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# Lecture 8 and 9 Notes Math 240

AnnMarie

GPA 3.028
Precalculus
Jonathan B Walters

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These notes contain the lecture material that was covered on Sept. 28 and Sept. 30.
COURSE
Precalculus
PROF.
Jonathan B Walters
TYPE
Class Notes
PAGES
12
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 12 page Class Notes was uploaded by AnnMarie on Wednesday September 30, 2015. The Class Notes belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 57 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.

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Date Created: 09/30/15
33 Dividing Polynomials Review Divide 10903 by 29 375 r28 mm 87 We could say 375 or 10903 37529 This concept introduces the division algorithm which is explained below Division Algorithm If px and dx are polynomials with dx 79 O then there exists unique polynomials qx and rx where rx is either 0 or of degree less than the degree of dx such that 1 100 or W doc x qltxgt M The polynomials px and dx are called the dividend and the divisor respectively qx is the quotient and rx is the remainder ExampleFind the quotient and remainder when dividing px 6x4 5x3 2x 4 by dx 2x2 1 Solution 2x20x 16x4 5x30x2 2x 4 You will notice that I included a space holder for x and x2 in the dividend and the divisor I did this to allow number to line up correctly 13x2 x 2x20x 1V6x4 5x30x2 2x 4 6x4 0 3x2 5x33x2 2x 4 5x3 0 x 3x2 x 4 3x2 0 l 2x 4 qx3x2x and rx x 6x4 5x3 2x 43x2 x2x2 1 x Remainder Theorem If a polynomial pX is divided by Xk then the remainder in the division is pk Synthetic Division A quick method of dividing polynomials it can be used when the divisor is of the form x c The coefficients of the polynomial are inserted into a straight line and then the constant in the divisor is put on the outside c I use the to symbolized that there were coefficients listed Example Use synthetic division to divide 3x3 7x2 5x 10 by x 3 Solution 3 l3 7 5 1O The quotient is in green 3x22x1and the remainder is in orange 7 The coefficients of the polynomial of the degree n1 if n is degree of the original polynomial degree Remainder Theorem XC is a zero of pX if and only if Xc is a factor of pX Example Determine if x1 is a zero of px x3 7x6 using synthetic division 11076 x1 is a zero and furthermore x3 7x6 x2x 6x 1 can be factored further to get x3 7x 6x 1x 2x 3 This shows that X 1 2 and 3 Example Find the polynomial of degree 4 having zeros X2 x0 x1 x3 and the coefficient of x3 is 4 Solution Guess px x 2x Ox 1x 3 x2 2xx 1x 3 x3 x2 2x3 2xx 3 x4 2x3 5x26x Multiply 2 through to get px 2x4 4x3 10x2 12x 34 Real Zeros of Polynomials Rational Zeros Theorem If p and q have no common factors and 1 is a zero of a polynomial fx anx alx do Where all coefficients are integers then 1 p is a factor of a0 2 q is a factor of an Example Find all zeros of fx2x3 5x3 Solution Possibilities for p i1 i3 Possibilities for q i1 i2 Possibilities for g 1 i2 ig Starting with whole numbers perform synthetic division 2x3 5x3 x 12x22x 3 Using the quadratic formula to get the zeros of 2x2l2x 3 2i 4 42 3 xzwsxzwsxz 23mm 13w Thus x1 x2 andx3 1 Example Find all real zeros of 2x4 19x29 Solution Possibilities for p i1 i3 i9 Possibilities for q i 1 i2 Possibilities for g i1 i3 i9 1 1 1 2 O 19 O 9 2 2 17 17 2 2 17 17 8 3201909 6183 9 26130 2x4 19x2 9 x 32x3 6x2 x 3 Possibilities for p i 1 i3 Possibilities for q i 1 i2 Possibilities for g i 1 i3 1 3 2 6 1 3 2x4 19x2 9 x 3x 32x2 1 Using factoring we get the last two zero values 2x2 1 O 35 Complex Zeros and Fundamental Theorem of Algebra Theorem of Algebra Every non constant polynomial With complex coefficients has at least one complex zero Complete Factorization Theorem If px is a polynomial of degree n with complex coefficients then it can be factored in the form px ax clx 62x cm with clcn complex numbers and a is the leading coefficient of px Basically this means With every nth order polynomial has exactly n zeros Note 1392 1 and Vj 139 Example Find all the zeros of px x3 3x2 x 3 Solution Using the rational zeros theorem 531 j3 3 1 3 1 3 pxx3x21 3 o 3 1 O 1 O Zeros of the polynomial are x 3 x i x i Conjugate Pairs Theorem If p is a polynomial with real coefficients and at a bi With b 75 O is a zero of p then a bi is also a zero ofp Example Find all the zeros px x4 4x3 7x2 50x 50 given 3 z39 is a zero Solution From CPT we know x 3iis also a zero Then px x 3 ix 3 iqx px x2 x3 i x3 x 3 i3 iqx 1936 x2 x3 i x3 i 9 3i 3i i2qx px x2 x3 i x3 i 9 1qx px x2 3x xi 3x xi 10qx px x2 6x 10qx x44x37x250x50 qx x2 6x 10 Using long division we find what qx is x2 2x 5 x2 6x 10x4 4x3 7x2 50x 50 x4 6x3 10x2 2x3 17x2 50x 2x3 12x2 20x 5x2 30x 50 5x2 30x 50 0 qxx22x 5 Now we can say that px x2 6x 10x2 2x 5 To find the other zeros use qf x 24 24l 5 x M xz 2 2 x liV6 Thus me x lt3 w lt3 igtgtltx lt 1 1m lt 1 16 Zeros of the polynomial are x 3 i 3 i 1 V5 and 1 V5 Example Find a polynomial with the given properties Degree 4 real coefficient zeros at 12i 1i and p11 Solution 1906 Z 6136 12ix 1 2ix 1 ix 1i px ax2 x1 2i x1 2i12i1 2ix2 x1i x 1 i1 i1i px ax2 x 2xi x2xi 1 2i22i 4i2x2 x xi xxi 1 i i i2 px ax2 2x 1 41 2x2 2x 1 i2 px ax2 2x2x2 2x5 p1 ax2 2x2x2 2x 5 1a1 221 25 1 a4 px x2 2x 2x2 2x 5 36 Rational Functions Recall y Transformation properties still apply L x2 Example Graph gx Solution If fx 916 then gx fx 2 Example Sketch hx Solution Try to put h into the form h bfx a c Where fx lc Using long division 8 5x 2Vm hx 5x 2 8x156 hxX5xI 2 hx 21 5 hx shifts t0 the right units re ects over the y axis shrinks vertically 2 15 and 3 shifts down The x intercept is g and y intercept is 5 its

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