Transcript Intermediate Algebra - Seminole State College

Intermediate Algebra
Exam 1 Material
Factoring and Applications
Factoring Numbers
• To factor a number is to write it as a product
of two or more other numbers, each of which
is called a factor
12 = (3)(4)
3 & 4 are factors
12 = (6)(2)
6 & 2 are factors
12 = (12)(1)
12 and 1 are factors
12 = (2)(2)(3)
2, 2, and 3 are factors
In the last case we say the 12 is “completely
factored” because all the factors are prime
numbers
Prime Numbers
• Numbers, not including 1, whose only
factors are themselves and 1
2, 3, 5, 7, 11, 13, 17, 19, 23, etc.
Factoring Groups of Numbers to
Find the Greatest Common Factor
• Factor each number completely
• Construct the Greatest Common Factor by
including all factors that are common to all
groups
• Example: Find the GCF of: 30, 12, and 18
30 = (2)(3)(5), 12= (2)(2)(3), 18 = (2)(3)(3)
GCF = (2)(3) = 6
Find the GCF of the terms:
18x 2 y, 27x 4 y 2 , 12x3 y 2
18x 2 y  2  3  3  x  x  y
27x 4 y 2  3  3  3  x  x  x  x  y  y
12x y  2  2  3  x  x  x  y  y
3
2
GCF 
2
3x y
Polynomial
• A finite sum of terms
– Term: a number, a variable, or any product of
numbers and variables
• Examples of polynomials:
6 x 2  x  15
5x y  xy 15x y  8
2
8x3  27 y 6
4
3
Factoring Polynomials
• To factor a polynomial is to write it as a product of two or more
other polynomials, each of which is called a factor
• In a sense, factoring is the opposite of multiplying polynomials:
We have learned that:
(2x – 3)(3x + 5) = 6x2 + x – 15
If we were asked to factor 6x2 + x – 15 we would write it as:
(2x – 3)(3x + 5)
So we would say that (2x – 3) and (3x + 5) are factors of
6x2 + x – 15
Prime Polynomials
• A polynomial is called prime, if it is not 1,
and if its only factors are itself and 1
• Just like we learn to identify certain
numbers as being prime we will learn to
identify certain polynomials as being
prime
• We will also completely factor
polynomials by writing them as a
product of prime polynomials
Importance of Factoring
• If you don’t learn to factor polynomials you
can’t pass college algebra or more
advanced math classes
• It is essential that you memorize the
following procedures and become
proficient in using them
5 Steps in Completely Factoring
a Polynomial
• (1) Write the polynomial in descending
powers of one variable (if there is more than
one variable, pick any one you wish)
2x + 3x2 – 1 would be written:
3x2 + 2x – 1
3xy2 + y3 + 4x2y – 3 could be written:
y3 + 3xy2 + 4x2y – 3 (powers of y)
4x2y + 3xy2 + y3 – 3 (powers of x)
5 Steps in Completely Factoring
a Polynomial
• (2) Look at each term of the polynomial to
see if every term contains a common factor
other than 1, if so, use the distributive
property in reverse to place the greatest
common factor outside a parentheses and
other terms inside parentheses that give a
product equal to the original polynomial
• In the previous two examples what was the
greatest common factor found in all terms:
3x2 + 2x – 1
y3 + 3xy2 + 4x2y – 3
1, so it's not necessaryto factorout theGCF
Factoring the Greatest Common
Factor from Polynomials
•
9y5
+
y2
What is the GCF?
y2
y 2(
)
y2(9y3 + 1)
• 6x2t + 8xt + 12t
2t(
)
2t(3x2 + 4x + 6)
What is the GCF?
2t
Factoring the Greatest Common
Factor from Polynomials
• 14m4(m + 1) – 28m3(m + 1) – 7m2(m + 1)
What is the GCF?
7m2 m  1
7m2(m + 1)(
)
7m2(m + 1)(2m2 – 4m – 1)
7m2(m + 1)(2m2 – 4m – 1)
5 Steps in Completely Factoring
a Polynomial
• (3) After factoring out the greatest
common factor, look at the new
polynomial factors to determine how
many terms each one contains
