The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra Every major branch of mathematics has a “fundamental theorem” which provides much of the theoretical underpinnings for that branch. Algebra is no exception and the theorem called The Fundamental Theorem of Algebra was so named because, at the time, finding solutions of polynomial equations had been in the forefront of mathematical research for quite some time. Had mathematicians’ interest been focused on other mathematical questions what became the Fundamental Theorem might well have been something very different than the theorem we see today. The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Quadratic Formula Examples 1. Solve: x2 – 2x + 1 = 0 x = 2a b2 – 4ac  ± –b = 2(1) (–2)2 – 4(1)(1)  ± +2 = 2  ± Quadratic Formula Examples The examples above illustrate the three possible outcomes for type and number of solutions. If the expression under the radical, called the discriminant, is zero, then as shown in Example 1, the equation will have one real solution. We note that in Example 1, the quadratic expression is actually a perfect-square trinomial, that is, the square of a binomial – in this case (x – 1)2 . If the discriminant is positive, then as shown in Example 2, the equation will have two real solutions. If the discriminant is negative, then as shown in Example 3, the equation will have two imaginary (complex) solutions. We generalize this in the next slide. = 1 Solution set: { 1 } Question: Is there an easier way ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Quadratic Formula Examples 2. Solve: x = 2a b2 – 4ac  ± –b x2 – 4x – 5 = 0 = 2(1) (–4)2 – 4(1)(–5)  ± +4 x = 2 36  ± 4 Quadratic Formula Examples The examples above illustrate the three possible outcomes for type and number of solutions. If the expression under the radical, called the discriminant, is zero, then as shown in Example 1, the equation will have one real solution. We note that in Example 1, the quadratic expression is actually a perfect-square trinomial, that is, the square of a binomial – in this case (x – 1)2 . If the discriminant is positive, then as shown in Example 2, the equation will have two real solutions. If the discriminant is negative, then as shown in Example 3, the equation will have two imaginary (complex) solutions. We generalize this in the next slide. = 5 –1 Solution set: { –1, 5 } Question: Is there an easier way ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Quadratic Formula Examples 3. Solve: x = 2a b2 – 4ac  ± –b x2 – 2x + 5 = 0 = (–2)2 – 4(1)(5) 2(1)  ± +2 x = 2 –16  ± Quadratic Formula Examples The examples above illustrate the three possible outcomes for type and number of solutions. If the expression under the radical, called the discriminant, is zero, then as shown in Example 1, the equation will have one real solution. We note that in Example 1, the quadratic expression is actually a perfect-square trinomial, that is, the square of a binomial – in this case (x – 1)2 . If the discriminant is positive, then as shown in Example 2, the equation will have two real solutions. If the discriminant is negative, then as shown in Example 3, the equation will have two imaginary (complex) solutions. We generalize this in the next slide. Solution set: = 1 ± 2i { 1 – 2i , 1 + 2i } Question: Is there an easier way ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Complex Solutions Complex solutions always occur in conjugate pairs Depends on discriminant of the quadratic formula : WHY ? x = 2a b2 – 4ac  ± –b Complex Solutions For quadratic equations we can see from the quadratic formula the form that all solutions will have. We note that the radical with the discriminant has ± appearing in front of it, a result of completing the square and using the square root property. This expression yields two solutions, for a discriminant that is not zero. If the discriminant is negative, both solutions are complex numbers, one with the positive sign and one with the negative sign, i.e. the solutions are complex conjugates. If the discriminant is positive, the two signs yield real solutions, and of course, if the discriminant is zero, the radical vanishes and we are left with a single real number as the solution. We can extend this notion to polynomial equations of degree higher than 2, noting that the principal root is always taken to be a real number. Thus, for example, a cubic equation will have a real solution (positive or negative) and possibly two complex solutions – complex conjugates. Similarly, for a quartic equation there can be two or four real solutions (though there may