Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 1.a) Factor: x x + 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: = 10 b)Use the zero product property to find the solutions and 5 5 = 25 (x + 5)(x + 5) = 0 Factored form Zero Product Property x + 5 = 0 – 5 Subtract x = − 5 © by S-Squared, Inc. All Rights Reserved.

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

2.a) Factor: x 2 – 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 Property x + 3 = 0 and x – 8 = 0 – 3 Subtract x = − Add x = 8

3.a) Factor: x 2 – 81 = 0 Difference of two perfect squares pattern * a 2 – b 2 = (a ─ b)(a + b) * Identify the a and the b by taking the square root of each term Notice, x2 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

3.Factor: x 2 – 81 = 0 b) Use the zero product property to find the solutions: (x – 9)(x + 9) = 0 Factored form Zero Product Property x – 9 = 0 and x + 9 = Add x = 9 – 9 Subtract x = − 9 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

a) Factor out the common monomial 4. Given: 4x 2 – 14x + 6 = 0 2(2x 2 – 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

b) Factor the resulting trinomial 4. Given: 4x 2 – 14x + 6 = 0 2(2x 2 – 7x + 3) = 0 −6−6 * 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 −1 −1 and Reduce −3 x 2x −1 −1 and * 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

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

h = height of the object at time t t = time in seconds s = initial height 5.Using the Vertical Motion Model h = − 16t 2 + s where 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 t = t (need to find) s = 784 * Identify h, t and s Substitute 0 = − 16t – 784 Subtract Isolate t 2 − 784 = − 16t 2 Divide −16 49 = t 2 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

h = height of the object at time t t = time in seconds s = initial height 5.Using the Vertical Motion Model h = − 16t 2 + s where 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 = t 2 + 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

− (2) 2(1) y = x 2 + 2x – 8 6.Complete the following given: a) Find the vertex x = − b 2a a = 1 b = 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

y = x 2 + 2x – 8 6.Complete the following given: a) Find the vertex a = 1 b = 2 c = − 8 x = − 1 * Substitute into the quadratic equation to find y y = x 2 + 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

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

6.Complete the following given: c)Let y = 0, factor and solve using the zero product property a = 1 b = 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 Property x + 4 = 0 and x – 2 = 0 – 4 Subtract x = − Add x = 2 0 = x 2 – 6x + 8 Let y = 0 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test y = x 2 + 2x – 8

6.Complete the following given: d)Identify the x – intercepts using the results from part c a = 1 b = 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 = x 2 + 2x – 8

x – intercepts: (− 4, 0) and (2, 0) 6.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 = x 2 + 2x – 8 You are a Math Super Star