 OR -78 ISN'T ALL THAT EASY,
SO LET'S TAKE A LOOK AT THE PRIME FACTORIZATION OF 78
BECAUSE THIS WILL HELP US FIND ALL OF THE FACTORS.
I KNOW 78 IS 2 x 39, AND 39 IS EQUAL TO 3 x 13.
SO WE CAN USE THE PRIME FACTORIZATION
TO LIST ALL THE FACTORS OF -78.
WELL, OF COURSE, +/-1 WOULD BE A FACTOR, +/-2, +/-3.
WELL, IF 2 AND 3 ARE FACTORS, THEN SO WOULD 6, SO +/- 6.
THEN WE HAVE +/- 13, AND WE'RE NOT DONE
BECAUSE 2 x 13 WOULD ALSO GIVE US FACTORS OF 78.
SO WE HAVE +/-26, AND THEN 3 x 13 WOULD BE 39,
SO WE ALSO HAVE FACTORS +/-39.
THEN, OF COURSE, +/-78.
THE ONLY FACTORS OF 1 ARE +/-1.
NOW, IF WE FORM A RATIO OF THESE FACTORS,
SINCE OUR DENOMINATOR ONLY CONTAINS THE FACTORS OF +/-1,
THE POSSIBLE RATIONAL ZEROS
WOULD JUST BE TE FACTORS OF - 78.
SO HERE'S THE LIST OF POSSIBLE RATIONAL ZEROS,
AGAIN, BECAUSE THE LEADING COEFFICIENT IS JUST ONE,
THIS LEAVES US WITH THE FACTORS OF -78.
NOW, AT THIS POINT WE COULD USE TRIAL AND ERROR
AND SUBSTITUTE THESE VALUES INTO THE FUNCTION
TO SEE WHICH WOULD MAKE THE FUNCTION EQUAL TO ZERO,
BUT THAT WOULD TAKE A LOT TIME.
I LIKE TO USE THE GRAPH OF THE FUNCTION
TO FIND THE REAL RATIONAL ZEROS.
AND BECAUSE WE'RE LOOKING FOR THE X VALUES
THAT MAKE THE FUNCTION OF Y EQUAL TO ZERO,
THE X INTERCEPTS OF THIS FUNCTION
WILL GIVE US THE REAL ZEROS,
AND THEN FROM THERE WE CAN FIND WHICH ONES ARE RATIONAL.
SO IF WE GRAPH THIS FUNCTION, IT'S GOING TO LOOK LIKE THIS,
AND THE FIRST THING WE SHOULD NOTICE IS,
THERE'S ONLY X INTERCEPT
WHICH MEANS THERE'S ONLY ONE REAL ZERO
AND NOTICE OUR X INTERCEPT IS +6,
THEREFORE, +6 IS A REAL RATIONAL ZERO
OF THE GIVEN FUNCTION.
SO LET'S GO AHEAD AND WRITE THAT DOWN.
WE KNOW WE'RE GOING TO HAVE THREE ZEROS,
AND WE NOW KNOW ONE OF THEM IS X = 6.
NOW, I SAY THREE ZEROS
BECAUSE IT'S A DEGREE THREE POLYNOMIAL.
WE KNOW IT WILL HAVE THREE ZEROS
AS LONG AS WE COUNT MULTIPLICITY
BECAUSE IT'S A DEGREE THREE POLYNOMIAL FUNCTION.
NOW, THE NEXT CONNECTION WE NEED TO MAKE
IS THAT IF X = 6 IS A ZERO OF THIS FUNCTION,
THEN X - 6 WOULD HAVE TO BE A FACTOR OF THE FUNCTION.
SO WE CAN REWRITE THE ORIGINAL FUNCTION
AS F OF X = THE FACTOR OF X - 6 x ANOTHER DEGREE TWO FACTOR
WHICH WE CAN FIND BY DIVIDING THE ORIGINAL POLYNOMIAL
BY X - 6,
OR WE CAN PERFORM SYNTHETIC DIVISION USING X = 6.
LET'S GO AHEAD AND USE SYNTHETIC DIVISION.
SO WE'LL HAVE A 6 OVER HERE IN THE CORNER.
WE'RE GONNA LIST THE COEFFICIENTS
OF THE ORIGINAL POLYNOMIAL FUNCTION
WHICH ARE 1, -12, 49, AND -78.
NOW, BECAUSE WE KNOW 6 IS A ZERO OF THE FUNCTION,
OR X - 6 IS A FACTOR, THE REMAINDER HERE SHOULD BE ZERO.
LET'S GO AHEAD AND GIVE IT A TRY.