be none) and, if only two or none, there must be at least two complex solutions. If there are no real solutions, then all solutions are complex, occurring in complex conjugate pairs. Similar results can be obtained for any nth-degree polynomial equation. Note ± radical 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Complex Solutions Quadratic formula : x = 2a b2 – 4ac  ± –b b2 – 4ac Discriminant = b2 – 4ac One real solution Complex Solutions For quadratic equations we can see from the quadratic formula the form that all solutions will have. We note that the radical with the discriminant has ± appearing in front of it, a result of completing the square and using the square root property. This expression yields two solutions, for a discriminant that is not zero. If the discriminant is negative, both solutions are complex numbers, one with the positive sign and one with the negative sign, i.e. the solutions are complex conjugates. If the discriminant is positive, the two signs yield real solutions, and of course, if the discriminant is zero, the radical vanishes and we are left with a single real number as the solution. We can extend this notion to polynomial equations of degree higher than 2, noting that the principal root is always taken to be a real number. Thus, for example, a cubic equation will have a real solution (positive or negative) and possibly two complex solutions – complex conjugates. Similarly, for a quartic equation there can be two or four real solutions (though there may be none) and, if only two or none, there must be at least two complex solutions. If there are no real solutions, then all solutions are complex, occurring in complex conjugate pairs. Similar results can be obtained for any nth-degree polynomial equation. b2 – 4ac > Two real solutions < b2 – 4ac Two complex conjugate solutions 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra A polynomial of degree n ≥ 1 has at least one complex zero Consequence: Every polynomial has a complete factorization WHY ? The Fundamental Theorem of Algebra This theorem was proved by Carl Friedrich Gauss at age 20 in 1797. The immediate consequence of this theorem is that every polynomial has a complete factorization, provided we include complex numbers as zeros. The reason for this consequence is shown in the illustration. From the Fundamental Theorem a polynomial f(x) has at least one (complex) zero k1 and so from the Factor Theorem (x – k1) is a factor of f(x). This gives us f(x) = (x – k1)Q1(x) for some quotient polynomial Q1(x) with deg Q1(x) < deg f(x). Applying the Fundamental Theorem again to Q1(x) we have Q1(x) = (x – k2)Q2(x) Thus, f(x) = (x – k1)(x – k2)Q2(x) Applying the Fundamental Theorem repeatedly, until all n zeros are accounted for, we have f(x) = (x – k1)(x – k2)... (x – kn)Cn where deg Cn(x) = 0; that is, Cn(x) is just a constant Cn. Thus we are led to the complete factorization of f(x). NOTE: Throughout the above discussion bear in mind that when we say complex number we are including the real numbers. Refer to the definition on the first slide of this module. For polynomial f(x) there exists a zero k1 such that f(x) = (x – k1)Q1(x) 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra For polynomial f(x) there exists a zero k1 such that f(x) = (x – k1)Q1(x) where deg Q1(x) = deg f(x) – 1 Apply the Fundamental Theorem to Q1(x) Q1(x) = (x – k2)Q2(x) so The Fundamental Theorem of Algebra This theorem was proved by Carl Friedrich Gauss at age 20 in 1797. The immediate consequence of this theorem is that every polynomial has a complete factorization, provided we include complex numbers as zeros. The reason for this consequence is shown in the illustration. From the Fundamental Theorem a polynomial f(x) has at least one (complex) zero k1 and so from the Factor Theorem (x – k1) is a factor of f(x). This gives us f(x) = (x – k1)Q1(x) for some quotient polynomial Q1(x) with deg Q1(x) < deg f(x). Applying the Fundamental Theorem again to Q1(x) we have Q1(x) = (x – k2)Q2(x) Thus, f(x) = (x – k1)(x – k2)Q2(x) Applying the Fundamental Theorem repeatedly, until all n zeros are accounted for, we have f(x) = (x – k1)(x – k2)... (x – kn)Cn where deg Cn(x) = 0; that is, Cn(x) is just a constant Cn. Thus we are led to the complete factorization of f(x). NOTE: Throughout the above discussion bear in mind that when we say complex number we are including the real numbers. Refer to the definition on the first slide of this module. f(x) = (x – k1)(x – k2)Q2(x) where deg Q2(x) = deg f(x) – 2 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra f(x) = (x – k1)(x – k2)Q2(x) where deg Q2(x) = deg f(x) – 2 For deg f(x) = n , apply n times : f(x) = (x – k1)(x – k2) • • • (x – kn)Cn where Cn is a nonzero constant The Fundamental Theorem of Algebra This theorem was proved by Carl Friedrich Gauss at age 20 in 1797. The immediate consequence of this theorem is that every polynomial has a complete factorization, provided we include complex numbers as zeros. The reason for this consequence is shown in the illustration. From the Fundamental Theorem a polynomial f(x) has at least one (complex) zero k1 and so from the Factor Theorem (x – k1) is a factor of f(x). This gives us f(x) = (x – k1)Q1(x) for some quotient polynomial Q1(x) with deg Q1(x) < deg f(x). Applying the Fundamental Theorem again to Q1(x) we have Q1(x) = (x – k2)Q2(x) Thus, f(x) = (x – k1)(x – k2)Q2(x) Applying the Fundamental Theorem repeatedly, until all n zeros are accounted for, we have f(x) = (x – k1)(x – k2)... (x – kn)Cn where deg Cn(x) = 0; that is, Cn(x) is just a constant Cn. Thus we are led to the complete factorization of f(x). NOTE: Throughout the above discussion bear in mind that when we say complex number we are including the real numbers. Refer to the definition on the first slide of this module. f(x) is now completely factored ! 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra The Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros From the Complete Factorization: f(x) = (x – k1)(x – k2)...(x – kn)Cn If all ki are distinct there are n distinct zeros ... otherwise one or more zeros are repeated zeros The Number of Zeros Theorem This theorem is an immediate consequence of the Fundamental Theorem. The degree of the polynomial determines an upper limit on the number of zeros the polynomial can have. Counting multiplicities and complex roots, the degree is the number of zeros. The examples illustrate these facts. 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra The Number of Zeros Theorem Examples 1. f(x) = 5x(x + 3)(x – 1)(x – 4) 2. f(x) = 4x2(x – 2)3(x – 7) How many distinct zeros ? The Number of Zeros Theorem This theorem is an immediate consequence of the Fundamental Theorem. The degree of the polynomial determines an upper limit on the number of zeros the polynomial can have. Counting multiplicities and complex roots, the degree is the number of zeros. The examples illustrate these facts. How many distinct zeros ? Total zeros ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra The Conjugate Zeros Theorem If f(x) is a polynomial with only real coefficients then its complex zeros occur in conjugate pairs Examples 1. f(x) = x3 + x = x(x2 + 1) = x(x + i)(x – i) The Conjugate Zeros Theorem This theorem deals with complex zeros and verifies that they must occur in conjugate pairs, as already indicated. In the first example we make a change of variable from x to u, replacing x2 by u, which alters the form of the functional expression from a fourth degree polynomial to a second degree polynomial. Factoring the second degree polynomial is relatively easy, giving two factors in the variable u. Replacing u with x2 yields the factors (x2 – 4) and (x2 + 9). The first of these is just a difference of squares which factors to the product of two binomial conjugates. The second factors to the product of two complex conjugates, thus giving four factors and hence four zeros, two real zeros and two complex zeros for the fourth degree polynomial function f(x). Remember: Complex solutions for polynomial equations always occur in complex conjugate pairs. In the second example we note that x is a factor of every term, and hence, factoring out the largest common factor, we rewrite the expression as x(x2 + 1). Factoring the second factor we obtain two complex conjugate factors, for a total of three factors altogether. Using the zero product property to find the zeros for the function gives us one real zero and two complex conjugate zeros. Zeros ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra The Conjugate Zeros Theorem If f(x) is a polynomial with only real coefficients then its complex zeros occur in conjugate pairs Examples 2. f(x) = x4 + 5x2 – 36 = u2 + 5u – 36 = (u – 4)(u + 9) = (x2 – 4)(x2 + 9) Let u = x2 The Conjugate Zeros Theorem This theorem deals with complex zeros and verifies that they must occur in conjugate pairs, as already indicated. In the first example we make a change of variable from x to u, replacing x2 by u, which alters the form of the functional expression from a fourth degree polynomial to a second degree polynomial. Factoring the second degree polynomial is relatively easy, giving two factors