review 1 - polynomial functions - Review Questions Polynomial Functions Multiple Choice 1 10 ll 12 13 The minimum information required to determine the

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Unformatted text preview: Review Questions — Polynomial Functions Multiple Choice 1. 10. ll. 12. 13. The minimum information required to determine the equation of a cubic function is: a all the zeros b. all the zeros and another point c. all the zeros, another point, and the sign of the ﬁrst coefﬁcient d. all the zeros, another point, the sign of the first coefﬁcient, and the degree of the polynomial The equation ﬁx) = k (x — SK): — 3:12 represents a cubic function: a. with two zeros c. with no turning point 13. with three zeros d. with three turning points . x5 + 5x + 2x3 + 3x2 + 4x4 + 6 arranged in descending order is: a. 6+5x+3xz+2x3+4x4+3:5 c. x5+4x4+2x3+3x2+5x+6 b. —(6 + 5x + 3x2 + 2x3 + 4x4 + x5) d. all of the above The division statement is: a. dividend x quotient + remainder = divisor b- divisor &gt;&lt; quotient + remainder = dividend 0- divisor &gt;&lt; dividend + remainder = quotient d. none of the above , A division has remainder 0. Which statement is true? a. The divisor and quotient are factors of the dividend. b. The divisor and remainder are factors of the quotient. c. The quotient and the dividend are factors of the divisor. d. The quotient and remainder are factors of the dividend. The divisor is x -« 3, the quotient is x + 2, and the remainder is l. The dividend is: a. x2—6 c. xz—x—S b. x2 — x — 6 d. 2x x + 2 is a factor of: a. x2+4x+4 C. x3+2x2—x—2 b. x2 — 4 d. all of the above A factor of x4 — 5x2 + 4 is: ' a. x — 2 C- 'a and b b. x w l d. neither a nor b The remainder is 1, the quotient is 2x + 3, and the dividend is 2x2 + 7x + 7. The divisor is: a. x~7 C. 2(x+7)+7 b. x+2 d. 2x2+7(x+1) The factors of x3 —- y3 are: a- ()6 ~y)(x2 + xy + yz) 0- (x ~y)(x + y)2 b- (x +y)(x2 — xy + yz) d- ()6 +y)(x —y)2 Which is a true statement regarding the factor theorem? a. x — k is a factor of ﬁx) if and only if f(k) = 0 b. if x — b is a factor of g(x), then g(b) = 0 c. if Me) = 0, then x w c is a factor of h(x) d. all of the above Which is a true statement regarding the remainder theorem? a. when ﬁx) is divided by x — k, then ﬂk) = the remainder b. when g(x) is divided by jx — k, then g(j/k) = the remainder 0. a and b d. neither 21 nor b The roots of (x + l)():2 — x — 6) = O are: a. —1,1,—6,0 0: 4,—2.3 14. 15. 16. 17. 18. 19. 20. 13. 1,-1,—6 d. l,2,—3 The root(s) of x3 = 8 is (are): a. 2 c. 2, 2i, —2i b. 2, —2 d. 2, —2, 21' The roots of 2x3 = 18x are: a. 0, 2, 18 0. 0,2, 3, —3 b. 0, 2, —1 8 d. 0, 3, —3 All the roots of a polynomial equation are 0, 1, 2, and 3. The general equation is: a. k(x)(x+1)(x+2)(x+3)=0 c. k(x+ l)(x+2)(x+3)=0 b. k(x)(x — 1)(x - 2)(x — 3) = 0 d. k(x + 1)(x —- 2)(x - 3) = 0 The solution to (x — 5)(x + 5) &gt; O is: a. [xl&gt;5 I C. —5&lt;x&lt;5 b. [x|&lt;5 d- —5£x35 The solution to (x2 ~ 1) S 0 is: a- 1x1&gt;1 c. lx[&lt;1 b» 1x121 d. mgr The solution to (x - 2)(x + 1) &gt; 0 is: a. x &lt; —1 C. x &gt; 2 b. —1&lt;x&lt;2 d. x&lt;—l andx&gt;2 The cubic function is greater than the linear function when: —3.5 &lt;x &lt;—0.5 andx&gt; 3 a. x&lt;—2 andx&gt; 1.43 c. b. —2&lt;x&lt;l.43 d. x&lt;—3.5 and—0.5&lt;x&lt;3 21. ﬁx) &lt; 0 where the graph of ﬁx) is: i a. above the x—axis c. between its zeros 13. below the x—axis d. outside its zeros Short Answer 22. Describe the interval(s) where each of the three functions is increasing. I 23.Match each graph with a description. a) increasing linear function b) constant linear function c) quadratic function with a local minimum d) quadratic function with a maximum e) cubic function with positive ﬁrst coefﬁcient f) cubic function with negative ﬁrst coefﬁcient 24. Describe the difference between a local maximum and a local maximum value. 