Transcript CT #15 Practice Test (Diamond Method)
Slide 1
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
1.
Simplify (Place answer in standard form):
(8x2 – 5) + (3x + 7) – (2x2 – 4x)
NOTE: The subtraction must be distributed to each term
6x2
+ 2
+ 7x
Place in standard form
6x2 + 7x + 2
© 2007- 09 by S-Squared, Inc. All Rights Reserved.
Slide 2
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
2.
Simplify (Place answer in standard form):
(2a3 – a + 7) ─ (5a2 ─ 2a + 7)
NOTE: The subtraction must be distributed to each term
2a3
+ a
+ 0
– 5a2
Place in standard form
2a3 – 5a2 + a
Slide 3
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
3.
Simplify (Place answer in standard form):
NOTE: To square a binomial, you must multiply it
by itself
(5x – 2)2
( 5x –
2 )( 5x
–
2)
Combine
25x 2 – 10x – 10x + 4
25x2 – 20x + 4
Slide 4
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
4.
Simplify (Place answer in standard form):
(m – 3)(7 – 2m2 + 5m)
7m – 2m3 + 5m2
– 15m
+ 6m2 – 21
– 8m – 2m3 + 11m2 – 21
− 2m3 + 11m2 – 8m – 21
Combine like terms
Place in standard form
Slide 5
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
5.
− 2t ( 3 + 7t ) + 6t
Given:
a)
Simplify and put in standard form.
− 2t ( 3 + 7t ) + 6t
− 6t – 14t2 + 6t
− 14t2
b)
Place in standard form
Identify the degree of the polynomial:
2
The degree is the largest
exponent of the polynomial
Slide 6
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
5.
Given:
c)
− 2t ( 3 + 7t ) + 6t
Name the polynomial based on the degree.
quadratic
d)
Since the polynomial is of
degree 2
Identify the type of polynomial based on the
number of terms.
monomial
There is one term in the
polynomial
Slide 7
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
a)
Factor:
x2 + 10x + 25 = 0
a = 1 , b = 10 , c = 25
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice:
and
5 • 5 = 25
5 + 5 = 10
25
f1
* Build factor fractions
x
5
,
x
5
5
5
10
b
* Build binomial factors
(x + 5)(x + 5) = 0
Or
(x + 5)2 = 0
f2
Slide 8
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
Factor:
b)
x2 + 10x + 25 = 0
Use the zero product property to find the
solutions
(x + 5)(x + 5) = 0
Factored form
x + 5 = 0
Zero Product Property
–5 –5
x = −5
Subtract
Slide 9
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
Factor:
c)
x2 + 10x + 25 = 0
Check your solutions
x = −5
Equation
x2 + 10x + 25 = 0
Substitute (− 5)2 + 10(− 5) + 25 = 0
Simplify
25 – 50 + 25 = 0
− 25 + 25 = 0
0 = 0
Check
Slide 10
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
7.
x2 – 5x – 24 = 0
Factor:
a = 1 , b = − 5, c = − 24
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: 3 • (− 8) = − 24
and 3 + (− 8) = − 5
− 24
* Build factor fractions
x
3
,
x
−8
* Build binomial factors
(x + 3)(x – 8) = 0
f1
−8
3
−5
b
f2
Slide 11
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
7.
x2 – 5x – 24 = 0
Factor:
b)
Use the zero product property to find the
solutions
(x + 3)(x – 8) = 0
x + 3 = 0
Subtract
–3 –3
x = −3
and
Factored form
x – 8 = 0
+8 +8
x = 8
Zero Product Property
Add
Slide 12
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
8.
x2 – 81 = 0
Factor:
a = 1 , b = 0 , c = − 81
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: 9 • (− 9) = − 81
−
81
and 9 + (− 9) = 0
* Build factor fractions
x
9
,
x
−9
* Build binomial factors
(x + 9)(x – 9) = 0
f1
−9
9
0
b
f2
Slide 13
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
8.
