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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.

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2.Simplify (Place answer in standard form): (2a 3 – 6a + 7) (5a 2 2a + 7) 2a 3 – 4a + 0 2a 3 – 5a 2 – 4a NOTE: The subtraction must be distributed to each term Place in standard form – 5a 2 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

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( ) 5x 2 – 2 – 10x– 25x 2 – 4 +10x Combine 3.Simplify (Place answer in standard form): 25x 2 – 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

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

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5.Given: a)Simplify and put in standard form. 2t ( 3 + 7t 2 ) 6t – 14t 3 14t 3 – 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 + 7t 2 )

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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 + 7t 2 ) cubic Since the polynomial is of Degree 3

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Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 6.a) Factor: x 2 + 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 Property x + 5 = 0 – 5 Subtract x = 5

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6.Factor: x 2 + 10x + 25 = 0 c) Check your solution x = 5 x 2 + 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

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7.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 = 3 + 8 Add x = 8

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8.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

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8.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 = 0 + 9 Add x = 9 – 9 Subtract x = 9 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

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a) Factor out the common monomial 9. 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

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b) Factor the resulting trinomial 9. Given: 4x 2 – 14x + 6 = 0 2(2x 2 – 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 1 and Reduce 3 x 2x 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

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

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h = height of the object at time t t = time in seconds s = initial height 10.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 2 + 784 – 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

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h = height of the object at time t t = time in seconds s = initial height 10.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

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(2) 2(1) y = x 2 + 2x – 8 11.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

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y = x 2 + 2x – 8 11.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

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11.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 = 0 + 0 – 8 y = 8 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

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11.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 = 4 + 2 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

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11.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

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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 = x 2 + 2x – 8 You are a Math Super Star

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