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Packet #9 Zeros of Polynomials
Math 160 Packet #9 Zeros of Polynomials
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Rational Zeros Theorem If π π₯ = π π π₯ π + π πβ1 π₯ πβ1 +β¦+ π 1 π₯+ π 0 has integer coefficients, then every rational zero of π is of the form π π where π is a factor of π 0 and π is a factor of π π .
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Ex 1. Use the Rational Zeros Theorem to list all possible rational zeros of π π₯ =4 π₯ 5 β5 π₯ 3 β6 π₯ 2 β2π₯+3.
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Ex 2. Factor and find all zeros of π π₯ =2 π₯ 3 + π₯ 2 β13π₯+6.
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Ex 3. Factor and find all zeros of π π₯ = π₯ 4 β5 π₯ 3 β5 π₯ 2 +23π₯+10.
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Notice that π₯ 2 +1=0 does not have any real number solutions
Notice that π₯ 2 +1=0 does not have any real number solutions. The imaginary number π= βπ was introduced to create a solution (note that π 2 =β1).
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Notice that π₯ 2 +1=0 does not have any real number solutions
Notice that π₯ 2 +1=0 does not have any real number solutions. The imaginary number π= βπ was introduced to create a solution (note that π 2 =β1).
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With π, you can create a number system (called the complex number system) that is larger than the real number system. Hereβs how itβs done: Every complex number can be written in the form π+ππ, where π is the real part and π is the imaginary part. Both π and π are real numbers. Note that any real number can be written as a complex number (just let π=0).
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With π, you can create a number system (called the complex number system) that is larger than the real number system. Hereβs how itβs done: Every complex number can be written in the form π+ππ, where π is the real part and π is the imaginary part. Both π and π are real numbers. Note that any real number can be written as a complex number (just let π=0).
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Here are some examples of complex numbers: β1+3π 2 5 β2π
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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3+5π + 4β2π =π+ππ 3+5π β 4β2π =βπ+ππ 3+5π 4β2π =12β6π+20πβ10 π 2 =12+14π+10 =ππ+πππ
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The conjugate of π+ππ is __________
The conjugate of π+ππ is __________. To divide complex #βs, we can use the conjugate to help. Ex π 1β2π = Ex π 4π =
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The conjugate of π+ππ is __________
The conjugate of π+ππ is __________. To divide complex #βs, we can use the conjugate to help. Ex π 1β2π = Ex π 4π = πβππ
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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3+5π 1β2π = 3+5π β
1+2π 1β2π β
1+2π = 3+6π+5π+10 π 2 1+2πβ2πβ4 π 2 = 3+11πβ = β7+11π 5 =β π π + ππ π π
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Fundamental Theorem of Algebra Every polynomial with complex coefficients has at least one complex zero. Complete Factorization Theorem Every polynomial can be broken down into linear factors like this: π π₯ =π π₯β π 1 π₯β π 2 β¦(π₯β π π ) (here, π, π 1 , π 2 , ... , π π are complex numbers)
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Fundamental Theorem of Algebra Every polynomial with complex coefficients has at least one complex zero. Complete Factorization Theorem Every polynomial can be broken down into linear factors like this: π π₯ =π π₯β π 1 π₯β π 2 β¦(π₯β π π ) (here, π, π 1 , π 2 , ... , π π are complex numbers)
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Ex 4. Find the complete factorization and all zeros of π π₯ = π₯ 3 β3 π₯ 2 +π₯β3. of π.
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Ex 5. Find the complete factorization and all zeros of π π₯ = π₯ 3 β2π₯+4
Ex 5. Find the complete factorization and all zeros of π π₯ = π₯ 3 β2π₯+4. Now, find the complete factorization of π.
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Zeros Theorem Every degree π polynomial has exactly π zeros (here, a zero of multiplicity π is counted π times). Ex 3. Find the complete factorization and all five zeros of the polynomial π π₯ =3 π₯ π₯ 3 +48π₯.
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Zeros Theorem Every degree π polynomial has exactly π zeros (here, a zero of multiplicity π is counted π times). Ex 6. Find the complete factorization and all five zeros of the polynomial π π₯ =3 π₯ π₯ 3 +48π₯.
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Conjugate Zeros Theorem Complex zeros come in conjugate pairs
Conjugate Zeros Theorem Complex zeros come in conjugate pairs. That is, if π+ππ is a zero, then πβππ is also a zero. (Note: This is only true for polynomials with real coefficients, which is mostly what you see anyway.) Ex 7. Find a degree 3 polynomial with zeros 2 and 1+3π.
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Conjugate Zeros Theorem Complex zeros come in conjugate pairs
Conjugate Zeros Theorem Complex zeros come in conjugate pairs. That is, if π+ππ is a zero, then πβππ is also a zero. (Note: This is only true for polynomials with real coefficients, which is mostly what you see anyway.) Ex 7. Find a degree 3 polynomial with zeros 2 and 1+3π.
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Conjugate Zeros Theorem Complex zeros come in conjugate pairs
Conjugate Zeros Theorem Complex zeros come in conjugate pairs. That is, if π+ππ is a zero, then πβππ is also a zero. (Note: This is only true for polynomials with real coefficients, which is mostly what you see anyway.) Ex 7. Find a degree 3 polynomial with real coefficients and zeros 2 and 1+3π.
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Intermediate Value Theorem (for Polynomials)
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Intermediate Value Theorem (for Polynomials)
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Note: Since polynomials are continuous (can be drawn without picking up your pencil), if you find two function values, π(π) and π(π), that have opposite signs, then π(π₯) must cross the π₯-axis at some π₯-value π between π and π. This is called the Intermediate Value Theorem (for Polynomials). The same thing is true for all continuous functions.
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