Algebra, Polyomial ad Ratioal Fuctios Page 1 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes 3.1 Zeros of a Fuctio Set the fuctio to zero ad solve for x. The fuctio is zero at these values of x. Zeros = Roots 3.1 Log Divisio of Polyomials Arrage polyomial i descedig order by degree ad use place holders (zero so all terms are represeted. 3.1 Sythetic Divisio Ca oly be used whe dividig by ( x a 3.1 Remaider Theorem & Factor Theorem. Arrage polyomial i descedig order by degree ad use place holders (zero so all terms are represeted. If ( x c the remaider is Pc (. Px is divided by ( Pc = The ( x c If ( 0 is a factor of Px. ( 3.1 Reduced Polyomials Whe a polyomial is divided by a factor, what remais Reduced Polyomial Origial Polyomial is a reduced polyomial. ( Factor Origial Polyomial = (Factor (Reduced Polyomial 3. Far-Ed Behavior of Polyomial Fuctios Graph Ed Behavior for Degree ( is Degree ( Eve is Odd Positive Negative 3. X ad Y Itercepts X-Itercept: Where f( x crosses X axis. Here, y = 0. Y-Itercept: Where f( x crosses Y axis. Here, x = 0. 3. Zero Locatio Theorem If Pa ( ad ( 3. Maximum ad Miimum Values Pb have differet sigs (oe is positive ad oe is egative the there is at least oe real zero betwee them. The maximum or miimum poits, for a polyomial, occur at the ed poits or at a turig poit. 3.3 Multiplicity of Liear Factors Whe a polyomial fuctio is writte as a set of liear factors, the umber of times a liear factor (or root occurs is its multiplicity. 3 4 Example: Px ( = ( x 1 ( x+ Roots & Multiplicity are: 1(m3, -(m4 Eve Mult. è Bouce, Odd Mult. è Go thru.
Algebra, Polyomial ad Ratioal Fuctios Page K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes 3.3 Example: Roots & Mult. are: 0(m, 1/(m3 Px ( = 4x ( x 1 3 Degree = 5, Eds: ^ ] 3.3 How may roots ca a Polyomial Fuctio have? 3.3 Not all Polyomials are i factored form. Too Bad!!! L A polyomial fuctio Px ( of degree has at most real roots if you cout multiplicity. I geeral, Px ( of degree, has exactly roots if you cout multiplicity ad imagiary roots. If a polyomial is writte as a set of liear factors, it is fairly easy to fid the roots ad graph it. If the polyomial is ot i factored form, we must do some work to fid the roots. Use Theorems from famous mathematicias to help!!! J 3.3 Ratioal Zero Theorem This theorem isused to make a list of all possible ratioal zeros of a polyomial fuctio, Px ( = ax +... + b All possible ratioal zeros = p factors of last term factors of b ± =± =± q factors of first term factors of a 3.3 Upper ad Lower-Boud Theorem This theorem helps us fid the rage where the real zeros will occur. It helps us fid the upper & lower bouds. 1 For a polyomial fuctio Px ( = ax + a x +... The upper ad lower bouds ca be foud by dividig it by ( x b usig sythetic divisio. b is Upper Boud Lower Boud If 1 Positive root ( b > 0 First term positive ( a > 0 Bottom row all positive Positive root ( b > 0 Negative first term ( a < 0 Bottom row all egative Negative root ( b < 0 Bottom row alterate sig
Algebra, Polyomial ad Ratioal Fuctios Page 3 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes 3.3 Descartes Rule of This rule ca help us fid the umber of real zeros. Sigs For a polyomial fuctio Px ( = ax + a 1 x +... 1 # Positive Real Zeros # Negative Real Zeros # positive real zeros = # sig chages of Px ( - OR that umber decreased by a eve iteger. The umber of egative real zeros is the umber of sig chages of P( x - OR that umber decreased by a eve iteger. 3.4 Fudametal Theorem of Algebra If Pxis ( a polyomial fuctio of degree 1 with complex coefficiets the Px ( has at least oe complex zero. Traslatio: Ay polyomial has at least oe zero. 3.4 Liear Factor Theorem If Pxis ( a polyomial fuctio of degree the it has exactly liear factors. Px ( = a( x c1( x c...( x c Where c1, c,... c are complex umbers. Note: Complex umbers iclude real umbers. Complex Number format: a + bi (b = 0 for real # s 3.4 Number of Zeroes of a Polyomial Fuctio Theorem 3.4 Cojugate Pair Theorem If Pxis ( a polyomial fuctio of degree the it has exactly complex zeros provided each zero is couted accordit to its multiplicity. Imagiary roots come i pairs. a+ bi is a zero of a polyomial fuctio, the the If ( cojugate ( a bi Example: x + 9= 0 x = 9 Px is also a zero. ( x 9 = + Has roots whe Px ( = 0 It has roots at x=± 9 =± 9 1=± 3i 3.4 Complex Numbers ad Imagiary Roots (geeral iformatio i i = 1 = 1 Complex Numbers are i the form a + bi where a is the real part ad bi is the imagiary part. If b is zero, the there is o imagiary part ad the the a part. complex umber is also a real umber ( If a is zero, the there is o real part ad the complex the bi part. umber is just the imagiary part ( Example: Suppose a polyomial, Px, ( has two complex roots ( 3 i ad ( 3i factors are: ( +. The, the liear ( x 3 i ad ( x ( 3i +.
