Activity 2: Practice with Polynomial Division and Zeros
Objectives:
 Understand the Fundamental Theorem of Algebra
 Understand that if a value, c, is a zero of a polynomial, the remainder of
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the polynomial divided by x-c is 0 and thus x – c is a factor of the
polynomial
Understand how to use synthetic division to further factor a polynomial.
Understand the Rational Zeros Theorem
Understand that if a polynomial has real coefficients non-real zeros must
also have their conjugates also as zeros.
Ability to write the factored form of a polynomial knowing its zeros.
Ability to sketch general graph of polynomial knowing its zeros and
leading coefficient. (Assume all real zeros for this activity)
Use quadratic formula to find zeros of a quadratic.
Understand how one can see the multiplicity in the graph
Understand how to find the degree of a polynomial even if in factored
form and then determine the end behavior of the graph.
Examples:
o Given f(x) = 2x4-3x3 –31x2 +48x –16 we will find all zeros and sketch the graph.
By the Fundamental Theorem of Algebra and the connected theorems we know in the
set of Complex Numbers there are 4 zeros counting multiplicities. Notice all of the
coefficients are integers. The Rational Zeros Theorem states that any rational zeros will
be among the list of ratios
1 2 4 8 16 1 2 4 8
16
 ,  ,  ,  ,  ,  ,  ,  ,  , and 
This list reduces to
1 1 1 1
1
2 2 2 2
2
1
 ,  1,  2,  4,  8, and  16 . Start at the beginning of the list and begin substituting the
2
values in for x to see if we obtain a zero.
4
3
2
1
1
1
1
 2   3   31   48 
2   3   31   48   16 =             16 
2
2
2
2
 16   8   4   2 
2 6 124 384
256
 

 16 
 16  16  16  0 . Right away we found a zero. If we
16 16 16 16
16
now divide f(x) by x – ½ we have two simpler factors of f(x) with which to find the
remaining zeros.
1
2 x 3  2 x 2  32 x  32
x  1 2 2 x 4  3x 3  31x 2  48x  16
 (2 x 4  x 3 )
f(x) = ( x – ½ ) (2x3 –2x2 –32x +32)
We already know x = ½ is a zero.
 2 x  31x
 (2 x 3  x 2 )
3
We can use (2x3 –2x2 –32x +32)
to find the remaining zeros of f(x).
2
 32 x 2  48 x
 (32x 2  16x)
32x 16
 32 x  16
0
3
2
2(-1) –2(-1) –32(-1)+32 = -2 –2 +32 + 32 = 60. Thus, x = -1 is not a zero.
But, 2(1)3-2(1)2-32(1)+32 = 2 – 2 – 32 + 32 = 0. Thus, with the simpler polynomial
factor we were able to find a second zero, x = 1.
Now divide 2x3 –2x2 –32x +32 by x –1.
2 x 2  0 x  32
x  1 2 x 3  2 x 2  32 x  32 We now know that f(x) = (x – ½ )(x – 1)(2x2 – 32) =

 2x 3  2x 2

0  32 x
 (0  0 x )
2(x2 –16)(x – ½)(x – 1)=
2(x – 4)(x + 4)(x – ½ )(x – 1)
The other two zeros are x = 4 and x = -4.
 32x  32
 32 x  32
0
Each of the four zeros has multiplicity 1. The x-intercepts are (0,-4), (0, ½ ), (0, 1), and
(0, 4). The y-intercept is (0, -16). These 5 points are shown on the graph of f(x) below.
o Given f(x) = x4-11x3 +34x2 -6x – 52 we will find all zeros and sketch the graph.
2
By the Fundamental Theorem of Algebra and the connected theorems we know in the
set of Complex Numbers there are 4 zeros counting multiplicities. Notice all of the
coefficients are integers. The Rational Zeros Theorem states that any rational zeros will
be among the list of ratios
1 2 4 13 26 52
 , , , , ,
. Starting at the beginning of the list, test if +1 is a zero.
1 1 1 1
1
1
(1)4 – 11(1)3 +34(1)2 – 6(1) –27 = 1 –11 +34 – 6 – 52 = -34. 1 is not a zero.
Try –1. (-1)4 – 11(-1)3 +34(-1)2 – 6(-1) –52 = 1 +11 + 34 + 6 – 52 = 0. x = -1 is a zero
of the polynomial.
Divide f(x) by x + 1 to obtain two simplified factors with which to find the other zeros.
x 3  12 x 2  46 x  52
x  1 x 4  11x 3  34 x 2  6 x  52 f(x) = ( x +1)( x 3  12 x 2  46 x  52 )
 ( x 4  x 3` )
We know x = -1 is a zero and the other
 12 x 3  34 x 2
 12 x 3 12 x 2

zeros are zeros of x 3  12 x 2  46 x  52 .
Test x = 2. (2)4 – 11(2)3 +34(2)2 – 6(2) –52=

