Transcript 1/3, and

Plowing Through Sec. 2.4b
with Two New Topics:
Synthetic Division
Rational Zeros Theorem
Synthetic Division
Synthetic Division is a shortcut method for the
division of a polynomial by a linear divisor, x – k.
Notes:
This technique works only when dividing by a
linear polynomial…
It is essentially a “collapsed” version of the long
division we practiced last class…
Synthetic Division – Examples:
2 x  3x  5 x  12
3
Evaluate the quotient:
Zero of
divisor:
2
x 3
Coefficients of dividend:
3
2
–3
6
2
3
–5 –12
9
12
4
0
Remainder
Quotient
2 x  3x  5 x  12
3
2
x 3
 2 x  3x  4
2
Synthetic Division – Examples:
4
2
Divide x  2 x  3 x  3 by x  2 and write a
summary statement in fraction form.
–2
1
0 –2
–2 4
3 –3
–4 2
1
–2
–1 –1
2
x  2 x  3x  3
1
3
2
 x  2x  2x 1 
x2
x2
4
2
Verify Graphically?
Rational Zeros Theorem
Real zeros of polynomial functions are either rational
zeros or irrational zeros. Examples:
f  x   4 x  9   2 x  3 2 x  3
2
The function has rational zeros –3/2 and 3/2


f  x  x  2  x  2 x  2
2
The function has irrational zeros – 2 and

2
Rational Zeros Theorem
Suppose f is a polynomial function of degree n > 1 of the form
f  x   an x  an1 x
n
n 1

 a0
with every coefficient an integer and a0  0. If x = p /q is
a rational zero of f, where p and q have no common integer
factors other than 1, then
• p is an integer factor of the constant coefficient
• q is an integer factor of the leading coefficient
a0 , and
an .
RZT – Examples:
Find the rational zeros of
f  x   x  3x  1
3
2
The leading and constant coefficients are both 1!!!
 The only possible rational zeros are 1 and –1…check them out:
f 1  1  3 1  1  1  0
3
2
f  1   1  3  1  1  3  0
3
2
So f has no rational zeros!!!
(verify graphically?)
RZT – Examples:
3
2
Find the rational zeros of f  x   3x  4 x  5x  2
Potential Rational Zeros:
Factors of –2
Factors of 3
1, 2
1, 3
1 2
1, 2,  , 
3 3
Graph the function to narrow the search…
Good candidates: 1, – 2, possibly –1/3 or –2/3
Begin checking these zeros, using synthetic division…
RZT – Examples:
3
2
Find the rational zeros of f  x   3x  4 x  5x  2
1
3
3
4 –5 –2
3
7
2
7
2
0
Because the remainder is zero,
x – 1 is a factor of f(x)!!!
f  x    3x  7 x  2   x  1
2
Now, factor the remaining quadratic…
f  x    3x  1 x  2 x 1
The rational zeros are 1, –1/3, and –2
RZT – Examples:
Find the polynomial function with leading coefficient 2 that has
degree 3, with –1, 3, and –5 as zeros.
First, write the polynomial in factored form:
f  x   2  x  1 x  3 x  5
Then expand into standard form:
f  x   2  x  2 x  3  x  5 
2
 2  x  5 x  2 x  10 x  3 x  15 
3
2
2
 2 x  6 x  26 x  30
3
2
RZT – Examples:
Using only algebraic methods, find the cubic function with the
given table of values. Check with a calculator graph.
x
–2
–1
1
5
f(x)
0
24
0
0
(x + 2), (x – 1), and (x – 5)
must be factors…
f  x   k  x  2 x 1 x  5
f  1  24 :
k  1  2 1 1 1  5  24  k  2
But we also have
f  x   2  x  2 x 1 x  5
 2  x  x  2   x  5   2 x3  8 x 2  14 x  20
2
A New Use for Synthetic
Division in Sec. 2.4:
Upper and
Lower Bounds
What are they???
A number k is an upper bound for the real zeros of
f if f (x) is never zero when x is greater than k.
A number k is a lower bound for the real zeros of
f if f (x) is never zero when x is less than k.
What are they???
Let’s see them graphically:
y  f  x
c
d
c is a lower bound
and d is an upper
bound for the real
zeros of f
Upper and Lower Bound Tests
for Real Zeros
Let f be a polynomial function of degree n > 1 with a positive
leading coefficient. Suppose f (x) is divided by x – k using
synthetic division.
• If k > 0 and every number in the last line is nonnegative
(positive or zero), then k is an upper bound for the real
zeros of f.
• If k < 0 and the numbers in the last line are alternately
nonnegative and nonpositive, then k is a lower bound
for the real zeros of f.
Cool Practice Problems!!!
Prove that all of the real zeros of the given function must lie in
the interval [–2, 5].
f  x   2x  7 x  8x  14x  8
4
3
2
The function has a positive leading coefficient, so we employ
our new test with –2 and 5:
5
2 –7
–8
14
10
15
35 245
3
7
49 253
2
8
This last line is all positive!!!
 5 is an upper bound
–2
2 –7
–4
–8
14
8
22 –28 28
2 –11 14 –14 36
This last line has
alternating signs!!!
 –2 is a lower bound
Let’s check these results graphically…
Cool Practice Problems!!!
Find all of the real zeros of the given function.
f  x   2x  7 x  8x  14x  8
4
3
2
From the last example, we know that all of the rational zeros
must lie on the interval [–2, 5].
Next, use the Rational Zero Theorem…potential rational zeros:
Factors of 8
Factors of 2
1
1, 2, 4, 8
1, 2, 4, 8, 
2
1, 2
Look at the graph to find likely candidates: Let’s try 4 and –1/2
Cool Practice Problems!!!
Find all of the real zeros of the given function.
f  x   2x  7 x  8x  14x  8
4
4
2 –7
8
–8 14 8
4 –16 –8
2
–4
1
–2
3
2
0
f  x    x  4  2x  x  4x  2
3
–1/2
2
1
–1
–4
0
–2
2
2
0
–4
0
2
1

2
f  x    x  4  x    2x  4
2

Cool Practice Problems!!!
Find all of the real zeros of the given function.
f  x   2x  7 x  8x  14x  8
4
3
2
1

2
f  x    x  4  x    2x  4
2

1 2

f  x   2  x  4  x   x  2
2

1

f  x   2  x  4  x   x  2 x  2
2






The zeros of f are the rational numbers 4 and –1/2 and the
irrational numbers are – 2 and 2
Cool Practice Problems!!!
Prove that all of the real zeros of the given function lie in the
5
2
interval [0, 1], and find them.
f  x   10 x  3x  x  6
Check our potential bounds:
0 10 0
0
0
0
–3
0
1
0
–6
0
10 0
0
–3
1
–6
 0 is a lower bound!!!
1 10 0 0 –3
10 10 10
10 10 10
7
1
7
–6
8
8
2
 1 is an upper bound!!!
Cool Practice Problems!!!
Prove that all of the real zeros of the given function lie in the
5
2
interval [0, 1], and find them.
f  x   10 x  3x  x  6
Possible rational zeros:
1 3 1 2 3 6
1
3
1, 2, 3, 6,  ,  ,  ,  ,  ,  ,  , 
2 2 5 5 5 5 10 10
Check the graph (with 0 < x < 1) to select likely candidates…
The function has no rational zeros on the interval!!!
Are there any zeros??? Lone Real Zero: 0.951