Post on 29-Jan-2016
Plowing Through Sec. 2.4bPlowing Through Sec. 2.4bwith Two New Topics:with Two New Topics:
Synthetic DivisionSynthetic Division
Rational Zeros TheoremRational Zeros Theorem
Synthetic DivisionSynthetic Division
Synthetic DivisionSynthetic Division is a shortcut method for the is a shortcut method for thedivision of a polynomial by a linear divisor, division of a polynomial by a linear divisor, xx – – kk..
Notes:Notes:
This technique works This technique works only only when dividing by awhen dividing by alinear polynomial…linear polynomial…
It is essentially a “collapsed” version of the longIt is essentially a “collapsed” version of the longdivision we practiced last class…division we practiced last class…
Synthetic Division – Examples:Synthetic Division – Examples:3 22 3 5 12x x x
3x Evaluate the quotient:Evaluate the quotient:
33 22 ––33 ––55 ––1212
22 33 44 00
66 99 1212
Coefficients of dividend:Coefficients of dividend:Zero ofZero ofdivisor:divisor:
RemaindeRemainderr
QuotientQuotient3 22 3 5 12x x x
3x 22 3 4x x
Synthetic Division – Examples:Synthetic Division – Examples:4 22 3 3x x x 2x Divide by and write a
summary statement in fraction form.
––22 11 00 ––22 33
4 23 22 3 3 12 2 1
2 2
x x xx x x
x x
––33
11 ––22 22 ––11 ––11
––22 44 ––44 22
Verify Graphically?Verify Graphically?
Rational Zeros TheoremRational Zeros Theorem
Real zeros of polynomial functions are either Real zeros of polynomial functions are either rationalrationalzeroszeros or or irrational zerosirrational zeros. Examples:. Examples:
24 9f x x 2 3 2 3x x The function has The function has rational zerosrational zeros –3/2 and 3/2 –3/2 and 3/2
2 2f x x 2 2x x The function has The function has irrational zerosirrational zeros – 2 and 2 – 2 and 2
Rational Zeros TheoremRational Zeros TheoremSuppose f is a polynomial function of degree n > 1 of the form
11 0
n nn nf x a x a x a
with every coefficient an integer and . If x = p /q isa rational zero of f, where p and q have no common integerfactors other than 1, then
0 0a
• p is an integer factor of the constant coefficient , and0a
• q is an integer factor of the leading coefficient .na
RZT – Examples:RZT – Examples:Find the rational zeros of 3 23 1f x x x The leading and constant coefficients are both 1!!!
The only possible rational zeros are 1 and –1…check them out:
3 21 1 3 1 1 3 0f 3 21 1 3 1 1 1 0f
So So ff has no rational zeros!!! has no rational zeros!!!(verify graphically?)(verify graphically?)
RZT – Examples:RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x
Potential Rational Zeros:
1, 2 Factors of –2
Factors of 3 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:RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x
1 3 4 –5 –2
3 7 2
3 7 2 0
Because the remainder is zero,x – 1 is a factor of f(x)!!!
23 7 2 1f x x x x Now, factor the remaining quadratic…
3 1 2 1f x x x x
The rational zeros are 1, –1/3, and –2The rational zeros are 1, –1/3, and –2
RZT – Examples:RZT – Examples:Find the polynomial function with leading coefficient 2 that hasdegree 3, with –1, 3, and –5 as zeros.
First, write the polynomial in factored form:
2 1 3 5f x x x x
Then expand into standard form:
22 2 3 5f x x x x
3 2 22 5 2 10 3 15x x x x x 3 22 6 26 30x x x
RZT – Examples:RZT – Examples:Using only algebraic methods, find the cubic function with thegiven table of values. Check with a calculator graph.
x
2 1 5f x k x x x
–2 –1 1 5
f(x) 0 24 0 0
(x + 2), (x – 1), and (x – 5)must be factors…
But we also have : 1 24f 1 2 1 1 1 5 24k 2k
2 2 1 5f x x x x
22 2 5x x x 3 22 8 14 20x x x
A New Use for SyntheticA New Use for SyntheticDivision in Sec. 2.4:Division in Sec. 2.4:
Upper andUpper andLower BoundsLower Bounds
What are they???
A number k is an upper bound for the real zeros off if f (x) is never zero when x is greater than k.
A number k is a lower bound for the real zeros off 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
cc is a is a lower bound lower bound and and dd is an is an upper upper boundbound for the real for the realzeros of zeros of f f
Upper and Lower Bound Tests for Real Zeros
Let f be a polynomial function of degree n > 1 with a positiveleading coefficient. Suppose f (x) is divided by x – k usingsynthetic 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!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function must lie inthe interval [–2, 5].
4 3 22 7 8 14 8f x x x x x The function has a positive leading coefficient, so we employour new test with –2 and 5:
5 2 –7 –8 14 8
10 15 35 245
2 3 7 49 253
–2 2 –7 –8 14 8
–4 22 –28 28
2 –11 14 –14 36
This last line is all positive!!! This last line hasalternating signs!!!
5 is an upper bound5 is an upper bound – –2 is a lower bound2 is a lower bound
Let’s check these results Let’s check these results graphicallygraphically……
Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.
4 3 22 7 8 14 8f x x x x x From the last example, we know that all of the rational zerosmust lie on the interval [–2, 5].
Next, use the Rational Zero Theorem…potential rational zeros:
Factors of 8
Factors of 2
1, 2, 4, 8
1, 2
11, 2, 4, 8,
2
Look at the graph to find likely candidates: Let’s try 4 and –1/2
Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.
4 3 22 7 8 14 8f x x x x x 4 2 –7 –8 14 8
8 4 –16 –8
2 1 –4 –2 0
3 24 2 4 2f x x x x x –1/2 2 1 –4 –2
–1 0 2
2 0 –4 0
214 2 4
2f x x x x
Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.
4 3 22 7 8 14 8f x x x x x
214 2 4
2f x x x x
212 4 2
2f x x x x
12 4 2 2
2f x x x x x
The zeros of f are the rational numbers 4 and –1/2 and theirrational numbers are – 2 and 2
Cool Practice Problems!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function lie in theinterval [0, 1], and find them. 5 210 3 6f x x x x Check our potential bounds:
0 10 0 0 –3 1
0 0 0 0
10 0 0 –3 1
0 is a lower bound!!!0 is a lower bound!!!
–6
0
–6
1 10 0 0 –3 1
10 10 10 7
10 10 10 7 8
1 is an upper bound!!!1 is an upper bound!!!
–6
8
2
Cool Practice Problems!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function lie in theinterval [0, 1], and find them. 5 210 3 6f x x x x Possible rational zeros:
1 3 1 2 3 6 1 31, 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!!!The function has no rational zeros on the interval!!!
Lone Real Zero: 0.951Lone Real Zero: 0.951Are there any zeros???