Rational Expressions Assignments - 2

Assignment 17 Represent the trinomial as a sum of two members, the one of which is a square of binomial:
A) x2 + 8x + 25
B) 4x2 – 4x + 10
C) 16x2 – 24x + 30
D) 81x2 + 6x + 32
Solution:
A) x2 + 8x + 25 = x2 + 2.x.4 + 42 + 9 = (x + 4)2 + 9
B) 4x2 – 4x + 10 = (2x)2 – 2.2x.1 + 1 + 9 = (2x – 1)2 + 9
C) 16x2 – 24x + 30 = (4x)2 – 2.4x.3 + 9 + 21 = (4x – 3)2 + 21
D) 81x2 + 6x + 32 = (9x)2 + 2.9.1/3x + 1/9 + 31.8/9 = (9x + 1/3)2 + 287/3

Assignment 18 Expand to multipliers:
A) 4ax – 2ay
B) a2 + ab – a - b
C) 9x + 9y + ax + ay
D) 28a2b3 – 4ab4
E) a4 + 2a2 + 1
F) (a + b)4 – (a - b)4
Solution:
A) 4ax – 2ay = 2a(2x – y)
B) a2 + ab – a – b = a(a + b) – (a + b) = (a + b)(a – 1)
C) 9x + 9y + ax +ay = 9(x + y) + a(x + y) = (x + y)(9 + a)
D) 28a2b3 – 4ab4 = (4ab3)(7a – b)
E)a4 + 2a2 + 1 = (a2)2 + 2.(a2).1 + 1 = (a2 + 1)2
F) (a + b)4 – (a - b)4 = [(a + b)2]2 – [(a - b)2]2 = [(a+b)2 + (a - b)2] . [(a + b)2 - (a - b)2] = (a2 + 2ab + b2 + a2 - 2ab + b2).(a2 + 2ab + b2 - a2 + 2ab - b2) = (2a2 + 2b2)(4ab) = 2(a2 + b2).4ab = 8ab(a2 + b2)

Assignment 19 Simplify the following expression:
A) (x-3)2 – x(x + 9)
B) (2a + 5)2 – 5(4a + 5)
C) b2 + 49 – (b - 7)2
D) (y - 10)(y - 6) – (y - 8)2
Solution:
А) (x - 3)2 – x(x + 9) = x2 – 6x + 9 – x2 – 9x = 9 – 15x
B) (2a + 5)2 – 5(4a + 5) = (2a)2 + 2.2a.5 + 52 – 5.4a – 5.5 = 4a2 + 20a + 25 - 20a - 25 = 4a2
C) b2 + 49 – (b - 7)2 = b2 + 49 - (b2 - 2b.7 + 72) = b2 + 49 - b2 + 14b - 49 = 14b
D) (y - 10)(y - 6) – (y - 8)2 = yy - 6y - 10y + 60 - y2 + 16y - 64 = -4

Assignment 20 Prove the identity:
A) (-a - b)2 = (a + b)2
B) a3 - 3ab(a - b) - b3 = (a - b)3
C) (x - a)(x - b) = x2 - (a + b)x + ab
D) (a2 + b2)(c2 + d2) = (ac - bd)2 + (bc + ad)2
Solution: In order to prove an identity, we must perform the operations in the one of its sides and to establish that it is equal to the other side.
А) (-a - b)2 = [-1(a + b)]2 = (-1)2(a + b)2 = 1.(a + b)2 = (a + b)2
B) (a - b)3 = a3 - 3a2.b + 3a.b2 - b3 = a3 - 3ab(a - b) - b3 C) (x - a)(x - b) = xx - xb - ax + ab = x2 - x(a + b) + ab
D) (ac - bd)2 + (bc + ad)2 = a2.c2 - 2acbd + b2.d2 + b2.c2 + 2bcad + a2.d2 = a2(c2 + d2) + b2(d2 + c2) = (c2 + d2)(a2 + b2)

