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**1. Simplify (Place answer in standard form):**

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 1. Simplify (Place answer in standard form): (8x2 – 5) + (3x + 7) – (2x2 – 4x) NOTE: The subtraction must be distributed to each term 6x2 + 2 + 7x Place in standard form 6x x © by S-Squared, Inc. All Rights Reserved.

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

2. Simplify (Place answer in standard form): (2a3 – 6a ) ─ (5a2 ─ 2a ) NOTE: The subtraction must be distributed to each term 2a3 – 4a + 0 – 5a2 Place in standard form 2a3 – 5a2 – 4a

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**3. Simplify (Place answer in standard form):**

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 3. Simplify (Place answer in standard form): NOTE: To square a binomial, you must multiply it by itself (5x – 2)2 ( ) ( ) 5x 2 – Combine 25x 2 – 10x – 10x + 4 25x2 – 20x

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

4. Simplify (Place answer in standard form): (m – 3)(7 – 2m m) 7m – 2m3 + 5m2 – 15m + 6m2 – 21 Combine like terms – 8m – 2m3 + 11m2 – 21 Place in standard form − 2m3 + 11m2 – 8m – 21

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

5. Given: a) Simplify and put in standard form. − 2t ( t2 ) − 2t ( t2 ) − 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

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

5. Given: c) Name the polynomial based on the degree. − 2t ( t2 ) cubic Since the polynomial is of Degree 3 d) Identify the type of polynomial based on the number of terms. There are two terms in the polynomial binomial

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

6. a) Factor: x x = 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 Notice: = 10 and • 5 = 25 (x + 5)(x + 5) = Or (x + 5)2 = 0 Use the zero product property to find the solutions (x + 5)(x + 5) = 0 Factored form x = 0 Zero Product Property – 5 – 5 Subtract x = − 5

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

6. Factor: x x = 0 c) Check your solution x = − 5 Equation x x = 0 Substitute (− 5) (− 5) = 0 Simplify 25 – = 0 − = 0 0 = 0 Check

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

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: (− 8) = − 5 and • (− 8) = − 24 (x + 3)(x – 8) = 0 Use the zero product property to find the solutions (x + 3)(x – 8) = 0 Factored form x = and x – 8 = 0 Zero Product Property Subtract – 3 – 3 + 8 + 8 Add x = − 3 x = 8

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

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 81 9 = b * Substitute into the difference of two perfect square pattern (x – 9)(x + 9) = 0

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

8. Factor: x2 – 81 = 0 b) Use the zero product property to find the solutions: (x – 9)(x + 9) = 0 Factored form x – 9 = and x = 0 Zero Product Property Add + 9 + 9 – 9 – 9 Subtract x = 9 x = − 9

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

Given: 4x2 – 14x = 0 a) Factor out the common monomial * The greatest common monomial is 2 2(2x2 – 7x + 3) = 0 Factor out a 2

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**b) Factor the resulting trinomial**

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test Given: 4x2 – 14x = 0 b) Factor the resulting trinomial 2(2x2 – 7x + 3) = 0 * 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 Notice: − (−1) = − 7 and − 6 • (− 1) = 6 * Build fractions using the leading term less one degree as your numerator and −6 and −1 as your denominators 1 2x 2x x 2x and and Reduce −6 −1 −3 −1 −3 * Write your final factorization using the two fractions 2(x – 3)(2x – 1) = 0

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

Factor: 4x2 – 14x = 0 c) Use the zero product property to find the solutions 2(x – 3)(2x – 1) = 0 Factored form x – 3 = and 2x – 1 = 0 Zero Product Property Add + 3 + 3 + 1 + 1 Add x = 3 2x = 1 Divide 2 2 1 x = 2

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

10. Using the Vertical Motion Model h = − 16t s where h = height of the object at time t t = time in seconds s = initial height 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 = 0 * Identify h, t and s t = t (need to find) s = 784 Isolate t2 0 = − 16t Substitute – 784 – 784 Subtract − = − 16t2 Divide −16 −16 49 = t2

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

10. Using the Vertical Motion Model h = − 16t s where h = height of the object at time t t = time in seconds s = initial height 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

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

11. Complete the following given: a = 1 y = x x – 8 b = 2 * Identify a, b and c c = − 8 * The vertex is the highest or lowest point on a parabola a) Find the vertex − b Formula to find x-value of vertex x = 2a − (2) Substitute 2(1) − 2 Simplify 2 x = − 1 x-value of vertex

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

11. Complete the following given: y = x x – 8 a = 1 b = 2 a) Find the vertex c = − 8 x = − 1 * Substitute into the quadratic equation to find y y = x x – 8 Equation y = (−1) (− 1) – 8 Substitute y = 1 – 2 – 8 Simplify y = − 1 – 8 y = − 9 Vertex (−1, − 9)

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

11. Complete the following given: a = 1 y = x x – 8 b = 2 c = − 8 b) Find the y - intercept * Where the graph crosses the y-axis, NOTE: x = 0 y = x x – 8 y = (0) (0) – 8 y = – 8 y = − 8 (0, − 8)

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

11. Complete the following given: a = 1 y = x x – 8 b = 2 c = − 8 c) Let y = 0, factor and solve using the zero product property * You are looking for two numbers that do the following Add up to give you b and multiply to give you c Notice: (− 2) = 2 and • (− 2) = − 8 0 = x2 – 6x + 8 Let y = 0 0 = (x + 4)(x – 2) Factored form x = and x – 2 = 0 Zero Product Property Subtract – 4 – 4 + 2 + 2 Add x = − 4 x = 2

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

11. Complete the following given: a = 1 y = x x – 8 b = 2 c = − 8 Identify the x – intercepts using the results from part c x = − 4 x = 2 From part c. * The zeros of the quadratic are also the x - intercepts * Notice the x – intercepts have a y – coordinate of 0 (− 4, 0) and (2, 0)

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**You are a Math Super Star**

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: y = x x – 8 You are a Math Super Star e) Graph * Plot the following ordered pairs: Vertex: (− 1, − 9) y – intercept: (0, − 8) x – intercepts: (− 4, 0) and (2, 0)

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1.a) Factor: x 2 + 10x + 25 = 0 * Identify a, b and c. a = 1, b = 10, c = 25 (x + 5)(x + 5) = 0 Or (x + 5) 2 = 0 * Place ac and b into the diamond. ac.

1.a) Factor: x 2 + 10x + 25 = 0 * Identify a, b and c. a = 1, b = 10, c = 25 (x + 5)(x + 5) = 0 Or (x + 5) 2 = 0 * Place ac and b into the diamond. ac.

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