WE BRING DOWN THE 1 AND START MULTIPLYING BY 6.
1 x 6 IS 6, AND WE ADD -6.
-6 x 6 IS -36, AND WE ADD.
THIS WOULD BE POSITIVE 13.
13 x 6 IS 78.
THIS IS GOOD NEWS.
NOTICE OUR REMAINDER IS ZERO,
AND THIS QUOTIENT ALSO GIVES US OUR DEGREE TWO FACTOR
OF THE ORIGINAL FUNCTION.
THIS 1 REPRESENTS THE COEFFICIENT
OF THE X SQUARED TERM,
SO EITHER FACTOR WOULD BE X SQUARED = 6X + 13.
AND AGAIN, WE'RE LOOKING FOR THE X VALUES
THAT MAKE THIS FUNCTION EQUAL TO ZERO.
WELL, WE ALREADY FOUND THE ZERO FROM THIS FACTOR.
NOW, WE WANT TO FIND THE ZERO FROM THIS QUADRATIC FACTOR.
NOW, LET'S GO AHEAD AND FIND THE ZEROS OF THIS
ON THE NEXT SLIDE.
SO WE'RE GOING TO SET X SQUARED = 6X + 13
EQUAL TO ZERO.
UNFORTUNATELY, THIS IS NOT FACTORABLE, AND WE KNOW THAT
BECAUSE WE ONLY HAVE ONE REAL RATIONAL ZERO.
SO TO FIND THE ZEROS HERE,
WE'LL HAVE TO USE THE QUADRATIC FORMULA,
WHICH I PUT UP HERE JUST IN CASE FOR REVIEW.
SO WE SHOULD RECOGNIZE THAT A IS EQUAL TO 1,
B IS EQUAL TO -6, AND C IS EQUAL TO +13.
SO AGAIN, APPLYING THE QUADRATIC FORMULA,
WE'RE GOING TO HAVE
X = -(-6) +/- THE SQUARE ROOT OF B SQUARED = 4AC,
-6 SQUARED - 4 x 1 x 13 ALL OVER 2 x A AND A IS 1.
SO NOW LET'S GO AHEAD AND SIMPLIFY THIS.
WE'RE GOING TO HAVE +6 +/- THE SQUARE ROOT OF --
WE NEED TO BE CAREFUL
SIMPLIFYING UNDERNEATH THE SQUARE ROOT.
THIS NUMBER HERE IS CALLED THE DISCRIMINANT.
IT WOULD BE +36 - THIS WOULD BE 52.
WELL, 36 - 52 IS EQUAL TO -16.
SO THE NUMBER UNDERNEATH THE SQUARE ROOT HERE IS -16,
AND THE DENOMINATOR IS JUST 2.
NOW, LET'S CONTINUE SIMPLIFYING.
WE HAVE 6 +/- --WELL, THE SQUARE ROOT OF -16
IS EQUAL TO THE SQUARE ROOT OF -1 x 16.
THE SQUARE ROOT OF -1 IS EQUAL TO I.
THE SQUARE ROOT OF 16 IS 4.
SO WE HAVE 6 +/- 4I DIVIDED BY 2.
NOW, THIS DOES SIMPLIFY, BUT WE DO NEED TO BE CAREFUL HERE.
WE CANNOT JUST SIMPLIFY THE 6 AND THE 2 HERE,
BECAUSE WE CAN SIMPLIFY ACROSS ADDITION OF SUBTRACTION.
SO MY SUGGESTION HERE TO AVOID AN ERROR
IS TO BREAK THIS UP INTO TWO FRACTIONS.
WE HAVE 6 DIVIDED BY 2 +/- 4I DIVIDED BY 2.
WELL, 6 DIVIDED BY 2 IS 3,
AND, OF COURSE, 4 DIVIDED BY 2 IS 2, SO WE HAVE 3 +/- 2I.
NOTICE HOW THE TWO REMAINING ZEROS ARE COMPLEX ZEROS.
AND GOING BACK TO OUR PREVIOUS SLIDE,
AGAIN, WE KNOW ONE ZERO IS X = 6,
WE KNOW ANOTHER ZERO IS NOT X = 3 + 2I,
AND THE OTHER ZERO IS X = 3 - 2I,
WHICH AGAIN, WE COULD HAVE WRITTEN AS JUST 3 +/- 2I
AS WE HAD ON THE PREVIOUS SLIDE.
SO THESE WOULD BE THE THREE ZEROS
TO BE GIVEN DEGREE THREE POLYNOMIAL FUNCTION.
OKAY. I HOPE THIS HELPS.