Assignment 21 Find the polynomial A, for which:
A) (a - b)2 + A = (a + b)2
B) x2 + x + 1 + A = (x - 1)2 + 1
C) 2z2 - 5z + 6 + A = (z + 1)2 + (z - 1)2
D) (z - 1)(z + 1) + A = (2z2) + 5
Solution: Since the polynomial A sought constitutes one of the addends in the sum given, we can find it if we subtract the other addend from the sum. In this way we will obtain:
A) A = (a + b)2 – (a - b)2 = a2 + 2ab + b2- (a2 - 2ab + b2) = a2 + 2ab + b2 – a2 +2ab – b2 = 4ab
B) A = (x - 1)2 + 1 - (x2 + x + 1) = x2 - 2x + 1 + 1 – x2 – x - 1 = 1 – 3x
C) A = (z + 1)2 + (z - 1)2 – (2z2 -5z + 6) = z2 + 2z + 1 + z2 - 2z + 1 - 2z2 + 5z - 6 = 2 + 5z - 6 = 5z - 4
D) A = 2z2 + 5 – (z - 1).(z + 1) = 2z2 + 5 – (z2 - 1) = 2z2 + 5 – z2 + 1 = z2 + 6

Assignment 22 The following expression is given: A = k – (x + 1)2
A) Find k, if the highest value of A is 4;
B) Substitute in А the value of k obtained in subcondition A) of this assignment and expand to multipliers the expression obtained.
Solution:
A) Since (x + 1)2 is greater or equal to 0 for whatever value of x, i.e. it is not a negative number, we will obtain the highest value of A, if we do not subtract anything from k, i.e. if (x + 1) = 0 Then the highest value of A = 4 according the condition A = k. Thus we found that k = 4
B) A = 4 – (x+1)2 = 4 - (x2 + 2x + 1) = 4 - x2 - 2x - 1 = 3 - x2 - 2x = 2 + 1 – x2 - 2x = (1 - x2) + 2(1 - x) = (1 - x)(1 + x) + 2(1 - x) = (1 - x)(1 + x + 2) = (1 - x)(x + 3)

Assignment 23 The following expression is given: x2 – 2kx + k +2
A) Find k, if the expressions, a value is 0 when x = 1
B) Substitute k with the number found and expand the expression obtained.
Solution:
A) Since when x = 1 the expression is equal to 0, then the equation 1 is valid x2 – 2k.1 + k + 2 = 0 consequently 3 – k = 0 consequently k = 3;
B) We substitute k with 3 and obtain the following x2 - 2.3.x + 3 + 2 = x2 – 6x + 5 = x2 – x - 5x +5 = x(x- 1)-5(x-1) = (x-1)(x-5)

Assignment 24 The following expression is given: A = [(2x – 3)2] + 5
A) Find the lowest value of A
B) Represent the following expression as a polynomial of x: A – (x – 3)2
C) Expand to multipliers: 9 – A
Answers: A) A = 5 B) 3x2 – 6x + 5 C) (2x – 1)(5 – 2x)

Assignment 25 Prove the identity:
A) (a2 + 2)2 – (a - 2)(a + 2)(a2 + 4) = 4(a2 + 5)
B) (a + b)3 = a(a - 3b)2 + b(b - 3a)2
C) 5(a + b)2 – 4a2 - 4ab = (a + b)(a + 5b)
Solution: We use the shortened/reduced multiplication formulas and obtain:
A) (a2 + 2)2 – (a - 2)(a + 2)(a2 + 4) = a4 + 4a2 + 4 – (a2 - 4)(a2 + 4) = a4 + 4a2 + 4 – (a4 -16) = 4a2 + 20 = 4(a2 + 5)
B) a(a - 3b)2 + b(b - 3a)2 = a(a2 - 6ab + 9b2) + b(b2 - 6ab + 9a2) = a3 – 6a2.b + 9a.b2 + b3 - 6ab2 + 9a2.b = a3 + 3a2.b +3ab2 + b3 = (a + b)3
C) 5(a + b)2 – 4a2 - 4ab = 5a2 + 10ab + 5b2 - 4a2 - 4ab = a2 + 6ab + 5b2 = a2 + ab + 5ab + 5b2 = a(a + b) + 5b(a + b) = (a + b)(a + 5b)

