do now pass out calculators. work on practice eoc week # 8
TRANSCRIPT
Quick Check:
1. (x2 – 3x + 5) + (-2x2 + 11x + 1)
2. (8y3 – 7y2 + y) – (9y2 – 5y + 7)
3. -3x2(x3 – 3x2)
4. (2r + 11)(r – 6)
5. (m + 3)(-2m2 + 5m – 1)
6. (5w + 9z)2
Answers:
1. –x2 + 8x +6
2. 8y3 – 16y2 +6y – 7
3. -3x5 + 9x4
4. 2r2 – r – 66
5. -2m3 – m2 +14m – 3
6. 25w2 90wz +81z2
Zero – Product Property:
• The zero-product property is used to solve an equation when one side is zero and the other side is two polynomials being multiplied.
• The solutions of an equations like are called roots.
Use the zero-product property
EXAMPLE 1
Solve (x – 4)(x + 2) = 0.
(x – 4)(x + 2) = 0 Write original equation.
x – 4 = 0 x = 4
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 4 and –2.
oror x + 2 = 0
x = – 2
Use the zero-product propertyEXAMPLE 1
CHECK Substitute each solution into the original equation to check.
(4 4)(4 + 2) = 0
0 6 = 0
0 = 0
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?
(2 4)(2 + 2) = 0
6 0 = 0
0 = 0
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?
GUIDED PRACTICE for Example 1
1. Solve the equation (x – 5)(x – 1) = 0.
ANSWER
The solutions of the equation are 5 and 1.
SOLUTION
EXAMPLE 2 Find the greatest common monomial factor
Factor out the greatest common monomial factor.
a. 12x + 42y
a. The GCF of 12 and 42 is 6. The variables x and y have no common factor. So, the greatest common monomial factor of the terms is 6.
ANSWER
12x + 42y = 6(2x + 7y)
EXAMPLE 2 Find the greatest common monomial factor
b. The GCF of 4 and 24 is 4. The GCF of x4 and x3 is x3. So, the greatest common monomial factor of the terms is 4x3.
ANSWER
4x4 + 24x3 = 4x3(x + 6)
SOLUTION
Factor out the greatest common monomial factor.
b. 4x4 + 24x3
GUIDED PRACTICE for Example 2
2. Factor out the greatest common monomial factorfrom 14m + 35n.
ANSWER
14m + 35n = 7(2m + 5n)
EXAMPLE 3 Solve an equation by factoring
Solve 2x2 + 8x = 0.
2x2 + 8x = 0
2x(x + 4) = 0
2x = 0
x = 0
or x + 4 = 0
or x = – 4
ANSWER
The solutions of the equation are 0 and – 4.
Solve for x.
Zero-product property
Factor left side.
Write original equation.
EXAMPLE 4 Solve an equation by factoring
Solve 6n2 = 15n.
6n2 – 15n = 0
3n(2n – 5) = 0
3n = 0
n = 0
2n – 5 = 0
n =52
or
or Solve for n.
Zero-product property
Factor left side.
Subtract 15n from each side.
ANSWER
The solutions of the equation are 0 and52 .
GUIDED PRACTICE for Examples 3 and 4
Solve the equation.
ANSWER
0 and – 5
3. a2 + 5a = 0
4. 3s2 – 9s = 0
ANSWER
0 and 3
5. 4x2 = 2x.
ANSWER
0 and 12
Vertical Motion:• A projectile is an object that is propelled into the air but has no power to keep itself in the air. A thrown ball is a projective, but an airplane is not. The height of a projectile can be described by the vertical motion model.
• The height h (in feet) of a projectile can be modeled by:
h = -16t2 + vt + x
t = time (in seconds) the object has been in the air
v = initial velocity (in feet per second)
s = the initial height (in feet)
ARMADILLO
EXAMPLE 5 Solve a multi-step problem
A startled armadillo jumps straight into the air with an initial vertical velocity of 14 feet per second. After how many seconds does it land on the ground?
SOLUTION
EXAMPLE 5 Solve a multi-step problem
STEP 1
Write a model for the armadillo’s height above the ground.
h = –16t2 + vt + s
h = –16t2 + 14t + 0
h = –16t2 + 14t
Vertical motion model
Substitute 14 for v and 0 for s.
Simplify.
EXAMPLE 5 Solve a multi-step problem
STEP 2Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t.
0 = –16t2 + 14t
0 = 2t(–8t + 7)
2t = 0
t = 0
–8t + 7 = 0
t = 0.875
or
or Solve for t.
Zero-product property
Factor right side.
Substitute 0 for h.
ANSWER
The armadillo lands on the ground 0.875 second after the armadillo jumps.
GUIDED PRACTICE for Example 5
6. WHAT IF? In Example 5, suppose the initial vertical velocity is 12 feet per second.After how many seconds does armadillo land on the ground?
ANSWER
The armadillo lands on the ground 0.75 second after the armadillo jumps.