(The fourth step will depend on the
number of terms in each of the factors)
5 Steps in Completely Factoring
a Polynomial
• (4) Use the method appropriate to the number of terms in the
polynomial:
4 or more terms: “Factor by Grouping”
3 terms: PRIME UNLESS they are of the form “ax2 + bx + c”. If of
this form, use “Trial and Error FOIL” or “abc Grouping”
2 terms: Always PRIME UNLESS they are:
“difference of squares”:
“difference of cubes”:
“sum of cubes”:
a2 – b2
a3 – b3
a3 + b 3
In each of these cases factor by a formula
5 Steps in Completely Factoring
a Polynomial
• (5) Cycle through step 4 as many times
as necessary until all factors are
“prime”
Factor by Grouping
(Used for 4 or more terms)
(1) Group the terms by underlining:
If there are exactly 4 terms try:
2 & 2 grouping, 3 & 1 grouping, or 1 & 3
grouping
If there are exactly 5 terms try:
3 & 2 grouping, or 2 & 3 grouping
Factoring by Grouping
(2) Factor each underlined group as if it were a factoring problem by
itself
(3) Now determine if the underlined and factored groups contain a
common factor,
if they contain a common factor, factor it out
if they don’t contain a common factor, try other groupings, if
none work, the polynomial is prime
(4) Once again count the terms in each of the new polynomial
factors and return to step 4.
Example of Factoring by
Grouping
Factor: ax + ay + 6x + 6y
(1) Group the terms by underlining (start with 2
and 2 grouping):
ax + ay + 6x + 6y
(2) Factor each underlined group as if it were a
factoring problem by itself:
a(x + y) + 6(x + y)
[notice sign between groups gets carried down]
Factoring by Grouping Example
Continued
(3)
Now determine if the underlined and factored groups contain a
common factor, if they do, factor it out:
a(x + y) + 6(x + y)
(x + y)(a + 6)
ax + ay + 6x + 6y = (x + y)(a + 6)
Now factored
(4) Once again count the terms in each of the new polynomial factors
and return to step 4.
Each of these polynomial factors contains two terms, return to
step 4 to see if these will factor (SINCE WE HAVE NOT YET
DISCUSSED FACTORING POLYNOMIALS WITH TWO TERMS
WE WILL NOT CONTINUE AT THIS TIME)
Example of Factoring by
Grouping
Factor:
2 x 2  3x  2 xy  3 y
(1) Group the terms by underlining (Try 2 and 2 grouping):
2 x  3x  2 xy  3 y
2
(2) Factor each underlined group as if it were a factoring
problem by itself:
x2 x  3  y2 x  3
[notice sign between groups gets carried down and you
have to be careful with this sign]
Factoring by Grouping Example
Continued
(3)
Now determine if the underlined and factored groups contain a
common factor, if they do, factor it out:
x2 x  3  y2 x  3
2 x  3x  y 
Now factored
(4) Once again count the terms in each of the new polynomial factors
and return to step 4.
Each of these polynomial factors contains two terms, return to
step 4 to see if these will factor (AGAIN WE HAVE LEARNED TO
FACTOR BINOMIALS YET, SO WE WON’T CONTINUE ON
THIS EXAMPLE)
Note on Factoring by Grouping
• It was noted in step 3 of the factor by grouping
steps that sometimes the first grouping, or the
first arrangement of terms might not result in
giving a common factor in each term – in that
case other groupings, or other arrangements of
terms, must be tried
• Only after we have tried all groupings and all
arrangement of terms can we determine whether
the polynomial is factorable or prime
Try Factoring by Grouping
Without First Rearranging
Factor: 12x  3 y  9 xy  4
(1) Group the terms by underlining (Try 2 and 2):
12x  3 y  9 xy  4
(2) Factor each underlined group as if it were a
factoring problem by itself:
3
  1

