CHAPTER 4 “Quiz” " Answers (if you r questions aren’t here, they are vey similar)
1.
P(x) = x6 + x4 + x2+ 1 must have six roots (zeroes).
None of the roots of P(x) = x6 + x4 + x2+ 1 can be positive real numbers.
Why no positive real roots:
There are no sign changes in this polynomial.
According to Descartes’ Rule of Signs, there can be no positive roots.
(This makes obvious sense: how can a sum of positives equal zero?)
Why no negative real roots:
P cannot have any negative real roots, either because
P(–x) = (–x)6 + (–x)4 + (–x)2+ 1 = x6 + x4 + x2+ 1
... no sign changes, so no positive roots for P(–x), so no negative roots for P(x).
2.
/5
List all theoretically possible rational roots of the polynomial:
4x4 + 20x3 + 23x2 — 5x — 6
Since this polynomial has all coefficients in the set of integers,
any rational roots must be of the form p/q
where p is a factor of the constant term (so p = ± 6, 3, 2 or 1) and
q is a factor of the leading coefficient (so q = ± 4, 2, or 1).
Thus candidates for rational roots of this polynomial are:
±6
±3
±2
±1
± 3/2
±1/2
± 3/4
± 6/2 & ± 2/2 are redundant!
3.
/7
±1/4
...likewise for ± 6/4 & ± 2/4
Use polynomial long division to fill in the blanks with polynomials of degree < 2:
6x4 + 17x2 + x
)))))))))))))
=
2x2 + 5
3x2 + 1
x—5
+ )))))))
2x2 +5
This polynomial and fraction
may be added together (easily) to
check the answer.
3x2 + 1
2x2 + 5

))))))))))))))))
4
2
6x + 17 x
+x
4
2
6x + 15 x
2 x2 + x
x2
+5
x—5
Note we could also make the statement:
6x4 + 17x2 + x = (3x2 + 1) ( 2x2 +5 ) + x — 5
Notice it is important to maintain
alignment of like terms; for instance
it is necessary to leave space for the
“missing” x term.
For #4: P(x) = 2x3 + 3x2 +2x – 2
4.
/7
USE synthetic division to locate a rational root of P. (Synthetic division MUST be shown.)
The potential rational roots (explanation÷ see #2) are ±1, ±2, ±½.
1
½
2
3
2
2
5 7 5
2
2
5.
/7
2 –2
5 7
–1
3 2 –2
1 2 2
4
2
3
–2
2 –2
–1 –1
2
1
1 –3
–½ 2
3
2
2
3 2 –2
4 14 32
–2
2 7 16 30
3
-4
2
2
–2
–8
2 –1
4
–10
–2
Given that 3 is a root of P(x) = x 3 — x2 — 4x – 6,
3
1
We divide x3 — x2 — 4x – 6 by x – 3 :
1
3
2
...and find that: ( x — x — 4x – 6 )
=
x =
–2 ± r%%%2%%%!%%%6
))))))))))))))) =
2
( x – 3) ( x + 2x + 2)
–2 ± r%%!%%4
)))))))))
2
Roots of P are 3 and 1 – i
Sketch the graph of y = P(x) =
–1 – 4 –6
3 6 6
2
2 0
2
x2 + 2x + 2 = 0 when ...
We find the remaining roots via the quadratic formula.
6.
/7
2
From this we know ½ is a root, & P(x) = (x – ½ ) (2x2 + 4x + 4)
û
4 0
2
=
–2 ± 2 i
)))))))))
2
=
–1 ± i
and 1 + i
1
2
)) ( x + 2)2 (x — 3)
Polynomial, domain is all of R, continuous, etc.
. . . . . . . . . 18 . . . . . . . .
. . . . . . . . . . . . . . . . .
y-intercept: P(0) = ½ (0+2) (0 – 3) = –6
. . . . . . . . . . . . . . . . .
x-intercept(s): P(x) = 0
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
only when
( x + 2)2 (x – 3) = 0
. . . when
x = – 2 (twice), x = 3
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
What happens to y=P(x) when x ÷ 4
. . . . . . . . . . . . . . . . .
(increases towards 4) ? Consider that
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
(x + 2)2 and (x — 3) both grow unboundedly
-2
3
large, both positive, as x ÷ 4. So
. . . . . . . . . . . . . . . . .
2
( ½ ) (x + 2) (x — 3) ÷ 4 as x ÷ 4.
