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COLLEGE ALGEBRA for ENGINEERING (ENGALG1) REVIEWER
Quiz Number: 1
I.
Multiple Choice
1. Which of the following is NOT a polynomial?
a. 2 + 5x2 – 3x
b. 9x6
c. √𝑥 2 + 2𝑥 + 2
d.
8
𝑥2
1
+ + 5x + 2
𝑥
2. The property of addition/multiplication in which the order of the addends/factors to be added/multiplied
is not important is called _______?
a. Associative Property
b. Closure Property
c. Commutative Property
d. Distributive Property
3. Which of the following is in factored form?
a. (5x + 3)x – 2(4x – 7)
b. 5y(2y – 5) – (7y + 2)(2y + 7)
c. (x – 2)(4x + 7)(5x + 3)
d. (2y + 7)(7y – 2)2y + 5
4. The variable x in
a.
b.
c.
d.
𝒂𝒎
𝒂𝒏
= 𝒂𝒙 will be positive if:
m>n
m<n
m≥n
m≤n
5. The factors of 9x2 – 12x – 45 are:
a. 3(x + 3)(3x + 5)
b. 3(x + 3)(3x – 5)
c. 3(x – 3)(3x + 5)
d. 3(x – 3)(3x – 5)
“We are built to build”
II.
1. Subtract the product of x2m + 2xmym – y2m and xm – ym from the product of xm + ym and x2m – y2m.
2. Divide 3x3y + 6x4 – 25y2 + 5x2y by 3x2 – 5y.
3. Remove the grouping symbols then simplify:
– 2xy {6x2y + [– 6x3 – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2}
III.
1.
2.
3.
4.
5.
Factor the algebraic expressions completely.
40x8zy2 + 625x2zy3m+2
9a2 + 25b2 – 4c2 – 9d2 – 30ab + 12cd
4x4 – 61x2y2 + 9y4
90x2 – 120xy + 40y2 – 21x + 14y – 12
(x + 3y)3 – x3 + 9xy2
Parts of the reviewer were obtained from
ENGALG1 Quiz Number 1 of Ms. W. Baraoidan
AY 2014 – 2015 Term 1
“We are built to build”
ANSWERS & SOLUTIONS
COLLEGE ALGEBRA for ENGINEERING (ENGALG1) REVIEWER
Quiz Number: 1
I.
Multiple Choice
8
1
1. d. 𝑥 2 + 𝑥 + 5x + 2
2.
3.
4.
5.
c. Commutative Property
c. (x – 2)(4x + 7)(5x + 3)
a. m > n
c. 3(x – 3)(3x + 5)
II.
1. Subtract the product of x2m + 2xmym – y2m and xm – ym from the product of xm + ym and x2m – y2m.
x2m + 2xmym – y2m
xm – ym
𝑥 3𝑚 + 2𝑥 2𝑚 𝑦 𝑚 − 𝑥 𝑚 𝑦 2𝑚
− 𝑥 2𝑚 𝑦 𝑚 − 2𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚
𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 3𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚
xm + ym
x2m – y2m
𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚
𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚
− 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚
− 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚
(xm + ym) (x2m – y2m)
= 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 𝑥 𝑚 𝑦 2𝑚 − 𝑦 3𝑚
− (x2m + 2xmym – y2m)(xm – ym) = 𝑥 3𝑚 + 𝑥 2𝑚 𝑦 𝑚 − 3𝑥 𝑚 𝑦 2𝑚 + 𝑦 3𝑚
=
2𝒙𝒎 𝒚𝟐𝒎 − 𝟐𝒚𝟑𝒎
2. Divide 3x3y + 6x4 – 25y2 + 5x2y by 3x2 – 5y.
𝟓𝒙𝒚𝟐
6x4 + 3x3y + 5x2y – 25y2 ÷ 3x2 – 5y = 2x2 + xy + 5y + 𝟑𝒙𝟐 – 𝟓𝐲
“We are built to build”
3. Remove the grouping symbols then simplify:
– 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2}
=
=
=
=
=
=
=
=
=
III.
