Download Help Examples for w10 First of all, let us set a few terms straight. For

Help Examples for w10
First of all, let us set a few terms straight. For a function y = f (x), the x-intercepts
of the function are the solutions to the equation f (x) = 0 (if you think of the points
of the graph that are on the x-axis). The roots of f (x), or the zeroes of f (x), are just
different names for the x-intercepts. In short, they all mean the solutions of the equation
f (x) = 0. Next, let me give an example to explain what the multiplicity of a root means
(remember that a root is just an x-intercept). Say f (x) = 3(x + 2)3 (x − 4)2 (x − 7), then
the roots of f (x) are just x = −2, x = 4 and x = 7. You get them if you set each factor
(x + 2)3 , (x − 4)2 and (x − 7) equal to zero. However, the multiplicity of -2 is 3, the
multiplicity of 4 is 2 and the multiplicity of 7 is 1 (look at the powers of the factors).
1. Find all zeros of the polynomial
P (x) = x3 + x2 − 9x − 9.
Note: if a root has multiplicity more than one, you have to list that root the same
number of times as its multiplicity.
Solution. If you list the possible rational zeros using the rational zero test, then the
list you get is ±{1, 3, 9}. Go through this list and you will find that in this case x = −1
is a zero. By the factor theorem, it means that the polynomial has (x − (−1)) = (x + 1)
as a factor. To find the other factor of P (x), we will simply divide P (x) by (x + 1) using
the synthetic division. You will find the quotient is x2 − 9. Thus P (x) = (x + 1)(x2 − 9).
But x2 − 9 = (x − 3)(x + 3), so P (x) = (x + 1)(x + 3)(x − 3). So the roots of P (x) are
−1, −3 and 3 (each has multiplicity one so you only list them once).
2. It is the same kind as 1. The only difference is that you are asked to find two
rational zeros, which will give you two factors. So you have to perform the synthetic
division twice in order to find the other factor (which is the quotient after you do the
division twice).
3. List all possible rational roots for the function
f (x) = 3x4 − 6x3 − 9x2 − 1x + 57.
Give your list in INCREASING order. Beside each possible rational root, type ”yes”
if it is a root and ”no” if it is not a root. Leave any unnecessary answer blanks empty.
Solution. The factors of 57 are 1, 3, 19 and 57 and the factors of 3 are 1 and 3.
So the list is ±{1, 3, 19, 57, 1/3, 19/3} (3/3 = 1 and 57/3 = 19 repeat so are not listed
again). But in the increasing order, you would have to list them (CAREFUL here):
−57, −19, , −19/3, −3, −1, −1/3, 1/3, 1, 3, 19/3, 19, 57. Plug them in the equation and
you will find that none of these is a real solution so the answer to the second part is no
for each of them.
4. It is like number one. Find one zero first, that gives you a factor, then using
division to find the other factor (which is a quadratic function that cannot be factored
since it has no real zeros).
5. (a) Enter a quadratic polynomial which has zeros at 9 and 2.
(b) Enter a quadratic polynomial which has zeros at -2 and 5.
(c) Enter a quadratic polynomial which touches the x-axis at -10.
Solution. (a) A zero at 9 gives you a factor (x − 9) and a zero at 2 gives you a factor
(x − 2). So you already have (x − 9)(x − 2), which is a quadratic function. You can
enter this as your answer. But anything such as 3(x − 9)(x − 2), −5(x − 9)(x − 2) and
4(x − 9)(x − 2) would all work.
(b) Similar to (a).
(c) Similar to (a). However a touch zero has even multiplicity. So you would have a
factor like (x + 10)2 or (x + 10)4 , and so on. Since you have a quadratic, the power can
only be two in this case though.
7. For the polynomial P (x) = 2x4 + 2x3 − 5x2 − x + 2, find both rational zeros using
the rational root test and synthetic division to both test and factor them out. After
factoring out both rational roots, you should be left with a quadratic. Its roots are the
final two of the four roots of P (x) = 2x4 + 2x3 − 5x2 − x + 2.
Solution. If we list all possible rational zeros, they are {±1, ±2, ±1/2}. Graph the
polynomial on your calculator and you should see easily that 1 and −2 are the only
rational zeros. This means that P (x) = 2x4 + 2x3 − 5x2 − x + 2 has two factors (x − 1)
and (x + 2) by the factor theorem. That is
P (x) = 2x4 + 2x3 − 5x2 − x + 2 = (x − 1)(x + 2)Q(x)
and we need to find out what Q(x) is. If we divide both sides of the above by (x − 1)
first, this is what we get:
2x4 + 2x3 − 5x2 − x + 2
= (x + 2)Q(x).
x−1
The left side is just a simple division and you can get it using the synthetic division
(YOU SHOULD KNOW HOW TO DO IT NOW!). Perform this division and you get
a zero remainder hence the left side is just the quotient, which is 2x3 + 4x2 − x − 2. So
now the equation becomes
2x3 + 4x2 − x − 2 = (x + 2)Q(x).
So what do you do now in order to get Q(x)? You divide both sides of the above by
(x + 2) this time using the synthetic division again:
2x3 + 4x2 − x − 2
= Q(x).
x+2
The remainder should be zero again so you will only have the quotient on the left side,
which is 2x2 − 1. That is, Q(x) = (2x2 − 1). In order
q to get the other zeros, you will
2
2
2
now just set 2x − 1 = 0: 2x = 1, x = 1/2 so x = ± 1/2. That is, the other two zeros
q
are ± 1/2. (Notice that they are not rational numbers.)
14. It is given that a polynomial P (x) of degree 4 has a root of multiplicity 2 at
x = 3 and roots of multiplicity 1 at x = 0 and x = −3. Its lead coefficient is 3. Find a
formula for P (x).
Solution. P (x) = 3x(x + 3)(x − 4)2 .
15. In a way this is like 14 above. The difference is that you have to read the
zeros from the graph. There are two crossing x-intercepts and one touching x-intercept.
These will give you two factors of multiplicity one (remember that the multiplicity of a
factor coming from a crossing x-intercept is odd) and one factor of multiplicity 2 (the
power is even since the x-intercept is a touching point). The multiplicities have to be
1, 1 and 2 since otherwise the degree would be more than 4. Say you see that the zeros
are -3 (crossing), 0 (touching) and 4 (crossing), then the factors with x in there would
be (x + 3), x2 and (x − 4). However the polynomial has a constant factor too (do not
automatically assume that it is 1) so the function looks like y = ax2 (x + 3)(x − 4).
To find out what a is, you need to plug in the additional condition given to you. For
example, if I am given that (5, −400) is a point on the graph of the function, then I
should plug in 5 for x and −400 for y:
−400 = a(52 )(5 + 3)(5 − 4) = a(25)(8)(1) = 200a,
So a = −400/200 = −2 and my function is y = −2x2 (x + 3)(x − 4).