x2 – 81 = 0
Factor:
b)
Use the zero product property to find the
solutions
(x – 9)(x + 9) = 0
x – 9 = 0
Add
+9 +9
x = 9
and
Factored form
x + 9 = 0
–9 –9
x = −9
Zero Product Property
Subtract
Slide 14
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
Given:
a)
4x2 – 14x + 6 = 0
Factor out the common monomial
* The greatest common monomial is 2
2(2x2 – 7x + 3) = 0
Factor out a 2
Slide 15
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
4x2 – 14x + 6 = 0
Given:
b)
Factor the resulting trinomial
2(2x2 – 7x + 3) = 0
a = 2 , b = − 7, c = 3
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
−1 • (− 6) = 6
− 1 + (− 6) = − 7
Notice:
and
* Build factor fractions
2x ,
−1
1
2x
−6−3
Reduce
* Build binomial factors
2(2x – 1)(x – 3) = 0
6
f1
−1
−6f
2
−7
b
Slide 16
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
4x2 – 14x + 6 = 0
Factor:
c)
Use the zero product property to find
the solutions
2(x – 3)(2x – 1) = 0
x – 3 = 0
Add
+3 +3
x = 3
and
Factored form
2x – 1 = 0
+1 +1
2x =
2
1
2
x =
1
2
Zero Product Property
Add
Divide
Slide 17
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
10.
Isolate t2
Using the Vertical Motion Model h = − 16t2 + s
where
h = height of the object at time t
t = time in seconds
s = initial height
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
* Identify h, t and s
t = t (need to find)
s = 784
0 = − 16t2 + 784
Substitute
– 784
Subtract
– 784
− 784 = − 16t2
−16
−16
49 = t2
Divide
Slide 18
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
10.
Using the Vertical Motion Model h = − 16t2 + s
where
h = height of the object at time t
t = time in seconds
s = initial height
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
Slide 19
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
a = 1
b = 2
* Identify a, b and c
c = −8
* The vertex is the highest or lowest point on a parabola
a)
Find the vertex
−b
Formula to find
x =
x-value of vertex
2a
y = x2 + 2x – 8
− (2)
2(1)
Substitute
−2
2
Simplify
x = −1
x-value of vertex
Slide 20
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
a)
Find the vertex
a = 1
b = 2
c = −8
x = −1
* Substitute into the quadratic equation to find y
y = x2 + 2x – 8
Equation
y = (−1)2 + 2(− 1) – 8
Substitute
y = 1 – 2 – 8
Simplify
y =
−1 – 8
y =
−9
Vertex
(−1, − 9)
Slide 21
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
b)
Find the y - intercept
* Where the graph crosses the
y-axis, NOTE: x = 0
y = x2 + 2x – 8
y = (0)2 + 2(0) – 8
y = 0 + 0 – 8
y = −8
(0, − 8)
a = 1
b = 2
c = −8
Slide 22
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
c)
Let y = 0, factor and solve using the zero
product property
a = 1, b = 2 , c = − 8
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: (4)(− 2) = − 8
And 4 + (− 2) = 2
* Build factor fractions
x
4
,
x
−2
* Build binomial factors
0 = (x + 4)(x – 2)
−8
f1
−2
4
2
b
f2
Slide 23
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
c)
Let y = 0, factor and solve using the zero
product property a = 1 b = 2 c = − 8
0 = (x + 4)(x – 2)
x + 4 = 0
Subtract
–4 –4
x = −4
and
x – 2 = 0
+2 +2
x = 2
Factored form
Zero Product Property
Add
Slide 24
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y =
d)
x2
a = 1
+ 2x – 8
b = 2
c = −8
Identify the x – intercepts using the results
from part c
x = −4
x = 2
From part c.
* The zeros of the quadratic are also the x - intercepts
* Notice the x – intercepts have a y – coordinate of 0
(− 4, 0) and
(2, 0)
Slide 25
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
e)
Graph
* Plot the following
ordered pairs:
You are a
Math
Super Star
Vertex:
(− 1, − 9)
y – intercept:
x – intercepts:
(− 4, 0) and
(0, − 8)
(2, 0)
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
1.
