2.5 Zeros of Polynomial Functions

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Presentation transcript:

2.5 Zeros of Polynomial Functions

If the f(x) has any rational zeros they must be on this list Rational Zeros Theorem: Let f be a polynomial function of degree 1 or higher of the form where each coefficient is an integer. If p/q in lowest term is a factor of f then p is a factor of a0 and q is a factor an ? If the f(x) has any rational zeros they must be on this list The integer factors of -7 are p The integer factors of 3 are q f(x) may not have any rational zeros and can have at most 3 because it is 3rd degree so not all the numbers on the list can be zeros If I make a list of all the possible p/q’s it will be

There is actually one rational zero 7/3 The other two zeros are irrational 7/3

Make a list of the possible rational zeros according to the rational zeros theorem Reduce the fractions and get rid of doubles

List the possible rational zeros Find the zeros of The equation is 3rd degree so it can have at most 3 zeros List the possible rational zeros Graph and find one of the numbers on the rational zeros list where the graph the graph crosses the x-axis Remember the calculator makes rounding errors 0 on the calculator comes up as 1.23 E -12 You can also use to find x-intercepts. If you don’t have a graphing calculator you can plug the values in until one equals zero or synthetically divide them until you get a remainder of zero.

1 5 -1 -2 2 -5 Find the zeros of List the possible rational zeros The equation is 3rd degree so it can have at most 3 zeros List the possible rational zeros Graph and find one of the numbers on the rational zeros list where the graph the graph crosses the x-axis Y-max= 10 The graph appears to cross at x = -1 so x+1 must be a factor. Use synthetic division to divide . X-max= 4 X-min= -5 1 5 -1 -2 2 -5 Y-min= -10 The zeros are Use the quadratic formula to solve the quadratic

List the possible rational zeros The equation is 3rd degree so it can have at most 3 zeros Find the zeros of List the possible rational zeros Graph and find one of the numbers on the rational zeros list where the graph the graph crosses the x-axis

1 2 -12 20 -6 10 Find the zeros of List the possible rational zeros The equation is 3rd degree so it can have at most 3 zeros Find the zeros of List the possible rational zeros Graph and find one of the numbers on the rational zeros list where the graph the graph crosses the x-axis y-max= 12 The graph appears to only cross at x = 2 so x-2 must be a factor. Use synthetic division to divide . 1 X-min= -1 X-max= 6 2 -12 20 -6 10 y-min= -12 The zeros are Since two of the zeros are imaginary the graph only crosses the x-axis once. Use the quadratic formula to solve the quadratic

3 2 -4 -6 The graph looks like it may cross at 2/3 ,-4/3 & 4/3 Find the zeros. List the possible rational zeros. The graph looks like it may cross at 2/3 ,-4/3 & 4/3 Using the trace feature it only crosses at 2/3 3 2 -4 -6 The zeros are

Solve over the complex number system: The graph only crosses the x-axis once so there is one real root and two complex roots that are not real 13/7 Lets make the list of possible rational roots

Solve over the complex number system: 13/7 7 -20 20 -13 13 -13 13 7 -7 7 0 13/7 The solutions are

Solve for exact solutions: P 356 #3, 5, 8 List possible rational zeros #9-15 odd List possible rational zeros Use the list and the graph to find an actual zero Use synthetic division to factor the polynomial Factor or use the quadratic formula on the remaining quadratic to find the other zeros. #19 Solve for exact solutions:

Find an exact value for the zeros of the polynomial function There can be at most 4 real zeros because f(x) is 4th degree The list of possible rational zeros is Y-max= 20 Next graph the equation 3 -1 X-min= -6 X-max= 6 Hit trace and plug in each number in the list of possible rational zeros to find which are really zeros 3 and -1 are the only rational zeros the other two are irrational Y-min= -60

Find the zeros 3 and -1 are the only rational zeros the other two are irrational Factor the equation by using synthetic division 1 1 -14 1 15 3 12 -6 -15 -1 1 4 -2 -5 0 Now divide -1 into the 3rd degree polynomial -1 -3 5 1 3 -5 0

-1 is a zero 3 is a zero The other two irrational zeros come from this factor. To find where it equals zero use the quadratic formula Solve using the quadratic formula Use trace on your calculator to make sure your zeros are correct The exact real zeros are

The graph crosses at 5 & -5 1 -1 -29 25 100 5 20 -45 -100 1 4 -9 -20 0 Find the zeros List possible rational zeros The graph crosses at 5 & -5 Y-max= 160 1 -1 -29 25 100 5 20 -45 -100 1 4 -9 -20 0 X-max= 10 X-min= -10 -5 -5 5 20 1 -1 -4 0 Y-min= -160

Find the zeros

Solve over the complex number system: List the possible rational roots The graph appears to touch at -3 therefore it must be a zero of even multiplicity 2

Solve over the complex number system: Since the graph just touches at -3 it is a zero of even multiplicity and the factor (x+3) occurs more than once so we will need to do synthetic division with it at least twice. -3 1 8 23 30 18 -3 -15 -24 -18 1 5 8 6 0 -3 -3 -6 -6 1 2 2 0

There are roots are -3, -1+i, -1-i Solve over the complex number system. Use the quadratic formula There are roots are -3, -1+i, -1-i You can test these by plugging them back in the equation on your calculator

Find a second degree polynomial that has zeros of 3 and -4, in addition f(0)=24

Find a third degree polynomial that has zeros of 2, -3 and 1, in addition f(3)=12

Find a third degree polynomial that has zeros of 2i and 1, in addition f(2)=24 If 2i is a zero its conjugate must also be a zero

If 2+i is a zero 2-i is also a zero Find a 3rd degree polynomial f with real coefficients with zeros -5 & 2+i where f(2)=14 If 2+i is a zero 2-i is also a zero

P 357 22-24 25, 27, 42, 46 List possible rational roots Use the list and the graph to find an actual root Use synthetic division to factor the polynomial Factor or use the quadratic formula on the remaining quadratic to find the other roots.