in the variable u. Replacing u with x2 yields the factors (x2 – 4) and (x2 + 9). The first of these is just a difference of squares which factors to the product of two binomial conjugates. The second factors to the product of two complex conjugates, thus giving four factors and hence four zeros, two real zeros and two complex zeros for the fourth degree polynomial function f(x). Remember: Complex solutions for polynomial equations always occur in complex conjugate pairs. In the second example we note that x is a factor of every term, and hence, factoring out the largest common factor, we rewrite the expression as x(x2 + 1). Factoring the second factor we obtain two complex conjugate factors, for a total of three factors altogether. Using the zero product property to find the zeros for the function gives us one real zero and two complex conjugate zeros. = (x + 2)(x – 2)(x + 3i)(x – 3i) Zeros ? 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 1. f(x) = x4 + 5x2 – 36    Solving Equations Using Zeros The standard form of any polynomial equation has a zero on the right hand side of the equation. The usual method of solution is to factor the expression on the left and apply the zero product property of the real numbers (which also holds for complex numbers). The examples are simply those we have just used in the previous slide to find the factors of the given polynomials. The only step left is to apply the zero product property to each of them, as illustrated. As indicated, the fourth degree polynomial equation has four solutions and the third degree polynomial equation has three solutions. Also, as indicated previously, for polynomials of any degree, the complex solutions always occur in conjugate pairs. Note that the solutions are numbers not equations. That is, one solution is 2, not the equation x = 2. = (x + 2)(x – 2)(x + 3i)(x – 3i) = 0 Thus x + 2 = 0 OR x – 2 = 0 OR x + 3i = 0 OR x – 3i = 0 Solution set: { –2 , 2 , –3i , 3i } 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem The Fundamental Theorem of Algebra Solving Equations Using Zeros Factor polynomial Set completely factored form equal to 0 Apply zero product property Examples 2. f(x) = x3 + x    Solving Equations Using Zeros The standard form of any polynomial equation has a zero on the right hand side of the equation. The usual method of solution is to factor the expression on the left and apply the zero product property of the real numbers (which also holds for complex numbers). The examples are simply those we have just used in the previous slide to find the factors of the given polynomials. The only step left is to apply the zero product property to each of them, as illustrated. As indicated, the fourth degree polynomial equation has four solutions and the third degree polynomial equation has three solutions. Also, as indicated previously, for polynomials of any degree, the complex solutions always occur in conjugate pairs. Note that the solutions are numbers not equations. That is, one solution is 2, not the equation x = 2. = x (x + i)(x – i) = 0 Thus x = 0 OR x + i = 0 OR x – i = 0 Solution set: { 0, –i, i } 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem of Algebra Think about it ! 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013

The Fundamental Theorem of Algebra Spare Parts The Fundamental Theorem of Algebra x2 = –1 i  – a ±  –1  – a  –1 = ± x ±    –1  i = –i – –1  =      Complex Numbers and the Imaginary i The complex numbers are sometimes divided into all-real and all-imaginary numbers. That is, a + bi is real if b = 0, and is all-imaginary if b ≠ 0 and a = 0. Some authors use the definition shown in the illustration allowing a to be zero for an imaginary number but not requiring it. This minor difference does not produce any different mathematics at the application level since operations with complex numbers do not depend on the naming of the numbers. The complex plane allows us to plot complex numbers in graphical fashion. The vertical axis is normally taken as the pure imaginary axis and is sometimes designated as such by the “iy” tag shown in the illustration. Some authors do not plot complex numbers at all and some use merely the “y” designation with the understanding that all y components are considered to be imaginary. Still other authors define a complex number as an ordered pair of real numbers, the first of which is the “real” component and the second is the “imaginary” component. All of these views of complex numbers lead to essentially the same mathematical results. 9/2/2013 Fundamental Theorem The Fundamental Theorem 9/2/2013