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1. Simplify (Place answer in standard form): (8x 2 5) + (3x + 7) (2x 2 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed to each term Place in standard form Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test © 2007- 09 by S-Squared, Inc. All Rights Reserved.

1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

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Page 1: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

1. Simplify (Place answer in standard form):

(8x2 – 5) + (3x + 7) – (2x2 – 4x)

6x2 + 2 + 7x

6x2 + 7x + 2

NOTE: The subtraction must be distributed to each term

Place in standard form

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

© 2007- 09 by S-Squared, Inc. All Rights Reserved.

Page 2: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

2. Simplify (Place answer in standard form):

(2a3 – 6a + 7) ─ (5a2 ─ 2a + 7)

2a3 – 4a + 0

2a3 – 5a2 – 4a

NOTE: The subtraction must be distributed to each term

Place in standard form

– 5a2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 3: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

( )( )5x25x – 2–

10x –25x2 – 4 +10x

Combine

3. Simplify (Place answer in standard form):

25x2 – 20x + 4

NOTE: To square a binomial, you must multiply it by itself

(5x – 2)2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 4: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

4. Simplify (Place answer in standard form):

(m – 3)(7 – 2m2 + 5m)

– 15m

7m – 2m3 + 5m2

– 21+ 6m2 Combine like terms

– 21– 8m + 11m2– 2m3

– 21− 2m3 – 8m+ 11m2

Place in standard form

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 5: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

5. Given:

a) Simplify and put in standard form.

− 2t ( 3 + 7t2 )

− 6t – 14t3

− 14t3 – 6t Place in standard form

b) Identify the degree of the polynomial:

The degree is the largest exponent of the polynomial

3

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

− 2t ( 3 + 7t2 )

Page 6: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

5. Given:

c) Name the polynomial based on the degree.

d) Identify the type of polynomial based on the number of terms.

There are two terms in the polynomial

binomial

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

− 2t ( 3 + 7t2 )

cubic Since the polynomial is ofDegree 3

Page 7: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

6. a) Factor: x2 + 10x + 25 = 0

* Identify a, b and c a = 1 , b = 10 , c = 25 * You are looking for two numbers that do the following

Add up to give you b and multiply to give you c

(x + 5)(x + 5) = 0 Or (x + 5)2 = 0

Notice: 5 + 5 = 10

b) Use the zero product property to find the

solutions

and 5 • 5 = 25

(x + 5)(x + 5) = 0 Factored form

Zero Product Propertyx + 5 = 0 – 5– 5 Subtract

x = − 5

Page 8: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

6. Factor: x2 + 10x + 25 = 0

c) Check your solution x = − 5

x2 + 10x + 25 = 0 (− 5)2 + 10(− 5) + 25 = 0

25 – 50 + 25 = 0

− 25 + 25 = 0

0 = 0

Check

Equation

Simplify

Substitute

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 9: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

7. a) Factor: x2 – 5x – 24 = 0

* Identify a, b and c a = 1, b = − 5 , c = − 24

* You are looking for two numbers that do the following

Add up to give you b and multiply to give you c

Notice: 3 + (− 8) = − 5 and 3 • (− 8) = − 24

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

(x + 3)(x – 8) = 0

b) Use the zero product property to find the

solutions(x + 3)(x – 8) = 0 Factored form

Zero Product Propertyx + 3 = 0 and x – 8 = 0– 3– 3Subtract

x = − 3

+ 8+ 8 Add

x = 8

Page 10: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

8. a) Factor: x2 – 81 = 0

Difference of two perfect squares pattern

* a2 – b2 = (a ─ b)(a + b)

* Identify the a and the b by taking the square root of each term

Notice,

x2 x = a

9 81

= b

(x – 9)(x + 9) = 0

* Substitute into the difference of two perfect square pattern

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 11: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

8. Factor: x2 – 81 = 0

b) Use the zero product property to find the solutions:

(x – 9)(x + 9) = 0 Factored form

Zero Product Propertyx – 9 = 0 and x + 9 = 0

+ 9+ 9Add

x = 9

– 9– 9 Subtract

x = − 9

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 12: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

a) Factor out the common monomial

9. Given: 4x2 – 14x + 6 = 0

2(2x2 – 7x + 3) = 0

* The greatest common monomial is 2

Factor out a 2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 13: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

b) Factor the resulting trinomial

9. Given: 4x2 – 14x + 6 = 0

2(2x2 – 7x + 3) = 0

−6

* Identify a, b and c a = 2, b = − 7 , c = 3

* You are looking for two numbers that do the following

Add up to give you b and multiply to give you a • c

2(x – 3)(2x – 1) = 0

Notice: − 6 + (−1) = − 7 and − 6 • (− 1) = 6

2x 2x −1

and Reduce −3 x 2x

−1and

* Write your final factorization using the two fractions

* Build fractions using the leading term less one degree as your numerator and −6 and −1 as your denominators

1

−3

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 14: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

c) Use the zero product property to findthe solutions

9. Factor: 4x2 – 14x + 6 = 0

2(x – 3)(2x – 1) = 0 Factored form

Zero Product Propertyx – 3 = 0 and 2x – 1 = 0

+ 3+ 3Add

x = 3

+ 1+ 1 Add

1

Divide 2 2

2

2x = 1

x =

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 15: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

h = height of the object at time tt = time in secondss = initial height

10. Using the Vertical Motion Model h = − 16t2 + swhere

You are standing on a cliff 784 feet high and drop your cell phone. How long will it take until your phone hits the ground below (h = 0)?

h = 0t = t (need to find)s = 784

* Identify h, t and s

Substitute0 = − 16t2 + 784

– 784

– 784

Subtract

Isolate t2

− 784 = − 16t2

Divide −16 −16

49 = t2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 16: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

h = height of the object at time tt = time in secondss = initial height

10. Using the Vertical Motion Model h = − 16t2 + swhere

You are standing on a cliff 784 feet high and drop your cell phone. How long will it take until your phone hits the ground below (h = 0)?

49 = t2+ Square root

7 = t+

* Time is always positive

7 = t

It will take 7 seconds for the phone to hit the ground

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 17: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

− (2)

2(1)

y = x2 + 2x – 8

11. Complete the following given:

a) Find the vertex

x =− b

2a

a = 1b = 2 c = − 8

* Identify a, b and c

− 2 2

x = − 1

Formula to find x-value of vertex

Simplify

x-value of vertex

Substitute

* The vertex is the highest or lowest point on a parabola

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 18: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

y = x2 + 2x – 8

11. Complete the following given:

a) Find the vertex

a = 1b = 2 c = − 8

x = − 1* Substitute into the quadratic equation to find y

y = x2 + 2x – 8

y = (−1)2 + 2(− 1) – 8

y = 1 – 2 – 8

y = − 1 – 8

y = − 9 Vertex (−1, − 9)

Equation

Simplify

Substitute

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 19: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

11. Complete the following given:

b) Find the y - intercept

y = x2 + 2x – 8 a = 1b = 2 c = − 8

* Where the graph crosses the y-axis, NOTE: x = 0

(0, − 8)

y = x2 + 2x – 8

y = (0)2 + 2(0) – 8

y = 0 + 0 – 8

y = − 8

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

Page 20: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

11. Complete the following given:

c) Let y = 0, factor and solve using the zero product property

a = 1b = 2 c = − 8

* You are looking for two numbers that do the following

Add up to give you b and multiply to give you c

0 = (x + 4)(x – 2)

Notice: 4 + (− 2) = 2 and 4 • (− 2) = − 8

Factored form

Zero Product Propertyx + 4 = 0 and x – 2 = 0– 4– 4Subtract

x = − 4

+ 2+ 2 Add

x = 2

0 = x2 – 6x + 8 Let y = 0

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

y = x2 + 2x – 8

Page 21: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

11. Complete the following given:

d) Identify the x – intercepts using the resultsfrom part c

a = 1b = 2 c = − 8

x = − 4 x = 2 From part c.

* The zeros of the quadratic are also the x - intercepts

(− 4, 0) and (2, 0)

* Notice the x – intercepts have a y – coordinate of 0

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

y = x2 + 2x – 8

Page 22: 1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x 2 + 2 + 7x 6x 2 + 7x + 2 NOTE: The subtraction must be distributed

x – intercepts: (− 4, 0) and (2, 0)

11. Complete the following given:

e) Graph

Vertex: (− 1, − 9)

* Plot the following ordered pairs:

y – intercept: (0, − 8)

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

y = x2 + 2x – 8

You are aMath Super Star