Download Practice Test (page 201) 1. A. This is not true because 64 has these

Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 3
Factors and Products
Practice Test
(page 201)
1. A. This is not true because 64 has these factors: 1, 2, 4, 8, 16, 32, and 64
So, A is the correct answer.
2. Expand each product until the trinomial matches the given trinomial.
A. (2x + 1)(x + 6) = 2x2 + 12x + 1x + 6
= 2x2 + 13x + 6
B. (2x + 2)(x + 3) = 2x2 + 6x + 2x + 6
= 2x2 + 8x + 6
C. (2x + 3)(x + 2) = 2x2 + 4x + 3x + 6
= 2x2 + 7x + 6
This trinomial matches the given trinomial, so C is the correct answer.
3. 20 = 2 · 2 · 5
45 = 3 · 3 · 5
50 = 2 · 5 · 5
The least common multiple is the product of the greatest prime factor in each set of factors:
22 · 32 · 52 = 900
The least common multiple is 900.
The greatest common factor is the factor that occurs in every set of factors: 5
The greatest common factor is 5.
4. a) i)
For a perfect square; its prime factors can be arranged into 2 equal groups.
Use the factors from question 3.
20 = 2 · 2 · 5
Multiply 20 by 5 to get 2 equal groups of factors: 2 · 2 · 5 · 5
20 · 5 = 100
100 is a perfect square.
To get more perfect squares, multiply 20 by 5 times any perfect square, such as
20(5)(4) to get 400; and 20(5)(9) to get 900.
45 = 3 · 3 · 5
Multiply 45 by 5 to get 2 equal groups of factors: 3 · 3 · 5 · 5
45 · 5 = 225
225 is a perfect square.
To get more perfect squares, multiply 45 by 5 times any perfect square, such as
45(5)(4) to get 900; and 45(5)(9) to get 2025.
50 = 2 · 5 · 5
Multiply 50 by 2 to get 2 equal groups of factors: 2 · 2 · 5 · 5
50 · 2 = 100
100 is a perfect square.
To get more perfect squares, multiply 50 by 2 times any perfect square, such as
50(2)(4) to get 400; and 50(2)(9) to get 900.
Practice Test
Copyright  2011 Pearson Canada Inc.
1
Pearson
Foundations and Pre-calculus Mathematics 10
ii)
Chapter 3
Factors and Products
For a perfect cube, its prime factors can be arranged in 3 equal groups.
Use the factors from question 3.
20 = 2 · 2 · 5
Multiply 20 by 2 · 5 · 5 = 50 to get 3 equal groups of factors: 2 · 2 · 2 · 5 · 5 · 5
20 · 50 = 100
1000 is a perfect cube.
To get more perfect cubes, multiply 20 by 50 times any perfect cube, such as
20(50)(8) to get 8000; and 20(50)(27) to get 27 000.
45 = 3 · 3 · 5
Multiply 45 by 3 · 5 · 5 = 75 to get 3 equal groups of factors: 3 · 3 · 3 · 5 · 5 · 5
45 · 75 = 3375
3375 is a perfect cube.
To get more perfect cubes, multiply 45 by 75 times any perfect cube, such as
45(75)(8) to get 27 000; and 45(75)(27) to get 91 125.
50 = 2 · 5 · 5
Multiply 50 by 2 · 2 · 5 = 20 to get 3 equal groups of factors: 2 · 2 · 2 · 5 · 5 · 5
50 · 20 = 1000
1000 is a perfect cube.
To get more perfect cubes, multiply 50 by 20 times any perfect cube, such as
50(20)(8) to get 8000; and 50(20)(27) to get 27 000.
b) There is more than one answer in each of part a because a perfect square can be generated
by multiplying any two perfect squares; and a perfect cube can be generated by
multiplying any 2 perfect cubes.
5. a) (2c + 5)(3c + 2)
Use algebra tiles to make a rectangle with width 2c + 5 and length 3c + 2.
The tiles that form the product are: 6 c2-tiles, 19 c-tiles, and ten 1-tiles
So, (2c + 5)(3c + 2) = 6c2 + 19c + 10
Practice Test
Copyright  2011 Pearson Canada Inc.
2
Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 3
Factors and Products
b) (9 + 4r)(8 + 6r)
Sketch a rectangle with dimensions 8 + 6r and 9 + 4r. Divide it into 4 smaller rectangles
and calculate the area of each.
From the diagram:
(9 + 4r)(8 + 6r) = 72 + 54r + 32r + 24r2
= 72 + 86r + 24r2
c) (4t – 5)(3t + 7)
Sketch a rectangle. Label its dimensions 4t – 5 and 3t + 7. Divide it into 4 smaller
rectangles and label each one.
From the diagram:
(4t – 5)(3t + 7) = 12t2 + 28t – 15t – 35
= 12t2 + 13t – 35
6. Use the distributive property.
a) (2p – 1)(p2 + 2p – 7) = 2p(p2 + 2p – 7) – 1(p2 + 2p – 7)
= 2p3 + 4p2 – 14p – p2 – 2p + 7
= 2p3 + 4p2 – p2 – 14p – 2p + 7
= 2p3 + 3p2 – 16p + 7
b) (e + 2f)(2f2 + 5f + 3e2) = e(2f2 + 5f + 3e2) + 2f(2f2 + 5f + 3e2)
= 2ef2 + 5ef + 3e3 + 4f3 + 10f2 + 6e2f
c) (3y + 2z)(y + 4z) – (5y – 3z)(2y – 8z)
= 3y(y + 4z) + 2z(y + 4z) – [5y(2y – 8z) – 3z(2y – 8z)]
= 3y2 + 12yz + 2yz + 8z2 – [10y2 – 40yz – 6yz + 24z2]
= 3y2 + 12yz + 2yz + 8z2 – 10y2 + 40yz + 6yz – 24z2
= 3y2 – 10y2 + 12yz + 2yz + 40yz + 6yz – 24z2 + 8z2
= –7y2 + 60yz – 16z2
7. a) f2 + 17f + 16
Find two numbers whose sum is 17 and whose product is 16.