Simplify (Place answer in standard form):
(8x2 – 5) + (3x + 7) – (2x2 – 4x)
NOTE: The subtraction must be distributed to each term
6x2
+ 2
+ 7x
Place in standard form
6x2 + 7x + 2
© 2007- 09 by S-Squared, Inc. All Rights Reserved.
Slide 2
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
2.
Simplify (Place answer in standard form):
(2a3 – a + 7) ─ (5a2 ─ 2a + 7)
NOTE: The subtraction must be distributed to each term
2a3
+ a
+ 0
– 5a2
Place in standard form
2a3 – 5a2 + a
Slide 3
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
3.
Simplify (Place answer in standard form):
NOTE: To square a binomial, you must multiply it
by itself
(5x – 2)2
( 5x –
2 )( 5x
–
2)
Combine
25x 2 – 10x – 10x + 4
25x2 – 20x + 4
Slide 4
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
4.
Simplify (Place answer in standard form):
(m – 3)(7 – 2m2 + 5m)
7m – 2m3 + 5m2
– 15m
+ 6m2 – 21
– 8m – 2m3 + 11m2 – 21
− 2m3 + 11m2 – 8m – 21
Combine like terms
Place in standard form
Slide 5
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
5.
− 2t ( 3 + 7t ) + 6t
Given:
a)
Simplify and put in standard form.
− 2t ( 3 + 7t ) + 6t
− 6t – 14t2 + 6t
− 14t2
b)
Place in standard form
Identify the degree of the polynomial:
2
The degree is the largest
exponent of the polynomial
Slide 6
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
5.
Given:
c)
− 2t ( 3 + 7t ) + 6t
Name the polynomial based on the degree.
quadratic
d)
Since the polynomial is of
degree 2
Identify the type of polynomial based on the
number of terms.
monomial
There is one term in the
polynomial
Slide 7
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
a)
Factor:
x2 + 10x + 25 = 0
a = 1 , b = 10 , c = 25
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice:
and
5 • 5 = 25
5 + 5 = 10
25
f1
* Build factor fractions
x
5
,
x
5
5
5
10
b
* Build binomial factors
(x + 5)(x + 5) = 0
Or
(x + 5)2 = 0
f2
Slide 8
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
Factor:
b)
x2 + 10x + 25 = 0
Use the zero product property to find the
solutions
(x + 5)(x + 5) = 0
Factored form
x + 5 = 0
Zero Product Property
–5 –5
x = −5
Subtract
Slide 9
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
6.
Factor:
c)
x2 + 10x + 25 = 0
Check your solutions
x = −5
Equation
x2 + 10x + 25 = 0
Substitute (− 5)2 + 10(− 5) + 25 = 0
Simplify
25 – 50 + 25 = 0
− 25 + 25 = 0
0 = 0
Check
Slide 10
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
7.
x2 – 5x – 24 = 0
Factor:
a = 1 , b = − 5, c = − 24
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: 3 • (− 8) = − 24
and 3 + (− 8) = − 5
− 24
* Build factor fractions
x
3
,
x
−8
* Build binomial factors
(x + 3)(x – 8) = 0
f1
−8
3
−5
b
f2
Slide 11
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
7.
x2 – 5x – 24 = 0
Factor:
b)
Use the zero product property to find the
solutions
(x + 3)(x – 8) = 0
x + 3 = 0
Subtract
–3 –3
x = −3
and
Factored form
x – 8 = 0
+8 +8
x = 8
Zero Product Property
Add
Slide 12
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
8.
x2 – 81 = 0
Factor:
a = 1 , b = 0 , c = − 81
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: 9 • (− 9) = − 81
−
81
and 9 + (− 9) = 0
* Build factor fractions
x
9
,
x
−9
* Build binomial factors
(x + 9)(x – 9) = 0
f1
−9
9
0
b
f2
Slide 13
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
8.
