Download Section 6.4

Chapter 6
Factoring
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CHAPTER
6
Factoring
6.1 Greatest Common Factor and
Factoring by Grouping
6.2 Factoring Trinomials
6.3 Factoring Special Products and
Factoring Strategies
6.4 Solving Equations by Factoring
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6.4
Solving Equations by Factoring
1. Use the zero-factor theorem to solve equations by
factoring.
2. Solve problems involving quadratic equations.
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Polynomial equation: An equation that equates
two polynomials.
Zero-Factor Theorem
If a and b are real numbers and ab = 0, then a = 0
or b = 0.
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Example
Solve. (x + 4)(x + 5) = 0
Solution
Since we have a product of two factors, x + 4 and
x + 5, equal to zero, one or both factors must equal 0.
x+4=0
or
x + 5 = 0 Solve each equation.
x = 4
x = 5
Check Verify that 4 and 5 satisfy the original
equation, (x + 4)(x + 5) = 0.
For x = 4:
For x = 5:
(4 + 4)(4 + 5) = 0
(5 + 4)(5 + 5) = 0
0(1) = 0
(1)(0) = 0
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Polynomial equation in standard form: P = 0, where
P is a polynomial in terms of one variable written in
descending order of degree.
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Example
Solve. 2x2 – 5x = 3
Solution
First, we need the equation in standard form.
2x2 – 5x – 3 = 0
(2x + 1)(x – 3) = 0 Factor.
2x + 1 = 0 or x – 3 = 0 Use the zero-factor theorem to solve.
2x  1
1
x
2
x 3
The solutions are 
1
and 3.
2
To check, we verify that the
solutions satisfy the original
equations.
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Quadratic equation in one variable: An equation that
can be written in the form ax2 + bx + c = 0, where a,
b, and c are all real numbers and a  0.
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Example
Solve. x3 – 4x2 = 21x
Solution
x3 – 4x2 – 21x = 0
Write the equation in standard form.
x(x2 – 4x – 21) = 0
x(x – 7 )(x + 3) = 0
x=0
or
x – 7 = 0 or
x7
x+3=0
x  3
The solutions are 0, 7, and 3.
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Cubic equation in one variable: An equation that can
be written in the form ax3 + bx2 + cx + d = 0, where
a, b, c, and d, are all real numbers and a  0.
Solving Polynomial Equations Using Factoring
To solve a polynomial equation using factoring,
1. Write the equation in standard form
(Set one side equal to 0 with the other side in
descending order of degree when possible).
2. Write the polynomial in factored form.
3. Use the zero-factor theorem to solve.
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Example
Solve. y(y – 7y) = –12
Solution
Write the equation in standard form.
y ( y  7)  12  0
y 2  7 y  12  0
( y  3)( y  4)  0
y 3  0
y 3
or
y4 0
y4
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Example
Solve 14x2 + 9x + 2 = 10x + 6.
Solution
14x2 + 9x + 2 = 10x + 6
14x2 + 9x  10x + 2  6 = 0
14x2  x  4 = 0
(7x  4)(2x + 1) = 0
7x  4 = 0 or 2x + 1 = 0
7x = 4 or
2x = 1
x = 4/7 or
x = 1/2
The solutions are 4/7 and 1/2.
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Example
The equation h(t) = 16t2 + vot + h0 describes the height,
h, in feet of an object t seconds after being thrown
upwards with an initial velocity of v0 feet per second
from an initial height of h0 feet. Suppose a cannonball is
fired from a platform that is 384 feet above the ground .
The initial velocity of the object is 160 feet per second.
How long will it take for the cannonball to land on the
ground?
Understand We are given a formula, the initial velocity
of 160 feet per second, and the initial
height of 384 feet. We are to find the time
it takes the cannonball to reach the ground,
which is at 0 feet.
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continued
Plan Replace the variables in the formula with given values for h0
and v0, then solve for t.
Execute
0 = –16t2 + 160t + 384
= –16(t2 – 10t – 24)
= –16(t – 12)(t + 2)
t – 12 = 0 or t + 2 = 0
t = 12 or
t = 2
Answer Our answer must describe the amount of time, so only the
positive value, 12 makes sense. The cannonball takes 12
seconds to reach the ground.
Check h = 16(12)2  160(12)  384
= 2304 – 1920 – 384 = 0
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The Pythagorean Theorem
Given a right triangle, where a and b represent the
lengths of the legs and c represents the length of
the hypotenuse, then a2 + b2 = c2.
c (hypotenuse)
a (leg)
b (leg)
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Example
The figure shows a garden to be created. Find the length, in feet,
of each side.
x+4
x+2
x
Understand We are given expressions for the lengths of the three
sides of a right triangle and we are to find those lengths.
Pan and Execute Use the Pythagorean theorem.
(x)2 + (x + 2)2 = (x + 4)2
x2 + x2 + 4x + 4 = x2 + 8x + 16
x2 – 4x – 12 = 0
(x + 2)(x – 6) = 0
x + 2 = 0 or x – 6 = 0
x = –2 or
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x=6
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continued
Answer Because x describes a length in feet, only the
positive solution is sensible so x must be 6 feet.
This means x + 2 is 8 feet and x + 4 is 10 feet.
Check
(6)2 + (8)2 = (10)2
36 + 64 = 100
100 = 100
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