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MATH 1005 – College Algebra
Quiz 7B-10 Points
Name: Key
Please read each question carefully, follow directions and clearly mark your solutions. Remember to
show work for full credit.
1. Determine whether the second polynomial is a factor of the first polynomial without dividing or using
synthetic division.
Solution: We want to figure out if
is a factor of
. We could do this by using long
division or synthetic division. If we chose one of one of those ways, we would know that
was a
factor if we got a remainder of zero. However, this problem tells us NOT to use dividing or synthetic
division. We need to use the Factor Theorem that says, if r is a zero of the polynomial
, then
is
a factor of
. Conversely, if
is a factor of
, then r is a zero of
Thus, we need to figure out if
. If it does, r is a factor. If not, r isn’t a factor.
2. Use the graph of P(x) to write the solution set for the following inequalities.
(A)
4
Solution: Here we need to remember what
means.
2
, we put in an x value, input, into the function. We take
of the input, and we end up with and we get
2
1
an output, y. Therefore, we are trying to find when the
outputs of the function are greater than or equal to zero.
First, we should find where
.
at x=-2, x=-1, x=0, x=1.
Now, list the intervals of the x values that give you a positive y-value.
which is the solution set for the inequality.
(B)
Solution: Here we are trying to find when the
outputs of the function are less than zero.
1
2
4
2
This time the zeros of
are not included. Those x-values are x=-2, x=-1, x=0, x=1.
Now, list the intervals of the x values that give you a negative y-value.
which is the solution set for the inequality.
3. Find the polynomial of lowest degree, with leading coefficient 1, for the following graph. Assume zeros
4
are integers. Write the polynomial as a product of linear factors.
Solution: Again, a first good step would be to list all the zeros of the function.
are the roots of this function.
So we have (in
2
form) the factors
2
1
1
2
All that is left to do is to find the multiplicity of each linear factor.
4
By looking at the graph, we see that x=-2 has odd multiplicity and so does x=0
6
because the function crosses over the x-axis (the y –values go from positive to negative) .
On the other hand, x=-1 is even because it doesn’t cross over the x-axis; the function touches the root and
stays on the same side of the x-axis. Thus,
.
4. Find the smallest positive integer and largest negative integer that, by Theorem 1 in 4.2, are upper and
lower bounds, respectively, for the real zeros of P(x).
Finding the upper bound, we do synthetic division with
until all the coefficients end up positive.
Here we find our upper bound to be 1.
Finding the lower bound, we do synthetic division with
until the coefficients alternate signs.
This tells us -3 is our lower bound.
2