34 x  y   19 xy  4
.
What's theproblemwith trying to continue?
No commonfactorin thetwo underlinedgroups!
Now Try Same Problem
by Rearranging
12x  3 y  9 xy  4
Factor:
Rearrange: 9 xy  12x  3 y  4
(1) Group the terms by underlining:
9 xy  12x  3 y  4
(2) Factor each underlined group as if it were a
factoring problem by itself:
3x
 1

3x3 y  4 13 y  4
Can we continuefactoringnow?
Yes, thereis a commonfactorin thetwo underlinedgroups!
.
Factoring by Grouping Example
Continued
(3)
Now factor out the common factor:
3x3 y  4 13 y  4
3 y  4

3 y  43x 1
(4)
Rearranging made factoringpossible!
DOESN' T ALWAYS HELP!
Once again count the terms in each of the new polynomial factors
and return to step 4.
Each of these polynomial factors contains two terms, return to
step 4 to see if these will factor (AGAIN WE TO WAIT UNTIL WE
LEARN TO FACTOR BINOMIALS BEFORE WE CAN
CONTINUE)
Additional Note on
Factoring by Grouping
• In all of our examples we tried this method on
polynomials with four terms and used only “2 and 2
grouping”
• Sometimes with four terms we must use “3 and 1
grouping” or “1 and 3 grouping”
• On polynomials with more than four terms other
combinations of groupings must be tried
• In any case, all combinations of grouping must be tried
before we can determine of the polynomial is factorable
or is prime
• We will not deal with any of these situations at this time
Homework Problems
• Section: 5.1
• Page: 334
• Problems: Odd: 1 – 5, 9 – 19, 23 – 57,
59 – 63, 67 – 85
• MyMathLab Homework Assignment 5.1 for
practice
• MyMathLab Quiz 5.1 for grade
More on Factoring
• In the overall scheme of “factoring polynomials
completely” we need to know how to factor
polynomials containing various numbers of
terms
• Thus far we have learned that regardless of the
number of terms, we should always attempt, as
a first step, to factor out the GCF
• We have also learned that for a polynomial with
four or more terms we can try the “factor by
grouping” method
• We next learn methods for factoring polynomials
with three terms (trinomials)
Factoring Trinomials by
Trial and Error FOIL
(Used for 3 terms of form ax2 + bx + c)
• Given a trinomial if this form, experiment to try to find two
binomials that could multiply to give that trinomial
• Remember that when two binomials are multiplied:
First times First = First Term of Trinomial
Outside times Outside + Inside times Inside = Middle Term of
Trinomial
Last times Last = Last Term of Trinomial
Steps in Using Trial and Error FOIL
•
Given a trinomial of the form:
ax  bx  c
2
•
Write two blank parentheses that will
each eventually contain a binomial

•


Use the idea that “first times first = first”
to get possible answers for first term of
each binomial



Continuing Steps in Trial and Error
FOIL
• Given a trinomial of the form:
ax  bx  c
2
• Next use the idea that “last times last =
last” to get possible answers for last term
of each binomial



Continuing Steps in Trial and Error
FOIL
• Given a trinomial of the form:
ax  bx  c
2
• Finally use the idea that “Outside times
Outside + Inside times Inside = Middle
Term of Trinomial” to get the final answer
for two binomials that multiply to give the
trinomial



Prime Trinomials
• A trinomial is automatically prime if it is not of the