The numbers are 1 and 16.
Practice Test
Copyright  2011 Pearson Canada Inc.
3
Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 3
Factors and Products
So, f2 + 17f + 16 = (f + 1)(f + 16)
I could use these algebra tiles: 1 f2-tile, 17 f-tiles, and sixteen 1-tiles to make a rectangle
with length f + 16 and width f + 1.
b) c2 – 13c + 22
Find two numbers whose sum is –13 and whose product is 22.
Since the constant term is positive and the c-term is negative, the numbers are negative.
The numbers are –2 and –11.
So, c2 – 13c + 22 = (c – 2)(c – 11)
I could use these algebra tiles: 1 c2-tile, 13 negative c-tiles, and twenty-two 1-tiles to
make a rectangle with length c – 11 and width c – 2.
c) 4t2 + 9t – 28
Use decomposition.
Multiply: 4(–28) = –112
Find factors of –112 that have a sum of 9.
Factors of –112 are: 1 and –112; –1 and 112; 2 and –56; –2 and 56; 4 and –28; –4 and 28;
7 and –16; –7 and 16
The factors of –112 that have a sum of 9 are –7 and 16.
So, 4t2 + 9t – 28 = 4t2 – 7t + 16t – 28
= t(4t – 7) + 4(4t – 7)
= (4t – 7)(t + 4)
I would not use algebra tiles to factor. I would need to use guess and check to find a
combination of positive and negative t-tiles to form a rectangle with 4 t2-tiles, and 28
negative 1-tiles.
d) 4r2 + 20rs + 25s2
This is a perfect square trinomial because:
the 1st term is a perfect square: 4r2 = (2r)(2r)
the 3rd term is a perfect square: 25s2 = (5s)(5s)
and the 2nd term is: 20rs = 2(2r)(5s)
So, 4r2 + 20rs + 25s2 = (2r + 5s)(2r + 5s), or (2r + 5s)2
I could not use algebra tiles to factor the given trinomial because I do not have tiles for
more than one variable. I could use tiles to factor 4r2 + 20r + 25, then include the variable
s when I write the factors.
e) 6x2 – 17xy + 5y2
Use decomposition.
Multiply: 6(5) = 30
Find factors of 30 that have a sum of –17.
Since the coefficient of y2 is positive and the xy-term is negative, the numbers are
negative.
List negative factors of 30: –1 and –30; –1 and –30; –2 and –15
The factors of 30 that have a sum of –17 are –2 and –15.
So, 6x2 – 17xy + 5y2 = 6x2 – 2xy – 15xy + 5y2
= 2x(3x – y) – 5y(3x – y)
= (3x – y)(2x – 5y)
I could not use algebra tiles to factor the trinomial because I do not have tiles for more
than one variable.
Practice Test
Copyright  2011 Pearson Canada Inc.
4
Pearson
Foundations and Pre-calculus Mathematics 10
Chapter 3
Factors and Products
f) h2 – 25j2
This is a difference of squares.
h2 = (h)(h)
25j2 = (5j)(5j)
So, h2 – 25j2 = (h + 5j)(h – 5j)
I could not use algebra tiles to factor the binomial because I do not have tiles for more
than one variable.
8. The remaining volume is the difference between the volume of the cube and the volume of
the prism.
The volume of the cube is: (2r + 1)3
The volume of the prism is: r(r)(2r + 1)
The remaining volume is:
(2r + 1)3 – r(r)(2r + 1) = (2r + 1)(2r + 1)2 – [r2(2r + 1)]
= (2r + 1)(4r2 + 4r + 1) – [2r3 + r2]
= 2r(4r2 + 4r + 1) + 1(4r2 + 4r + 1) – 2r3 – r2
= 8r3 + 8r2 + 2r + 4r2 + 4r + 1 – 2r3 – r2
= 8r3 – 2r3 + 8r2 + 4r2 – r2 + 2r + 4r + 1
= 6r3 + 11r2 + 6r + 1
The volume that remains is: 6r3 + 11r2 + 6r + 1
9. All the trinomials that begin with 8t2 and end with +3 have t-terms with coefficients that are
the sum of the factors of 8(3) = 24.
The factors of 24 are: 1 and 24; –1 and –24; 2 and 12; –2 and –12; 3 and 8; –3 and –8;
4 and 6; –4 and –6
The sums of the factors are: 25, –25; 14, –14; 11; –11; 10; –10
So, the possible trinomials are:
8t2 + 25t + 3; 8t2 – 25t + 3; 8t2 + 14t + 3; 8t2 – 14t + 3; 8t2 + 11t + 3; 8t2 – 11t + 3;
8t2 + 10t + 3; 8t2 – 10t + 3
I have found all the trinomials because there are no other pairs of factors of 24.
Practice Test
Copyright  2011 Pearson Canada Inc.
5