x2 – 81 = 0
Factor:
b)
Use the zero product property to find the
solutions
(x – 9)(x + 9) = 0
x – 9 = 0
Add
+9 +9
x = 9
and
Factored form
x + 9 = 0
–9 –9
x = −9
Zero Product Property
Subtract
Slide 14
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
Given:
a)
4x2 – 14x + 6 = 0
Factor out the common monomial
* The greatest common monomial is 2
2(2x2 – 7x + 3) = 0
Factor out a 2
Slide 15
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
4x2 – 14x + 6 = 0
Given:
b)
Factor the resulting trinomial
2(2x2 – 7x + 3) = 0
a = 2 , b = − 7, c = 3
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
−1 • (− 6) = 6
− 1 + (− 6) = − 7
Notice:
and
* Build factor fractions
2x ,
−1
1
2x
−6−3
Reduce
* Build binomial factors
2(2x – 1)(x – 3) = 0
6
f1
−1
−6f
2
−7
b
Slide 16
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
9.
4x2 – 14x + 6 = 0
Factor:
c)
Use the zero product property to find
the solutions
2(x – 3)(2x – 1) = 0
x – 3 = 0
Add
+3 +3
x = 3
and
Factored form
2x – 1 = 0
+1 +1
2x =
2
1
2
x =
1
2
Zero Product Property
Add
Divide
Slide 17
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
10.
Isolate t2
Using the Vertical Motion Model h = − 16t2 + s
where
h = height of the object at time t
t = time in seconds
s = initial height
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
* Identify h, t and s
t = t (need to find)
s = 784
0 = − 16t2 + 784
Substitute
– 784
Subtract
– 784
− 784 = − 16t2
−16
−16
49 = t2
Divide
Slide 18
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
10.
Using the Vertical Motion Model h = − 16t2 + s
where
h = height of the object at time t
t = time in seconds
s = initial height
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
Slide 19
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
a = 1
b = 2
* Identify a, b and c
c = −8
* The vertex is the highest or lowest point on a parabola
a)
Find the vertex
−b
Formula to find
x =
x-value of vertex
2a
y = x2 + 2x – 8
− (2)
2(1)
Substitute
−2
2
Simplify
x = −1
x-value of vertex
Slide 20
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
a)
Find the vertex
a = 1
b = 2
c = −8
x = −1
* Substitute into the quadratic equation to find y
y = x2 + 2x – 8
Equation
y = (−1)2 + 2(− 1) – 8
Substitute
y = 1 – 2 – 8
Simplify
y =
−1 – 8
y =
−9
Vertex
(−1, − 9)
Slide 21
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
b)
Find the y - intercept
* Where the graph crosses the
y-axis, NOTE: x = 0
y = x2 + 2x – 8
y = (0)2 + 2(0) – 8
y = 0 + 0 – 8
y = −8
(0, − 8)
a = 1
b = 2
c = −8
Slide 22
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
c)
Let y = 0, factor and solve using the zero
product property
a = 1, b = 2 , c = − 8
* Identify a, b and c
* Place ac and b into the diamond
* Identify the two factors
ac
Notice: (4)(− 2) = − 8
And 4 + (− 2) = 2
* Build factor fractions
x
4
,
x
−2
* Build binomial factors
0 = (x + 4)(x – 2)
−8
f1
−2
4
2
b
f2
Slide 23
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
c)
Let y = 0, factor and solve using the zero
product property a = 1 b = 2 c = − 8
0 = (x + 4)(x – 2)
x + 4 = 0
Subtract
–4 –4
x = −4
and
x – 2 = 0
+2 +2
x = 2
Factored form
Zero Product Property
Add
Slide 24
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y =
d)
x2
a = 1
+ 2x – 8
b = 2
c = −8
Identify the x – intercepts using the results
from part c
x = −4
x = 2
From part c.
* The zeros of the quadratic are also the x - intercepts
* Notice the x – intercepts have a y – coordinate of 0
(− 4, 0) and
(2, 0)
Slide 25
Algebra I Concept Test # 15 – Solving Quadratic Equations by Factoring
Practice Test
11.
Complete the following given:
y = x2 + 2x – 8
e)
Graph
* Plot the following
ordered pairs:
You are a
Math
Super Star
Vertex:
(− 1, − 9)
y – intercept:
x – intercepts:
(− 4, 0) and
(0, − 8)
(2, 0)