form: ax2  bx  c
• However, a trinomial of this form is also prime if
all possible combinations of “trial and error
FOIL” have been tried, and none have yielded
the correct middle term
2
• Example: Why is this prime? x  5 x  3
• The only possible combinations that give the
correct first and last terms are:
x  3x  1
and x  3x 1
• Neither gives the correct middle term:
x  2 x  3 and x  2 x  3
2
2
Example of Factoring by
Trial and Error FOIL
•
•
Factor:
12x2 + 11x – 5
Using steps on previous slides, we see all the possibilities that give the
correct first and last terms on the left and the result of multiplying them on
the right (we are looking for the one that gives the correct middle term):
(12x + 1)(x – 5) = 12x2 – 59x – 5
(12x – 1)(x + 5) = 12x2 + 59x – 5
(12x + 5)(x – 1) = 12x2 – 7x – 5
(12x – 5)(x + 1) = 12x2 + 7x – 5
(6x + 1)(2x – 5) = 12x2 – 28x – 5
(6x – 1)(2x + 5) = 12x2 +28x – 5
(6x + 5)(2x – 1) = 12x2 + 4x – 5
(6x – 5)(2x + 1) = 12x2 – 4x – 5
(4x + 1)(3x – 5) = 12x2 – 17x – 5
(4x – 1)(3x + 1) = 12x2 + x – 5
Only CorrectFactoring
(4x + 5)(3x – 1) = 12x2 +11x – 5
(4x – 5)(3x + 1) = 12x2 -11x – 5
Factoring
“Perfect Square Trinomials”
• A trinomial is a “perfect square trinomial” if it has
resulted from squaring a binomial:
a  b2  a2  2ab  b2 and a  b2  a2  2ab  b2
• Perfect square trinomials have the characteristic
that the first and last terms are perfect squares,
and the middle term is twice the product of the
square roots of the first and last terms
• When this characteristic is seen in a trinomial,
we automatically factor it as a binomial squared
being careful to place the correct sign in the
middle
Example of Factoring
“Perfect Square Trinomials”
• Which of these is a perfect square trinomial?
4 x 2  5x  3 or 9 x 2  12x  4
• Only the second has first and last terms that are
perfect squares and a middle term that is twice
the product of the square roots of the first and
last terms
• The second can be factored by writing a
binomial, with a square on it, whose first and last
terms are the square roots of the first and last
terms, that has a middle sign that matches the
sign of the middle term of the trinomial
9x 12x  4  3x  2 
2
2
Homework Problems
• Section: 5.2
• Page: 340
• Problems: Odd: 11 – 39, 43 – 59,
63 – 71
• MyMathLab Homework Assignment 5.2 for
practice
• MyMathLab Quiz 5.2 for grade
A Second Method of Factoring
Trinomials
• While the “Trial and Error FOIL” method
can always be used in attempting to factor
trinomials, and is usually best when first
and last terms have “small coefficients,”
there is a second method that is usually
best to use when first and last coefficients
are “larger”
• We call the second method:
“abc grouping”
Factoring Trinomials by
abc Grouping
(Used for 3 terms of form ax2 + bx + c)
•
(1)
(2)
(3)
(4)
(5)
When a polynomial is of this form:
ax2 + bx + c
Identify “a”, “b”, and “c”
Multiply “a” and “c”
Find two numbers “m” and “n”, that multiply to
give “ac” and add to give “b” (If this can not be
done, the polynomial is already prime)
Rewrite polynomial as: ax2 + mx + nx + c
Factor these four terms by 2 and 2 grouping
Example of Factoring by
abc Grouping
•
(1)
Factor:
12x2 + 11x – 5
Identify “a”, “b”, and “c”
a = 12, b = 11, c = - 5
ac  60

