1. Simplify (Place answer in standard form):
(8x2 – 5) + (3x + 7) – (2x2 – 4x)
6x2 + 2 + 7x
6x2 + 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
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2. Simplify (Place answer in standard form):
(2a3 – 6a + 7) ─ (5a2 ─ 2a + 7)
2a3 – 4a + 0
2a3 – 5a2 – 4a
NOTE: The subtraction must be distributed to each term
Place in standard form
– 5a2
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
( )( )5x25x – 2–
10x –25x2 – 4 +10x
Combine
3. Simplify (Place answer in standard form):
25x2 – 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 – 2m2 + 5m)
– 15m
7m – 2m3 + 5m2
– 21+ 6m2 Combine like terms
– 21– 8m + 11m2– 2m3
– 21− 2m3 – 8m+ 11m2
Place in standard form
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
5. Given:
a) Simplify and put in standard form.
− 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
− 2t ( 3 + 7t2 )
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 + 7t2 )
cubic Since the polynomial is ofDegree 3
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
(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 Propertyx + 5 = 0 – 5– 5 Subtract
x = − 5
6. Factor: x2 + 10x + 25 = 0
c) Check your solution x = − 5
x2 + 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
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
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 Propertyx + 3 = 0 and x – 8 = 0– 3– 3Subtract
x = − 3
+ 8+ 8 Add
x = 8
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
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
8. Factor: x2 – 81 = 0
b) Use the zero product property to find the solutions:
(x – 9)(x + 9) = 0 Factored form
Zero Product Propertyx – 9 = 0 and x + 9 = 0
+ 9+ 9Add
x = 9
– 9– 9 Subtract
x = − 9
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
a) Factor out the common monomial
9. Given: 4x2 – 14x + 6 = 0
2(2x2 – 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
9. Given: 4x2 – 14x + 6 = 0
2(2x2 – 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 2x −1
and Reduce −3 x 2x
−1and
* 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 findthe solutions
9. Factor: 4x2 – 14x + 6 = 0
2(x – 3)(2x – 1) = 0 Factored form
Zero Product Propertyx – 3 = 0 and 2x – 1 = 0
+ 3+ 3Add
x = 3
+ 1+ 1 Add
1
Divide 2 2
2
2x = 1
x =
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
h = height of the object at time tt = time in secondss = initial height
10. Using the Vertical Motion Model h = − 16t2 + swhere
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 = 0t = t (need to find)s = 784
* Identify h, t and s
Substitute0 = − 16t2 + 784
– 784
– 784
Subtract
Isolate t2
− 784 = − 16t2
Divide −16 −16
49 = t2
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
h = height of the object at time tt = time in secondss = initial height
10. Using the Vertical Motion Model h = − 16t2 + swhere
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
− (2)
2(1)
y = x2 + 2x – 8
11. Complete the following given:
a) Find the vertex
x =− b
2a
a = 1b = 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 = x2 + 2x – 8
11. Complete the following given:
a) Find the vertex
a = 1b = 2 c = − 8
x = − 1* Substitute into the quadratic equation to find y
y = x2 + 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
11. Complete the following given:
b) Find the y - intercept
y = x2 + 2x – 8 a = 1b = 2 c = − 8
* Where the graph crosses the y-axis, NOTE: x = 0
(0, − 8)
y = x2 + 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
11. Complete the following given:
c) Let y = 0, factor and solve using the zero product property
a = 1b = 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 Propertyx + 4 = 0 and x – 2 = 0– 4– 4Subtract
x = − 4
+ 2+ 2 Add
x = 2
0 = x2 – 6x + 8 Let y = 0
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring Practice Test
y = x2 + 2x – 8
11. Complete the following given:
d) Identify the x – intercepts using the resultsfrom part c
a = 1b = 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 = x2 + 2x – 8
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 = x2 + 2x – 8
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