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solutions of the equation g(x) = 0. The equation g(x) = 0 has a third SUIuiU of f(x) = 0; what is this third solution? 6.31 Let f(x) = x2 -3x + 3 and g(x) 3x-5 6.32 Show that if a polynomial f(x) leaves a remainder of the form px + q when it is divided by (x-a)(x-b)(x- c), where a, b, and c are all distinct, then (b-c)f (a) + (c- a)f(b) + (a - b)f(c) 0. Hints: 20 6.33 How can we use synthetic division to divide x- 3x +4x- 11x - 9 by x-3x + 2? Hints: 162 6.34* When P(x) = x81 + Lx57 + G41 + Hx19+2x +1 is divided by x - 1, the remainder is 5, and when P(x) is divided by x - 2, the remainder is -4. However, x Lx Gx Hx Kx R is exactly divisible by (x-1)(x-2). If L, G, H, K, and R are real, compute the ordered pair (K, R). (Source: NYSML) 19 6.35 Let P(x) = (x - 1)(x - 2)(x - 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P(Q(x)) = P(x). R(x)? (Source: AMC 12) Hints: 231, 356, 224 6.36* Find the remainder when the polynomial x81 x49 +x25 +x is not the exact same problem as Problem 6.19! Hints: 317 +x is divided by x + x. Note: This 6.37* Find a polynomial f(x) of degree 5 such that f(x) - 1 is divisible by (x - 1)s and f(x) is itself divisible by r. Hints: 286, 267 Extra! The ancient Greeks made many geometric discoveries, but their algebraic accomplish- ments were much more limited. One of the main reasons they didn't make much progress with algebra is that they had not developed sophisticated notation like the notation we now use. Indeed, much of their algebraic reasoning was done in geometric terms, such as the following translation of Euclid's Elements: To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to the given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of a the defect

solutions of the equation g(x) = 0. The equation g(x) = 0 has a third SUIuiU of f(x) = 0; what is this third solution? 6.31 Let f(x) = x2 -3x + 3 and g(x) 3x-5 6.32 Show that if a polynomial f(x) leaves a remainder of the form px + q when it is divided by (x-a)(x-b)(x- c), where a, b, and c are all distinct, then (b-c)f (a) + (c- a)f(b) + (a - b)f(c) 0. Hints: 20 6.33 How can we use synthetic division to divide x- 3x +4x- 11x - 9 by x-3x + 2? Hints: 162 6.34* When P(x) = x81 + Lx57 + G41 + Hx19+2x +1 is divided by x - 1, the remainder is 5, and when P(x) is divided by x - 2, the remainder is -4. However, x Lx Gx Hx Kx R is exactly divisible by (x-1)(x-2). If L, G, H, K, and R are real, compute the ordered pair (K, R). (Source: NYSML) 19 6.35 Let P(x) = (x - 1)(x - 2)(x - 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P(Q(x)) = P(x). R(x)? (Source: AMC 12) Hints: 231, 356, 224 6.36* Find the remainder when the polynomial x81 x49 +x25 +x is not the exact same problem as Problem 6.19! Hints: 317 +x is divided by x + x. Note: This 6.37* Find a polynomial f(x) of degree 5 such that f(x) - 1 is divisible by (x - 1)s and f(x) is itself divisible by r. Hints: 286, 267 Extra! The ancient Greeks made many geometric discoveries, but their algebraic accomplish- ments were much more limited. One of the main reasons they didn't make much progress with algebra is that they had not developed sophisticated notation like the notation we now use. Indeed, much of their algebraic reasoning was done in geometric terms, such as the following translation of Euclid's Elements: To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to the given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of a the defect

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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solutions of the equation g(x) = 0. The equation g(x) = 0 has a third SUIuiU of f(x) = 0; what is this third solution? 6.31 Let f(x) = x2 -3x + 3 and g(x) 3x-5 6.32 Show that if a polynomial f(x) leaves a remainder of the form px + q when it is divided by (x-a)(x-b)(x- c), where a, b, and c are all distinct, then (b-c)f (a) + (c- a)f(b) + (a - b)f(c) 0. Hints: 20 6.33 How can we use synthetic division to divide x- 3x +4x- 11x - 9 by x-3x + 2? Hints: 162 6.34* When P(x) = x81 + Lx57 + G41 + Hx19+2x +1 is divided by x - 1, the remainder is 5, and when P(x) is divided by x - 2, the remainder is -4. However, x Lx Gx Hx Kx R is exactly divisible by (x-1)(x-2). If L, G, H, K, and R are real, compute the ordered pair (K, R). (Source: NYSML) 19 6.35 Let P(x) = (x - 1)(x - 2)(x - 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P(Q(x)) = P(x). R(x)? (Source: AMC 12) Hints: 231, 356, 224 6.36* Find the remainder when the polynomial x81 x49 +x25 +x is not the exact same problem as Problem 6.19! Hints: 317 +x is divided by x + x. Note: This 6.37* Find a polynomial f(x) of degree 5 such that f(x) - 1 is divisible by (x - 1)s and f(x) is itself divisible by r. Hints: 286, 267 Extra! The ancient Greeks made many geometric discoveries, but their algebraic accomplish- ments were much more limited. One of the main reasons they didn't make much progress with algebra is that they had not developed sophisticated notation like the notation we now use. Indeed, much of their algebraic reasoning was done in geometric terms, such as the following translation of Euclid's Elements: To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to the given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of a the defect
Transcribed Image Text:solutions of the equation g(x) = 0. The equation g(x) = 0 has a third SUIuiU of f(x) = 0; what is this third solution? 6.31 Let f(x) = x2 -3x + 3 and g(x) 3x-5 6.32 Show that if a polynomial f(x) leaves a remainder of the form px + q when it is divided by (x-a)(x-b)(x- c), where a, b, and c are all distinct, then (b-c)f (a) + (c- a)f(b) + (a - b)f(c) 0. Hints: 20 6.33 How can we use synthetic division to divide x- 3x +4x- 11x - 9 by x-3x + 2? Hints: 162 6.34* When P(x) = x81 + Lx57 + G41 + Hx19+2x +1 is divided by x - 1, the remainder is 5, and when P(x) is divided by x - 2, the remainder is -4. However, x Lx Gx Hx Kx R is exactly divisible by (x-1)(x-2). If L, G, H, K, and R are real, compute the ordered pair (K, R). (Source: NYSML) 19 6.35 Let P(x) = (x - 1)(x - 2)(x - 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P(Q(x)) = P(x). R(x)? (Source: AMC 12) Hints: 231, 356, 224 6.36* Find the remainder when the polynomial x81 x49 +x25 +x is not the exact same problem as Problem 6.19! Hints: 317 +x is divided by x + x. Note: This 6.37* Find a polynomial f(x) of degree 5 such that f(x) - 1 is divisible by (x - 1)s and f(x) is itself divisible by r. Hints: 286, 267 Extra! The ancient Greeks made many geometric discoveries, but their algebraic accomplish- ments were much more limited. One of the main reasons they didn't make much progress with algebra is that they had not developed sophisticated notation like the notation we now use. Indeed, much of their algebraic reasoning was done in geometric terms, such as the following translation of Euclid's Elements: To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to the given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of a the defect
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