Assignment 26 Simplify the expression:
A) 4(a - 6) – a2.(2 + 3a) + a.(5a - 4) + 3a2.(a - 1)
B) (a2 - 1)2 – (a - 1)(a2 + 1)(a + 1)
C) (a + 1)3 + (a - 1)3 – 2a(a + 1)(a - 1)
D) (a + 5)3 – a(a - 5)2 -25(a + 1)2
Solution:
A) 4(a - 6) – a2.(2 + 3a) + a.(5a -4) + 3a2.(a - 1) = 4a -24 - 2a2 - 3a3 + 5a2 -4a + 3a3 - 3a2 = -24
B) (a2 - 1)2 – (a - 1)(a2 + 1)(a + 1) = a4 - 2a2 + 1 – (a - 1)(a + 1)(a2 +1) = a4 -2a2 + 1 – (a2 - 1)(a2 + 1) = a4 - 2a2 + 1 - (a4 - 1) = a4 - 2a2 + 1 – a4 + 1 = 2 - 2a2
C) (a + 1)3 + (a - 1)3 – 2a(a + 1)(a - 1) = a3 + 3a2 + 3a + 1 + a3 - 3a2 + 3a - 1 - 2a(a2 - 1) = 2a3 + 6a - 2a3 + 2a = 8a
D) (a + 5)3 – a(a - 5)2 - 25(a + 1)2 = a3 + 3a2.5 + 3a.52 + 53 – a(a2 - 10a + 25) - 25(a2 + 2a + 1) = a3 + 15a2 + 75a + 125 – a3 + 10a2 - 25a - 25a2 - 50a - 25 = (15 + 10 - 25)a2 + (75 - 25 - 50)a + 125 - 25 = 100

Assignment 27 Prove the identity:
A) (a + b + c)(a + b - c) = a2 + b2 – c2 + 2ab
B) (a2 – ab + b2)(a2 + ab + b2) = a4 + a2.b2 + b4
Solution:
A)(a + b + c)(a + b - c) = [(a + b) + c].[(a + b) – c] = (a + b)2 – c2 = a2 + 2ab + b2 – c2
B)(a2 – ab + b2)(a2 + ab + b2) = [(a2 + b2) – ab].[(a2 + b2) + ab] = (a2 + b2)2 – (ab)2 = a4 + b4 +2a2 . b2 – a2.b2 = a4 + b4 + a2.b2

Assignment 28 Expand to multipliers the following expression:
A) x2.y – y + x.y2 - x
B) x - 1 + 2ax - 2a – y + xy
C) 4x2 - 12xy + 9y2 - 4x + 6y
D) x2 + 6x – y2 - 4y +5
Solution:
A) x2.y – y + x.y2 – x = xy(x + y) – (x + y) = (x + y)(xy - 1)
B) x - 1 + 2ax - 2a – y + xy = (x-1) +2a(x - 1) + y(x - 1) = (x - 1)(1 + 2a + y)
C) 4x2 - 12xy + 9y2 - 4x + 6y = (4x2 -6xy) - 6xy + 9y2 -4x + 6y = 2x(2x -3y) – 3y(2x - 3y) - 2(2x - 3y) = (2x - 3y)(2x -3y -2)
D) x2 + 6x – y2 - 4y + 5 = x2 + 5x + x – y2 - 5y + y + 5 + xy – xy = x(x + y + 5) – y(x + y + 5) + x + y + 5 = (x + y + 5)(x – y + 1)

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