1.Simplify (Place answer in standard form): (8x 2 – 5) + (3x + 7) – (2x 2 – 4x) 6x x 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 © by S-Squared, Inc. All Rights Reserved.

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

( ) 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

4.Simplify (Place answer in standard form): (m – 3)(7 – 2m 2 + 5m) – 15m 7m – 2m 3 + 5m 2 – m 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

5.Given: − 2t ( 3 + 7t ) + 6t − 6t – 14t 2 − 14t 2 Place in standard form. b)Identify the degree of the polynomial: The degree is the largest exponent of the polynomial. 2 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test − 2t ( 3 + 7t ) + 6t + 6t a)Simplify and put in standard form.

5.Given: d)Identify the type of polynomial based on the number of terms. There is one term in the polynomial. Monomial Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test − 2t ( 3 + 7t ) + 6t Quadratic Since the polynomial is of Degree 2. c)Name the polynomial based on the degree.

6.a) Factor: x x + 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 f1f1 b f2f x 5 * Build factor fractions., x 5 Notice: 5 5 = 25 and = 10 * Build binomial factors. * Identify the two factors. Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

b)Use the zero product property to find the solutions. (x + 5)(x + 5) = 0 Factored form Zero Product Property x + 5 = 0 – 5 Subtract x = − 5 6.Factor: x x + 25 = 0 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

6.Factor:x x + 25 = 0 c) Check your solutions. 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

7.Factor: x 2 – 5x – 24 = 0 * Identify a, b and c. a = 1, b = − 5, c = − 24 * Place ac and b into the diamond. ac f1f1 b f2f2 − 24 − 5 − 8 3 x 3 * Build factor fractions., x − 8 Notice: 3 (− 8) = − 24 and 3 + (− 8) = − 5 * Build binomial factors. (x + 3)(x – 8) = 0 * Identify the two factors. Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

x = 8 7.Factor:x 2 – 5x – 24 = 0 (x + 3)(x – 8) = 0 Factored form Zero Product Property x + 3 = 0 and x – 8 = 0 – 3 Subtract x = − Add Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test b)Use the zero product property to find the solutions.

(x + 9)(x – 9) = 0 8.Factor: x 2 – 81 = 0 * Identify a, b and c. a = 1, b = 0, c = − 81 * Place ac and b into the diamond. ac f1f1 b f2f2 − 81 0 − 9 9 x 9 * Build factor fractions., x − 9 Notice: 9 (− 9) = − 81 and 9 + (− 9) = 0 * Build binomial factors. * Identify the two factors. Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

8.Factor: x 2 – 81 = 0 (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 b)Use the zero product property to find the solutions.

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

b) Factor the resulting trinomial. * Identify a, b and c. a = 2, c = 3 2(2x – 1)(x – 3) = 0 * Place ac and b into the diamond. ac f1f1 b f2f2 6 − 7− 7 − 6 − 1 2x − 1 * Build factor fractions., 2x − 6 Notice: −1 (− 6) = 6 and − 1 + (− 6) = − 7 * Build binomial factors. * Identify the two factors. Reduce 1 − 3 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 9. Given: 4x 2 – 14x + 6 = 0 2(2x 2 – 7x + 3) = 0, b = − 7

2x = 1 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 Propertyx – 3 = 0 and 2x – 1 = Add x = Add 1 Divide 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 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 – 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 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

* Identify a, b and c. a = 1, b = 2, c = − 8 * Place ac and b into the diamond. ac f1f1 b f2f2 − 8 2 − 2 4 x 4 * Build factor fractions., x − 2 Notice: (4)(− 2) = − 8 And 4 + (− 2) = 2 * Build binomial factors. * Identify the two factors. 0 = (x + 4)(x – 2) 6.Complete the following given: a)Let y = 0, factor and solve using the zero product property. y = x 2 + 2x – 8 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

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

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test 6.Complete the following given: b)Identify the x – intercepts. x = − 4 x = 2 From part (a). * 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. y = x 2 + 2x – 8

− 2 2 y = x 2 + 2x – 8 6.Complete the following given: c) Find the vertex. x = − x = − 1 The x-value of vertex is the mid-point of the x-intercepts. x-value of vertex Simplify * The vertex is the highest or lowest point on a parabola. Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test (− 4, 0) and (2, 0)

Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test y = x 2 + 2x – 8 6.Complete the following given: c) Find the vertex. 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

6.Complete the following given: d)Find the y – intercept. y = x 2 + 2x – 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

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