College Algebra  MAT 161 Page: 1 Copyright 2009 Killoran


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1 College Algebra  MAT 6 Page: Copyright 2009 Killoran Zeros and Roots of Polynomial Functions Finding a Root (zero or xintercept) of a polynomial is identical to the process of factoring a polynomial. Theorems used to find the Zero s (or Roots) of a Polynomial Function: nzero s Theorem: If p.x/ is of Degree n, then it has at most n Zero 0 s The Remainder Theorem: If The Factor Theorem: If p.x/ D q.x/ C r then p.c/ D r x c p.x/ D q.x/ C 0 then p.x/ D.x c/ q.x/ x c Also since p.c/ D 0 (by the remainder theorem) we also have that x D c is an xintercept of p.x/ Intermediate Value Theorem: If p.a/ p.b/ D where p.x/ is a continuous function on the interval of x 2.a; b/ then there exists some x D c where c 2.a; b/ and p.c/ D 0 New Theorem s Theorem Rational Root (zero) Theorem Given that p.x/ D a n x n C a n x n C C ax C a 0 where all a k are Real, then any Rational Roots of p.x/ will be of the form of (factors of a 0 /.factors of a n / This theorem is only useful if the Roots (zeros) of the polynomial are rational to begin with. If they are irrational (or imaginary roots) this theorem will not help us, and we will have to use another "trick" to locate them. The rationale can be inductively shown by solving a simple polynomial equation: 4x 4 37x 2 C 9 D 0 To factor this we need to find the Factors of.4/.9/ D 36 that add up to be 36 and : 37: Which would 0 D 4x 4 36x 2 x 2 C 9 0 D 4x 2 x 2 9 x D x 2 9 4x 2 thus: 0 D.x 3/.x C 3/.2x /.2x C / x D f3; 3; 2 ; 2 g
2 College Algebra  MAT 6 Page: 2 Copyright 2009 Killoran Setting up all the factors of a 0 D 9 ; 3; 9 we get D f; 3; 9; =2; 3=2; 9=2; =4; 3=4; 9=4g: a n 4 ; 2; 4 Notice that the roots found are from this list./ : Example 2 List the possible Rational Roots of f.x/ D 2x 3 2x 2 C 49x C 30 First write a 0 a n D 30 2 and their respected factors D ; 2; 3; 5; 6; 0; 5; 30 ; 2 Next write all combinations of these factors: ; 2 ; 3 ; 5 ; 6 ; 0 ; 5 ; 30 2 ; 2 2 ; 3 2 ; 5 2 ; 6 2 ; 0 2 ; 5 2 ; 30 2 After simplifying and removing repeats we get our final list f; 2; 3; 5; 6; 0; 5; 30; =2; 3=2; 5=2; 5=2g If this function had a rational root, it will have to be one of these numbers, positive or negative. 2 Descarte s Rule of Sign: To help us find the possible rational roots it would be helpful to know how many of them might be positive or negative. Using Descarte s Rule will help. To find what is the Maximum number of Positive Roots in a Polynomial (with no missing powers), one just has to count the number of Sign Changes: Example 3 How many possible positive roots exist for f.x/ D 6x 5 C x 4 33x 3 33x 2 C x C 6 There are two sign changes here, Cx 4 to 33x 3 and 33x 2 to Cx; thus there are at most TWO Positive Roots. Knowing that the Number of Roots is equal to the Degree of the polynomial we can deduce that there will be at most THREE Negative Roots.
