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 6x2 + 7x + 2 © 2007- 09 by S-Squared, Inc. All Rights Reserved.

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 2. Simplify (Place answer in standard form): (2a3 – 6a + 7) ─ (5a2 ─ 2a + 7) NOTE: The subtraction must be distributed to each term 2a3 – 4a + 0 – 5a2 Place in standard form 2a3 – 5a2 – 4a

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 + 4

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 4. Simplify (Place answer in standard form): (m – 3)(7 – 2m2 + 5m) 7m – 2m3 + 5m2 – 15m + 6m2 – 21 Combine like terms – 8m – 2m3 + 11m2 – 21 Place in standard form − 2m3 + 11m2 – 8m – 21

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 5. Given: a) Simplify and put in standard form. − 2t ( 3 + 7t2 ) − 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 5. Given: c) Name the polynomial based on the degree. − 2t ( 3 + 7t2 ) 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

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

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 6. Factor: x2 + 10x + 25 = 0 c) Check your solution x = − 5 Equation x2 + 10x + 25 = 0 Substitute (− 5)2 + 10(− 5) + 25 = 0 Simplify 25 – 50 + 25 = 0 − 25 + 25 = 0 0 = 0 Check

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

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

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 = 0 and x + 9 = 0 Zero Product Property Add + 9 + 9 – 9 – 9 Subtract x = 9 x = − 9

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 9. Given: 4x2 – 14x + 6 = 0 a) Factor out the common monomial * The greatest common monomial is 2 2(2x2 – 7x + 3) = 0 Factor out a 2

b) Factor the resulting trinomial Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 9. Given: 4x2 – 14x + 6 = 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: − 6 + (−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

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 9. Factor: 4x2 – 14x + 6 = 0 c) Use the zero product property to find the solutions 2(x – 3)(2x – 1) = 0 Factored form x – 3 = 0 and 2x – 1 = 0 Zero Product Property Add + 3 + 3 + 1 + 1 Add x = 3 2x = 1 Divide 2 2 1 x = 2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 10. Using the Vertical Motion Model h = − 16t2 + 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 = − 16t2 + 784 Substitute – 784 – 784 Subtract − 784 = − 16t2 Divide −16 −16 49 = t2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 10. Using the Vertical Motion Model h = − 16t2 + 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

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: a = 1 y = x2 + 2x – 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

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: y = x2 + 2x – 8 a = 1 b = 2 a) Find the vertex c = − 8 x = − 1 * Substitute into the quadratic equation to find y y = x2 + 2x – 8 Equation y = (−1)2 + 2(− 1) – 8 Substitute y = 1 – 2 – 8 Simplify y = − 1 – 8 y = − 9 Vertex (−1, − 9)

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: a = 1 y = x2 + 2x – 8 b = 2 c = − 8 b) Find the y - intercept * Where the graph crosses the y-axis, NOTE: x = 0 y = x2 + 2x – 8 y = (0)2 + 2(0) – 8 y = 0 + 0 – 8 y = − 8 (0, − 8)

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: a = 1 y = x2 + 2x – 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: 4 + (− 2) = 2 and 4 • (− 2) = − 8 0 = x2 – 6x + 8 Let y = 0 0 = (x + 4)(x – 2) Factored form x + 4 = 0 and x – 2 = 0 Zero Product Property Subtract – 4 – 4 + 2 + 2 Add x = − 4 x = 2

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: a = 1 y = x2 + 2x – 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)

You are a Math Super Star Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 11. Complete the following given: y = x2 + 2x – 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)