– 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – (3x – 2y)] – 3x2 + 2x2y}] – 3x2}
– 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – x[3x2y – 3x + 2y] – 3x2 + 2x2y}] – 3x2}
– 2xy {6x2y + [– 6x3y – 2x2y – 2{2xy – 3x3y + 3x2 – 2xy – 3x2 + 2x2y}] – 3x2}
– 2xy {6x2y + [– 6x3y – 2x2y – 2{– 3x3y + 2x2y}] – 3x2}
– 2xy {6x2y + [– 6x3y – 2x2y + 6x3y – 4x2y] – 3x2}
– 2xy {6x2y + [– 6x3y – 6x2y + 6x3y] – 3x2}
– 2xy {6x2y + [– 6x2y] – 3x2}
– 2xy {6x2y – 6x2y – 3x2}
– 2xy {– 3x2}
6x3y
Factor the algebraic expressions completely.
1. 40x8zy2 + 625x2zy3m+2
= 5x2zy2 (8x6z + 125y3m)
= 5x2zy2 (2x2z + 5ym)(4x4z – 10x2zym + 25y2m)
2. 9a2 + 25b2 – 4c2 – 9d2 – 30ab + 12cd
= 9a2 – 30ab + 25b2 – 4c2 + 12cd – 9d2
= (9a2 – 30ab + 25b2) – (4c2 – 12cd + 9d2)
= (3a – 5b)2 – (2c – 3d)2
= [3a – 5b + (2c – 3d)][3a – 5b – (2c – 3d)]
= (3a – 5b + 2c – 3d)(3a – 5b – 2c + 3d)
3. 4x4 – 61x2y2 + 9y4
4x4 = (2x2)2
9y4 = (3y2)2
 (2x2)( 3y2) = 12x2y2
= 4x4 – 61x2y2 + 49x2y2+ 9y4 – 49x2y2
= (4x4 – 12x2y2 + 9y4) – 49x2y2
= (2x2 – 3y2)2 – 49x2y2
= (2x2 – 3y2)2 – (7xy)2
= (2x2 – 3y2 + 7xy) (2x2 – 3y2 – 7xy)
= (2x2 + 7xy – 3y2) (2x2 – 7xy – 3y2)
Take out common factors
Pattern: (x3 + y3) = (x + y)(x2 – xy + y2)
Factoring by Grouping
Perfect Square Trinomial
Pattern: a2 – b2 = (a +b)(a – b)
Middle Term of Resultant Perfect Square Binomial
Add and subtract a perfect square monomial
Pattern: a2 – b2 = (a +b)(a – b)
“We are built to build”
4. 90x2 – 120xy + 40y2 – 21x + 14y – 12
= (90x2 – 120xy + 40y2) – (21x – 12y) – 12
= 10(9x2 – 12xy + 4y2) – 7(3x – 2y) – 12
= 10(3x – 2y)2 – 7(3x – 2y) – 12
= [2(3x – 2y) – 3] [5(3x – 2y) + 4]
= (6x – 4y – 3)(15x – 10y + 4)
Factoring by Grouping
Take out common factors
Perfect Square Trinomial
5. (x + 3y)3 – x3 + 9xy2
(Instruction: Do not expand)
3
3
2
= (x + 3y) – (x – 9xy )
Factoring by Grouping
3
2
2
= (x + 3y) – x(x – 9y )
Pattern: a2 – b2 = (a +b)(a – b)
= (x + 3y)3 – x(x + 3y)(x – 3y)
Take out common factors
2
= (x + 3y)[(x + 3y) – x(x – 3y)]
= (x + 3y)[x2 + 6xy + 9y2 – x2 + 3xy]
= (x + 3y)[9xy + 9y2]
= 9y(x + y)(x + 3y)
Prepared by: April Oilec Deanielle B. Pacayra
Academics Officer
Civil Engineering Society
“We are built to build”