Algebra, Polyomial ad Ratioal Fuctios Page 4 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes *** Geeral Guidelies for fidig Zeros Fid the umber of positive & egative real zeros by usig Descartes Rule of Sigs. Fid a list of possible real zeros by usig the Ratioal Root Theorem. Check the umbers from your list. o Use the Remaider Theorem to see if a umber, o your list, is a zero. o Whe you fid a zero (say it s at x = c the use sythetic divisio (divide by (x c ad check the bottom row. o Use Upper & Lower boud theorem to rule out other umbers from your list. Work with reduced polyomials. Every time you fid a zero, you obtai a reduced polyomial. If a reduced polyomial is degree, try to factor it or use the quadratic formula. b± b 4ac x = If you fid a imagiary root, use the Cojugate Pair Theorem to fid aother imagiary root. Keep lookig for roots util you fid them all. Use the a Liear Factor Theorem to fid the total umber of roots. (You should fid roots if Px ( is degree. Use the roots to write Px ( as a set of liear factors. Use Far-Ed Behavior, multiplicity, ad fid x & y- itercepts to sketch a graph. EXAMPLE: Step How to factor a polyomial usig Theorems i sectios 3.1 to 3.4. #1 4 3 Completely factor Px ( = x x 6x + x 15 This meas we wat to write Px ( as a product of liear factors. Degree = 4, so The Liear Factor Theorem says it will have exactly 4 liear factors, but some of the associated roots may be complex roots. # Use Descartes Rule of Sigs. To fid the umber of Real Positive zeros, look at the sig chages of the terms of Px. ( There are 3 sig chages so there are 3 or 1 positive real roots. To fid the umber of Real Negative zeros, look at the sig chages of the terms of 4 3 4 3 P( x = ( x ( x 6( x + ( x 15 = x + x 6x x 15 There is oe sig chage so there is oe egative real root. #3 Use Ratioal Zero Theorem. All possible ratioal zeros = factors of 15 1, 3, 5, 15 factors of 1 =± ± ± ±
Algebra, Polyomial ad Ratioal Fuctios Page 5 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Step How to factor a polyomial usig Theorems i sectios 3.1 to 3.4. #4 Use Remaider ad Factor Theorem Let s look for the oe egative real root. Try the possible egative zeros. 4 3 P( 1 = ( 1 ( 1 6( 1 + ( 1 15 = 1+ 6 15 = 4 0 L 4 3 P( 3 = ( 3 ( 3 6( 3 + ( 3 15 = 81+ 54 54 66 15 = 0 = 0 J The oe ad oly egative real zero is( 3. The associated factor is ( x + 3. #5 Use Sythetic Divisio to fid the reduced polyomial. Px x x x x 4 3 ( = 6 + 15-3 1 - -6-15 -3 15-7 15 1-5 9-5 0 #6 Now, we have P( x = ( x+ 3( x 3 5x + 9x Use the above approach to factor the reduced polyomial, R( x ( x 3 5x 9x Start by fidig the possible ratioal roots of R( x. factors of 5 1, 5 = +. factors of 1 =± ± We kow there are o more egative zeros so possible ratioal roots are 1, 5. 3 Use Remaider Theorem. ( ( ( ( x = 1 is aother root so ( x 1 is aother factor. Use sythetic divisio to reduce R( x. 3 R( x = ( x 5x + 9x 1 1-5 9-5 1-4 5 1-4 5 0 R 1 = 1 5 1 + 9 1 5= 0 #7 Now, we have P( x = ( x+ 3( x 1( x 4x+ Use the quadratic formula to fid the roots for ( x x+ b± b ac ( 4 ± ( 4 4(1( ± 4 4 16 0 x = = = a (1 4± 4 4± 4 1 4± i x= = = = ± i Cojugate pair of roots at ( + i ad ( i So we have two liear factors: ( + ad They ca also be writte as: ( x i ad ( x + i #8 Px ( ( x 3( x 1( x i( x i ( x i ( x ( i = + + Doe! J The root is egative ad the bottom row sigs alterate so -3 is a lower boud, accordig to the Upper ad Lower-Boud Theorem. But, we already kew that because there is oly oe egative real root. 4 5 Here a = 1, b = -4, c = 5
Algebra, Polyomial ad Ratioal Fuctios Page 6 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Check: ( ( ( ( Px ( = x+ 3 x 1 x i x + i ( x 3( x 1 ( x i( x i = + + ( ( = x+ 3 x 1 x x+ xi x+ 4 i xi+ i i ( ( = x+ 3 x 1 x 4x+ 4 i ( ( = x+ x x x+ 3 1 4 4 ( 1 ( ( = x+ x x x+ 3 1 4 5 ( x 3 3( x 4x 5x x 4x ( x 3 3( x 5x 9x = + + + = + + 4 3 3 = x 5x + 9x 5x+ 3x 15x + 4 3 = x x x + x 6 15 7x 15 This is the same as the origial Px ( See Step #1. Here s the fuctio, graphed o a TI-83 Graphic Calculator. It s a 4 th 4 degree fuctio ad the x term has a positive coefficiet so the ed behavior is ^Z. Real roots at x = 3 ad 1.