. . . . . . . . . . . . . . . . .
(i.e. grows unboundedly large)
. . . . . . . . .— 6 . . . . . . . .
. . . . . . . . . . . . . . . . .
2
. . . . . . . . . . . . . . . . .
... And when x ÷ -4, (x + 2) & (x — 3) both
grow unboundedly large, but (x—3) is large
. . . . . . . . . . . . . . . . .
negative, so the product is large & negative. . . . . . . . . . . . . . . . . .
( ½ ) (x + 2)2 (x — 3) ÷ -4 as x ÷ -4.
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
Plotting a few additional points, such as
. . . . . . . . . . . . . . . . .
(–5,–36) & (–4,–14) & (1, –9) & (2, –8) & (4,18) adds accuracy & serves as a check on the analysis.
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7.
4 — 2x
Sketch the graph of y = ))))))) Label all the intercepts & asymptotes.
3—x
/10
Rational function, therefore:
Domain: y is defined for all values
except where denominator = 0: x = 3
As x approaches 3, 4 — 2x ÿ –2,
while 3 — x ÿ 0,
so the quotient ÿ ± 4
Thus there is a vertical
asymptote at x = 3.**
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y=3 .
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As x ÿ 4 y ÿ 3
.
As x ÿ ! 4 y ÿ 3
.
Thus the function has a
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horizontal asymptoteu at y = 3.
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y-intercept: when x = 0
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y = 4/3
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x-intercept: when x = 2, y = 0
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** More on the Vertical Asymptote:
As x approaches 3 from below
e.g. take x = 2.5 ,
y grows unboundedly large, negative:
y = –1 ,
.5
2.8 , 2.9 , 2.99 , 2.9999 . . . ÷ 3
–1.6 , –1.8 , –1.98 , –1.9998
– .2
– .1
– .01 – .0001 ÷ 4
Similarly, computing function values for x that are just above 3,
demonstrates that the function values are positive and grow unboundedly large
(÷ 4) as x-values approach 3 from above.
u Another way to perceive the horizontal asymptote:
...divide & conquer:
–2x + 4
–x+3
)))))))
=
2x – 4
x– 3
)))))))
=
2
2 + )))))))
x—3
& f(x) ÿ 2
+ 0
As x ÿ
as xÿ
4
, the latter portion ÿ 0
4
And on the NEXT page... you can see the questions I didn’t ask....
CHAPTER 4 QUESTIONS omitted from the QUIZ
1.
2.
Which of the following statements are true? (If true, circle the letter.)
(In each statement, assume any function called “P” or “Q” is a polynomial function.)
a.
If P(4) > 0 and P(5) < 0, then P (r) must be 0 for some number r between 4 and 5.
b.
If 6 is a zero of P, then (x – 6) is a factor of P(x).
c.
If 6 is a zero of P, then P(x) = (x – 6)Q(x) for some polynomial Q.
d.
If P(x) = (x – 6)Q(x) + 2, for some polynomial Q, then P(6) = 2.
e.
If P is of degree 7, then P has 7 roots, taking into account multiplicities.
f.
If (x – 6)2 divides P(x), but (x – 6) does not, then we say: “6 is a zero of multiplicity 2 for P”, and we
count 6 as two of the zeroes (roots) of P.
g.
If P(x) = x3 — 2x2 — 8x
For #2-4, P(x) = 2x3 – 5x2 – 9
Complete the last step of this synthetic division, P ÷ D, where D(x) = x + 3
–3
3.
then the roots of P are 4 and —2.
2
–5
–6
2 –11
0
33
–9
What does this tell you about P(–3)?
Complete this synthetic division, for P(x) ÷ (x – 3).
3
2
–5
0
–9
4.
Find all the roots of P. List them here:
5.
How can we know that P(x) = x4 + 5x2 + 7 has no real roots, without a lot of work?
6.
Look out below:
List ALL the theoretically possible rational roots of P(x) = 2x3 – ½ x2 – 32 x + 8 :
7.
Find all the roots of the polynomial given in #6. They are:
Any surprises?
8.
List all the theoretically possible rational roots of P(x) = 2x3 – 5 x2 – 3x :
9.
Find all the roots of P(x) = 2x3 – 5 x2 – 3x . List them here:
10.
Find all the roots of P(x) = x4 – x2 – 20. List them here:
11.
Why does the function:
12.
