2. 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- 10 by S-Squared, Inc. All Rights Reserved.

3. 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

4. ( ) 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

5. 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

6. 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.

7. 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.

8. 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

9. 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

10. 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

11. 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

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

13. 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

14. b) Factor the resulting trinomial. 9. 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

15. 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

16. 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

17. 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

18. 11.Complete the following given: a)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

19. 11.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. Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test y = x 2 + 2x – 8

20. − 2 2 c) Find the vertex. x = − 4 + 2 2 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) 11.Complete the following given: y = x 2 + 2x – 8

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

22. 11.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 = 0 + 0 – 8 y = − 8 Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test

23. 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