15  4
m  n  11
(2)
Multiply “a” and “c”
ac = - 60
(3)
Find two numbers “m” and “n”, that multiply to give “ac” and add to give
“b” (If this can not be done, the polynomial is already prime)
m = 15 and n = - 4, because mn = -60 and m + n = 11
(4)
Rewrite as four terms:
12x2 + 15x – 4x – 5
(5)
Factor by grouping:
12x2 + 15x – 4x – 5
3x(4x + 5) – 1(4x + 5)
(4x + 5)(3x – 1)
Example of Factoring by
abc Grouping (with two variables)
•
(1)
35x2 – 12y2 – 13xy
35x2 – 13xy – 12y2 (descending powers of x)
Identify “a”, “b”, and “c”
(Ignore y variable)
a = 35, b = - 13, c = - 12
Factor:
(2)
Multiply “a” and “c”
ac = - 420
(3)
Find two numbers “m” and “n”, that multiply to give “ac” and add to give “b” (If this
can not be done, the polynomial is already prime)
m = 15 and n = - 28, because mn = - 420 and m + n = - 13
(4)
Rewrite as four terms:
35x2 + 15xy – 28xy – 12y2
(5)
Factor by grouping:
35x2 + 15xy – 28xy – 12y2
5x(7x + 3y) – 4y(7x + 3y)
(7x + 3y)(5x – 4y)
Another Comment on Prime
Trinomials
• A trinomial is prime if it is not of the form:
ax  bx  c
2
• A trinomial of this form is also prime, if it
can not be factored by the “abc grouping
method”
Homework Problems
• Section: 5.3
• Page: 347
• Problems: Odd: 21 – 83
• MyMathLab Homework Assignment 5.3 for
practice
• MyMathLab Quiz 5.3 for grade
More on Factoring
• In the overall scheme of “factoring polynomials
completely” we need to know how to factor polynomials
containing various numbers of terms
• Thus far we have learned that regardless of the number
of terms, we should always attempt, as a first step, to
factor out the GCF
• We have also learned that for a polynomial with four or
more terms we can try the “factor by grouping” method
• We have also learned that we should try to factor
trinomials of the form ax2+bx+c by either “trial and error
FOIL” or “abc grouping”
• We next learn methods for factoring binomials
Note on Factoring Binomials
• Binomials are factorable only if they are a:
– Difference of Squares:
– Difference of Cubes:
– Sum of Cubes:
a 2  b2
3
3
a b
3
3
a b
• In each of these cases, factoring is done by
means of a formula that needs to be memorized
• All other binomials are prime (In saying this, we
assume that any GCF has already been factored
out)
Factoring Binomials by Formula
• Factor by using formula appropriate for the binomial:
“difference of squares”:
a2 – b2 = (a – b)(a + b)
“difference of cubes”:
a3 – b3 = (a – b)(a2 + ab + b2) Trinomial is prime
“sum of cubes”:
a3 + b3 = (a + b)(a2 – ab + b2) Trinomial is prime
• If none of the formulas apply, the binomial is prime
BINOMIALS ARE PRIME UNLESS THEY ARE ONE OF THESE
Example of Factoring Binomials
• Factor: 25x2 – 9y2
• Note that this binomial is a difference of
squares:
(5x)2 – (3y)2
• Using formula gives:
(5x – 3y)(5x + 3y)
Example of Factoring Binomials
• Factor: 8x3 – 27
• Note that this is a difference of cubes:
(2x)3 – (3)3
• Using formula gives:
(2x – 3)(4x2 + 6x + 9)
Example of Factoring Binomials
• Factor: 4x2 + 9
• Note that this is not a difference of
squares, difference of cubes, or sum of
cubes, therefore it is prime
• (4x2 + 9)
• To show factoring of a polynomial that is
prime, put it inside parentheses
Homework Problems
• Section: 5.4
• Page: 356
• Problems: Odd: 7 – 29, 59 – 81
• MyMathLab Homework Assignment 5.4 for
practice
• MyMathLab Quiz 5.4 for grade
Completely Factoring
Polynomials
•
We now have all the skills necessary to either factor
polynomials completely, or to determine if they are prime
• Summary of steps:
– Arrange polynomial in descending powers of one variable
– Factor out the GCF (also factor out a negative if highest degree
term has a negative coefficient)
– For each polynomial factor in the expression, try to factor it by
using the method appropriate for the number of terms it has
– Continue factoring each new polynomial factor until all
polynomial factors are prime
• We now apply this procedure in completely factoring
polynomials
Example of Factoring
Polynomials Using Five Steps
• Factor: 2x3 – 8x + 2x6 – 8x4
• (1) Write the polynomial in descending
powers of one variable (if there is more than
one variable, pick any one you wish)
2x6 – 8x4 + 2x3 – 8x
• (2) Look at each term of the polynomial to
see if every term contains a common factor,
if so, use the distributive property in reverse
to place the greatest common factor outside
a parentheses
2x(x5 – 4x3 + x2 – 4)
Example of Factoring
Polynomials Using Five Steps
2x(x5 – 4x3 + x2 – 4)
• (3) After factoring out the greatest common factor, look at the
new polynomial factors to determine how many terms each one
contains
• The polynomial in parentheses has 4 terms
• (4) Use the method appropriate to the number of terms in the
polynomial
Since there are 4 terms we will try “factor by grouping”:
x5 – 4x3 + x2 – 4
x3(x2 – 4) + 1(x2 – 4)
(x2 – 4)(x3 + 1)
So far we have factored the original polynomial as:
2x(x2 – 4)(x3 + 1)
Example of Factoring
Polynomials Using Five Steps
2x(x2 – 4)(x3 + 1)
• (5) Cycle through step 4 as many times as
necessary until all factors are “prime” (count
terms and use appropriate method)
The first binomial is a difference of squares, and
the second is a sum of cubes so they must be
factored by formulas to get the final complete
factoring of:
• 2x(x – 2)(x + 2)(x + 1)(x2 – x + 1)
COMPLETELY FACTORED!
Homework Problems
• Section: Summary Exercises on Factoring
• Page: 358
• Problems: Odd: 1 – 79
• MyMathLab Homework Factoring
Summary Assignment for practice
• MyMathLab Quiz Factoring Summary for
grade
Solving Equations
• You have previously learned to identify
and solve linear equations
• The next objective is to learn to identify
and solve “quadratic equations”
• Before beginning the new goal, we will
quickly review identifying and solving
linear equations
Linear Equations in One Variable
• Linear Equation: any polynomial equation in one
variable where, after parentheses are gone, the
highest degree term is “1”
• Examples:
3x  5  2 x  x  1
3
4 x   x  1.7   x
4
Solving Linear Equations
• Get rid of parentheses
• Get rid of fractions and decimals by multiplying
both sides by LCD
• Collect like terms
• Decide which side will keep variable terms and
get rid of variable terms on other side
• Get rid of non-variable terms on variable side
• Divide both sides by the coefficient of variable
Solve the Equation
2
1