Sketch* the graph of
f(x) =
3x + 5
y = )))))) .
2 – x
x2 – 4x + 4
NOT have a vertical asymptote at x = 2 ?
2x – 8
))))))))))
2
Label all the intercepts & asymptotes.
CHAPTER 4 QUIZ— omitted questions— THE ANSWERS
1.
2.
Which of the following statements are true? (In each, any function called “P” is a polynomial function.)
a.
If P(4) > 0 and P(5) < 0, then P (r) must be 0 for some number r between 4 and 5. TRUE
b.
If 3 is a zero of P, then (x – 3) is a factor of P(x).
c.
If 3 is a zero of P, then P(x) = (x – 3)Q(x) for some polynomial Q. TRUE: r root ± (x – r) factor
d.
If (x – 3)2 divides P(x), then we say: “3 is a zero of multiplicity 2 for P”, and we count 3 as two of the
zeroes (roots) of P. TRUE : (x – r) factor ± r root
e.
If P is of degree 7, then P has 7 roots (taking into account multiplicities). TRUE
f.
If P(x) = (x – 3)Q(x) + 2, for some polynomial Q, then P(3) = 2. TRUE (try computing P(3)!)
g.
If P(x) = x3 – 2x2 – 4x + 8 = (x – 2) (x2 – 4), then the roots of P are 2 & 2 & –2. TRUE
For #2-4, P(x) = 2x3 – 5x2 – 9
Complete the last step of this synthetic division, P ÷D, where D(x) = x + 3
–3
3.
TRUE: r root ± (x – r) factor
2
–5
–6
2 –11
0
33
33
–9
What does this tell you about P(–3)? ...that it is –108, and NOT 0
-99
( So –3 is not a root of P.)
-108 ... NOT 0 !
Complete this synthetic division, for P(x) ÷ (x – 3).
3
2
2
–5
6
1
0
3
3
–9
9
0
This tells us that P(x) = (x – 3)(2x2 + x + 3)
Roots are:
3
3
& x = — 1 ± r%%1%%%!%2%%%4
))))))))))))))
4
— 1 ± r%%2%%3 i
))))))))))
4
4.
Find all the roots of P. List them here:
5.
How can we know that P(x) = x4 + 5x2 + 7 has no real roots, without a lot of work?
P(x) has 0 sign changes, so, by Descartes’ Rule of Signs: P(x) has at most 0 real positive roots.
P(–x) = x4 + 5x2 + 7 also (since all the powers are even)
P(–x) has 0 sign changes, so, by Descartes’ Rule of Signs: P(x) has at most 0 real negative roots.
6.
List all the theoretically possible rational roots of P(x) = 2x3 – ½ x2 – 32 x + 8
±16, 8, 4, 2, 1
))))))))))
3
2
=( ½ )(4x – x – 64 x + 16)
4, 2, 1
(THE RATIONAL ZEROES THEOREM REQUIRES INTEGER COEFFICIENTS .)
7.
± 1,2,4,8,16, ½, ¼
Find all the roots of the polynomial given in #6 They are: ¼ & ± 4
Any surprises?
8.
List all the theoretically possible rational roots of P(x) = 2x3 – 5 x2 – 3x : 0, ±3, 1
)))) ¸ ±3, 1, ½ , 3/2
= x ( 2x2 – 5 x – 3 )
1, 2
(THE RATIONAL ZEROES THEOREM REQUIRES NON -ZERO CONSTANT TERM .
9.
Find all the roots of P(x) = 2x3 – 5 x2 – 3x . List them here: 0, – ½ , 3
Use synthetic division, test the
rational candidates
10.
Find all the roots of P(x) = x4 – x2 – 20. List them here: ±r 5 , ±2i
<Factoring is the way to do it>
11.
x2 – 4x + 4
NOT have a vertical asymptote at x = 2 ?
2x – 8
Why does the function:
f(x) =
))))))))))
2
...BECAUSE...
f(x) =
))))))))))
:
( x – 2) 2
=
2(x+2)(x – 2)
x–2
2 (x + 2)
)))))))
...a function that has no
vertical asymptote at x = 2.
These two functions are identical, except that f is undefined at x = 2.
12.
Sketch* the graph of
3x + 5
y = )))))) .
2 – x
Label all the intercepts & asymptotes.
x=2
(0,5/2)
(–5/3,0)
y= —3
Provided x is
NOT 2