Question

Question #1, 3, 5, 7 & 9.

ero, then - i must also be Example 3. Let We could attempt to solve this problem in the same manner as we did problems in the previous si However, we can arrive at the solution much faster by using the given hint. Ifi is a zero, then-im a zero. The Division Algorithm and the Factor Theorem tell us that there is a polynomial, Q(x), such Rx) = 6 - Xr+DQ(x) = (x2 - 1(x) =(x + 1)(x) Dividingſ) by x! + 1 using long division will give us Ox) = x2 - 4x + 4. Nx) = (2+ 1)2 - 4x + 4) = (y2 + 16 - 2 The zeros are / and -1, each with a multiplicity of 1, and 2 with a multiplicity of 2. Section 3.5 Exercises Find a polynomial function of the specified degree with real coefficients that has the given numbers as its zeros. 1. 2-3i, 2 + 3i; degree 2 2. 1-2i, degree 2 3. 2,3 + i; degree 3 4. -3,1-1, degree 3 5. -1, 1, 2 + i; degree 4 6. 1 -21,3i; degree 4 Find the zeros: 7. x)=x*- 16 8. x) = xl-3x -4 9. Xx) x--4r - 6 10. Ax) = x+3x2 + 5x + 3 11. Ax) = x - 5x? + 8x - 6 (hint: 1 - 1 is a zero) 12. Ax)=x - 3x + 6r? - 12x + 8 (hint:21 is a zero) 13. Nx) = x + 2r-7X-18-18 (hint: -1 +i is a zero) 14. Ax) = x - 2r-x + 6x - 6 (hint: 1 +i is a zero) 15. Rx) + 4 (hint: 1 + i is a zero)

Answer 1

النوم2) -3) (3-1)-x) = (عا ( 2 - 3-X) (32+ 2 -ه) - + + +21- 22 - 23 - x2) - اند- و - . (را-ابرو- + لاہا ۔ م) = دا دا د 13 + ماد ما -« - Rool 2, 3 ti Third roots will be conjugate of a ti مدار ز- 3 -1 (: +3-) لذ-3-3) (2-3) - (1) ران) × (2) + 3-2- 2- 3 - 33 ه) - (A + 3-ه) ( 1 - 3 اله + 2 + مدخ- S2 - د ) - خلکد-+ 3- ) ولهذا + ماد xاعر) - کد 28-531-15 ده -خ32x2- 3دد نه + ۱۹ مها عندنا - 228 - 10 اده -3

,', ari awal also ad are لند2-) :- 2-1 ) (-ها(11) = ( )۲ دردسر ) (1-2) - [* ۹a - ۹ة) (۱) - - (1-1) (2-315) ا . 1 (5 + مداها (1-2) - که همه دكه ما له - وسدد بود + - کده -ه = ہے ااهه = (1) 0 = ماه م = لاد ( م 5 = (ه ها (2-10 (2+3) ذع - ,ذ2 2- ر2= 2 Zoo 6 = 6 -داد-د = (2) ه= 6-23+ مهما 25 ه .2 0= (3-داع لد-داد2 + (3-123 .و (2-3) (2+2+2) = 0 5=2+ادع + ردته ا- = 1 +د اد ا- - 2لا+1) ن = 1+1 -1- ,- 1- =لا خ - اسرا+1- ,3 = 2