3 College Algebra  MAT 6 Page: 3 Copyright 2009 Killoran 3 The Search for Zeros Example 4 Find all the Zeros of f.x/ D 2x 4 7x 3 4x 2 C 27x 8 Using everything that we have available, it would be better to analyze this first to know what we are looking for nzeros Theorem. f.x/ has at most 4 real roots Descarte s Rule of Sign: f.x/ has Three Changes in Sign, thus there are Three Positive roots, and One Negative Root, for the total of 4 zero 0 s Rational Root Theorem: List Possible Roots a 0 a n D 8 2 D ; 2; 3; 6; 9; 8 ; 2 D f; 2; 3; 6; 9; 8; 2 ; 3 2 ; 9 2 g The Remainder Theorem: Find f.c/ using the above numbers and synthetic division zero is of interest! j j Factor Theorem: Since the remainder is Zero, that means three things:. x D 2 is a Zero of f.x/ 2..x C 2/ is a Factor of f.x/ 3. The last line q.x/ D 2x 3 x 2 C 8x 9 is also a factor of f.x/ Now we can use Step 4 and 5 over again using q.x/ to find more factors. But we don t need to search for any more Negative Zero s because we found the ONE that existed (Descarte)
4 College Algebra  MAT 6 Page: 4 Copyright 2009 Killoran. j x D is a Zero,.x / is a Factor, and q.x/ is now: 2x 2 9xC9 (quadratic=factor!) 3. 2x 2 9x C 9 2x 2 6x 3x C 9 2x.x 3/ 3.x 3/.x 3/.2x 3/ x D f3; 3 2 g Putting it all together: Zeros: x D 2; ; 3; 3 2 Factored: f.x/ D 2x 4 7x 3 4x 2 C 27x 8 D.x C 2/.x /.x 3/.2x 3/ Corollary 5 Given p.x/ is an odd degree polynomial then there exist at least one real root. This is just a combination of Polynomial End Behavior and Intermediate Value Theorem. Now I have been stating over and over that the Real Roots are AT MOST so many. reason for this is simple: there may be complex or Imaginary roots involved. The Example 6 Find the Roots of p.x/ D 2x 3 x 2 C 8x 9 given that is a zero of p.x/ 2 Here we can start with x D and "reduce" our polynomial into one less degree
5 College Algebra  MAT 6 Page: 5 Copyright 2009 Killoran So upon setting up q.x/ D 0 we get: q.x/ D 2x 2 C 0x C 8 0 D 2x 2 C 8 2x 2 D 8 x 2 D 9 x D f3i; 3ig So the Roots of p.x/ D 2x 3 x 2 C 8x 9 are x D ; 3i; 3i : 2 Notice that there are Three Roots and the Degree of the Polynomial is Three... Theorem 7 Complex Root Theorem If p.x/ is a polynomial function with Real Coefficients and a C bi is a Root, then a must also be a Root of p.x/ : bi This theorem in short states that if the Coefficients are real, then all Complex Roots come in Conjugate Pairs! If there is one, there will be another, its conjugate. Example 8 If p.x/ is a polynomial with real coefficients and its roots are f find a polynomial f.x/ that is equal to a constant multiple of p.x/. ; 3; 2g then NOTE: f.x/ D k p.x/ ; but with out more information we will never find the value for k; but we must account for its existence. By the factor theorem we can take each root.c/ and write he binomial associated with it.x.c// : x D becomes.x C / D 0 x D 3 becomes.x 3/ D 0 x D 2 becomes.x 2/ D 0 So p.x/ D k.x C /.x 3/.x 2/ Example 9 Given p.x/ is a polynomial function with real coefficients and p.x/ 0 s roots are f 3i; 2 C ig ; find f.x/ D k p.x/ : Since both of these roots are not real, they must match up with a Conjugate Pair: f 3i; 3i; 2 C i; 2 ig
6 College Algebra  MAT 6 Page: 6 Copyright 2009 Killoran in which the Factors would be: f.x/ D.x. 3i//.x.3i//.x.2 C i//.x.2 i// f.x/ D.x C 3i/.x 3i/.x.2 C i//.x.2 i// Multiplying out the first two factors (remembering rule of imaginary numbers) we get: h i f.x/ D x 2 C 9.x.2 C i//.x.2 i// now the other two complex factors are a little more tricky, so I will use a simple substitution to help. Let A D.2 C i/ and B D.2 i/ then:.x.2 C i//.x.2 i// becomes.x A/.x B/ x 2 Bx Ax C AB x 2.A C B/ x C AB So all we need to find is the value of.a C B/ and.a B/ A C B D.2 C i/ C.2 i/ D 4 A B D.2 C i/.2 i/ D 4 C D 5 Thus.x.2 C i//.x.2 i// x 2.A C B/ x C AB x 2.4/ x C.5/ and h i f.x/ D x 2 C 9.x.2 C i//.x.2 i// becomes f.x/ D x 2 C 9 x 2 4x C 5 It should be clear that any Polynomial can be broken into exactly the same number of Unique Product of Linear Factors (real and complex) as the degree of the polynomial. Thus any p.x/ D.x c n /.x c n /.x c n 2 /.x c / k where the c s can be any complex number and n is the degree of p.x/. Remember that a complex number is also Real: 5 D 5 C 0i
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