25. Describe the behaviour of a function around a local minimum point. 26. Write f(x) = —2(x— 1)2(x+ 3) in expanded form. 27. Sketch the graph of ﬁx) = (x — 3)(x + 2)(x + 5) using the zeros and end behaviours. 28. Determine the remainder r for (x + l)(x —- 2) + r = x2 —x + 7. 29. a) Divide x4 + x3 + 2x2 + l by x + 1 using synthetic division. b) Write the division statement. 30. st—l afactor ofx3 +2x2—6x+3? 31. Determine the remainder when x3 + 23c2 — 6x + l is divided by x + 2. 32. Use long division to divide 2x3 + 5x2 — 4x — 5 by 2x + l. 33. Use synthetic division to divide 2x3 + 5x2 — 4x — 5 by 2x + l 34. Factor x3 - 5x2 — x + 5. 35. Factor 4x3+4x2—x— l. 36. Factor x3 — 1000. 37. Factor x4 — 7x2 — 6x fully. 38. Solvex3—3x2—x+3=0. 39. Algebraically solve x(2x + l)(x — 4) &gt; 0. 40. Determine the equation of the function with zeros at :1, —2, and y—intercept of ~6. Sketch the function using this information. 41. a) Without dividing, calculate the remainder when 4x4 —x3 + 2x2 — l is divided by 2x — l. b) Is 2x —1 a factor of4x4 ~x3 + 2x2 —— l? 42. Solve x3 — 3x2 + 5x — 3 = 0. Review Questions - Polynomial Functions (Units 1&amp;2) Answer Section MULTIPLE CHOICE 1. B 2. A 3. C 4. B 5. A 6. C 7. D 8. C 9. B 10. A 11. D 12. A 13. C 14. C 15. D 16. B 17. A 18. D 19. D 20. C 21. B SHORT ANSWER 22. ﬁx) increases for xeR; g(x) increases for x &lt; 3; and h(x) increases for x &lt; —1 and x &gt; 2.5. ’23- ﬂx) - 3); 3’06) ~ d); 1106) - e) 24. A local maximum is a point at which a curve changes from increasing to decreasing, whereas the local maximum value is the y-coordinate of that point. 25. When a function changes from decreasing to increasing, a local minimum occurs. 26. f(x)=—2x3—2x2+IUx-ﬁ 27. Zerosare3,—2,and—5.k&gt;0:&gt;asx—&gt;oo,f(x)—&gt;oo 28. r = 9 29. a) x + 1 is the divisor :&gt; k = —1 1x4+1x3+2x2+0x+1 i i J, J, i (remove literal coefﬁcients) 1 l 2 0 l i :1 Q —_2 Z (multiply number at lower left by k) 1 0 2 3 (add) if it i 12 i (insert literal coefﬁcients) 1x3 + 0x2 + 2x— 2 remainder 3 b) (x+ 1)(1x3+2x—2)+3 =x4 +x3+2x2 +1 30. yes 31. The remainder is 13. 2 x +2x—3 ) 3 2 32. 2x+1 2x +5x —4x—5remainder~2 l 33. 2x+l isthedivisor:&gt;k=—— 2 29+ 5x2—4x—5 i J, \L i (remove literal coefﬁcient) 2 5 —4 —5 i —_l —_2 ; (multiply number at lower left by k) 2 4 6 ~2 (add) i J! J/ i (+ 2, coefﬁcient ofx in divisor) 1 2 —3 l i \L \L i (insert literal coefﬁcients) lx2 + 2x — 3 remainder —2 34. x3—5x2—x+5=(x—l)(xm5)(x+1) 35. 4x3 +4x2—x— 1 =(2x— l)(2x+ l)(x+ 1) 36. x3 — 1000 =(x—10)(x2 +10x +100) 37. x4—7x2~6x =x(x+ l)(x—3)(x+ 2) 38. x=—1,l,3 39. x(2x+l)(x—4)&gt;0 l The zeros of the corresponding equation are —— ,,0 and 4 —1_ sign of (2x + 1) sign 2of (x— 4) sign of product x(2x+1)(x~4)&gt;0when—-2- &lt;x&lt;0andx&gt;4. 40. The given zeros: f(x) = k(x — l)(x + l)(x + 2) y-intercept: ﬂO) = —6 ﬂO) = k(0 —1)(0 + l)(O + 2) —6 = k(-2) 3 = k ﬁx) = 3(x -— l)(x + l)(x + 2) is the required function. I 41. a) When ﬁx) is divided by 2x — 1, the remainder = ({5} ﬁx)=4x4—x3+2x2—1 [I] [I] [I] [I] 1 1 I 3 f— =4 — — - +2 — —1=——-—+-—1=—— 2 16 8 4 4 8 2 8 3 Remainder = _E . 4 3 2 1 b)If2x—1 isafactor0f4x —x +2x —1,thenf :2: = 1 3 - 4 3 2 However,f E =—E,502x—1Isnotafactorofélx —x +2x ——1. 42. f(x)=x3—3x2+5x—3 f(1)=1—3+5—3=0=&gt;x—1isafactor x2—2x+3 x— 1 ix} — 3x:’1 + 535— 3 remainder 0 (x—1)(x2—2x+3)=0 x—1=00rx2—2x+3=0 4:: «fez—4m: 2a i X=10TX=—““—&quot;—&quot; «hi—2f —4t1&gt;(3&gt; _——_.____—_ 2(1) I! ._.. 1+ N x=1,1+z’ 2 ...
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