46 x 2  6 x
 46 x  46
16 - 88 +136 –12 – 52 = 0. We were lucky
again. We found another zero right away.
 52x  52 Divide x 3  12 x 2  46 x  52 by x - 2.
  52 x  52 
0
x 2  10 x  26
x  2 x 3  12 x 2  46 x  52

 x 3  2x 2

We can then find the remaining two zeros by
 10 x 2  46 x
  10 x 2  20 x

Now we know that f(x) = (x +1)(x –2)( x 2  10 x  26 )

either factoring x 2  10 x  26 or by using
the quadratic formula. Deciding to use the quadratic
26 x  52 formula
 26 x  52
5
100  4(1)( 26)
 (10)
4
=

 5
2(1)
2(1)
2
2i
 5i.
2
0
Each of the four zeros has multiplicity 1. The two x-intercepts are (0,-1), (0, 2). The two
complex zeros are 5  i and 5  i . The y-intercept is (0, -52). The intercepts are shown on
the graph of f(x) below.
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Supplementary Exercises:
1) Given f(x)= x4 –22x3 +39x2 +14x+120
complete the following.
Follow the examples in the enrichment to
A) Counting multiplicities, how many zeros are there for this polynomial in the set of
complex numbers ? ___________________
B) Following the Rational Zeros Theorem, state the list of potential rational zeros for
f(x). ___________________________
C) Divide x4 –22x3 +39x2 +14x+120 by x – 20 .
The quotient is ____________________. The remainder is _______________.
What does the remainder tell us about the status of x = 20 as a zero of f(x)?
________________
D) Now that you have completed the polynomial division in C), finish filling in the
two factors of f(x). x4 –22x3 +39x2 +14x+120 = (x – 20)(__________________)
E) Using your list of potential zeros in A), find another real zero of f(x) by finding a
zero of the quotient you found in A). Another zero is _______________
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F) Now divide (the quotient from A) by (x – zero you found in E)). This new quotient
from this second division should be quadratic in form. ax2 + bx + c.
G) Use the quadratic formula on the quotient from C) to find the other non-real zeros
of f(x).
non-real zeros are __________________
H) Write f(x) in completely factored form. f(x) = ___________________________
I) Sketch f(x) showing the x-intercepts, the y-intercept and any turning points that it
has. Label the intercepts.
2) Given f(x)= 3x3+9x2-135x-525 and the fact that one of the zeros is x = 7
A) The degree of f(x) is ___________________. The number of complex zeros
including multiplicities will be __________________.
B) Following the ideas in the enrichment and in exercise 1), find all zeros of f(x).
The zeros are ___________________
C) Sketch a graph of f(x) labeling all intercepts.
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3) A) Write a 6th degree polynomial with only real coefficients and leading coefficient -2
that has x = 0, x = 5280, and x = 7 – 5i as zeros. There is more than one correct
answer.
f(x) = ________________________
4) Write a 5th degree polynomial with only integer coefficients that has x = -1, x =
and x = i 2 as zeros. There is more than one correct answer.
2,
f(x) = ___________________________
B) Sketch a graph of your polynomial showing any real zeros and any turning points.
C) State the y-intercept of the polynomial function you obtained. (0, _______)
5) A) Given f(x) = 5.5x2 –11x- 440 use the quadratic formula to find the zeros of this
polynomial.
B) Understanding that the x-coordinate of the vertex is halfway in between the zeros,
find the x-coordinate of the vertex and then plug this number into f(x) to find the ycoordinate of the vertex.
C) Sketch a graph of f(x) showing the zeros and the vertex.
D) State the y-intercept of f(x). (0, __________)
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6) A) Given f(x) = .01(x + 15)2 + 2 Sketch the graph of f(x).
B) From the graph state why you know the graph has only non-real complex zeros.
C) Solve 0 = .01(x + 15)2 + 20 to find these zeros. You could also find the zeros by
multiplying f(x) out into standard or general form and then using the quadratic
formula.
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