2 x    .7 x  x 
3
2

• Identify the type of equation:
It is linear!
• Get rid of parentheses:
2x 
4
1
 .7 x  x 
3
2
• Get rid of fractions and decimals by multiplying
both sides by LCD: LCD of 3,10,and 2 is : 30
4
1



30 2 x   .7 x   30 x  
3
2



60 x  40  21x  30 x  15
Example Continued
60 x  40  21x  30 x  15
• Collect like terms:
39 x  40  30 x  15
• Decide which side will keep variable terms and
get rid of variable terms on other side:
If youchose to keep variableson left you willget rid of thoseon theright
9 x  40  15
• Get rid of non-variable terms on variable side:
9 x  55
• Divide both sides by coefficient of variable:
55
x
9
Identifying Quadratic Equations
• Technical Definition: any equation in one variable that
can be written in the form:
ax2 + bx + c = 0 where “a”, “b”, and “c” are real and a ≠ 0
(This form is called the “standard form”)
• Practical Definition: any polynomial equation in one
variable where, after parentheses are gone, the highest
degree term is “2”
• Examples:
5x2 + 7 = – 4x
9x2 = 4
2x(x – 3) = x – 1
Are any of thesein standardform?
No, but all could be put in standardform.
Solving Quadratic Equations by
Zero Factor Method
• Put equation in standard form (one side zero
other side in descending powers)
• Factor non-zero side
(If it won’t factor this method won’t work!)
• Use zero factor property that says, “if two
numbers multiply to get zero, one of them is
zero:”
ab = 0 if and only if a = 0 or b = 0
• Set each factor equal to zero
• Solve resulting equations
Solving by
Zero Factor Method
x2 x  5  3
Put in standard form:
2 x 2  5x  3
2 x 2  5x  3  0
Factor non-zero side:
2x 1x  3  0
Apply zero factor principle:
x3 0
2x 1  0
OR
Solve the equations:
x  3
2x  1
1
x
2
More Notes on
Solving Quadratic Equations
• All quadratic equations can be solved, but
the “zero factor method” works only when
the non-zero side of the quadratic
equation can be factored
• In college algebra we will learn methods of
solving quadratic equations that can not
be solved by the “zero factor method”
• Example: Why can’t this be solved by the
2
zero factor method? x  5x  2  0
x 2  5x  2 is primeand can't be factored
Solving Polynomial Equations
with Degree Higher than 2
• We have said that a quadratic equation is a
second degree polynomial equation and that
such an equation can sometimes be solved by
the “Zero Factor Method”
• It is also true that some higher degree
polynomial equations may be solved by this
method
• To attempt this method:
– Make one side zero
– Factor other side
– Apply Zero Factor Property (abcd=0 means a = 0,
b = 0, c = 0, or d = 0)
Example
3x  5 x  2 x
• Solve:
3
2
3x 3  5 x 2  2 x  0


x 3x  5x  2  0
2
x3x 1x  2  0
x  0 or 3x  1  0 or x  2  0
x  2
3x  1
1
x
3
Homework Problems
• Section: 5.5
• Page: 366
• Problems: Odd: 11 – 23, 27 – 81
• MyMathLab Homework Assignment 5.5 for
practice
• MyMathLab Quiz 5.5 for grade
Application Problems Involving
Quadratic Equations
• In previous algebra courses you have
learned to solve basic application (word)
problems
• In this course we learn that some
application problems translate to quadratic
equations
• Before introducing those type of problems,
we will review the basic approach to
solving application problems
Application Problems
• General methods for solving an applied (word)
problem:
1. Read problem carefully taking notes, drawing pictures, thinking
about formulas that apply, making charts, etc.
2. Read problem again to make a “word list” of everything that is
unknown
3. Give a variable name, such as “x” to the “most basic unknown” in
the list (the thing that you know the least about)
4. Give all other unknowns in your word list an algebraic expression
name that includes the variable, “x”
5. Read the problem one last time to determine what information
has been given, or implied by the problem, that has not been
used in giving an algebra name to the unknowns and use this
information to write an equation about the unknowns
6. Solve the equation and answer the original question
Solve the Application Problem
A 31 inch pipe needs to be cut into three pieces
in such a way that the second piece is 5 inches
longer than the first piece and the third piece is
twice as long as the second piece. How long
should the third piece be?
1. Read the problem carefully taking notes,
drawing pictures, thinking about formulas that
apply, making charts, etc.
Perhaps draw a picture of a pipe that is labeled
as 31 inches with two cut marks dividing it into
3 pieces labeled first, second and third
1st
2nd
3rd
31
Example Continued
2. Read problem again to make a “word list” of
everything that is unknown
What things are unknown in this problem?
The length of all three pieces (even though the
problem only asked for the length of the third).
Word List of Unknowns:
Length of first
Length of second
Length of third
Example Continued
3. Give a variable name, such as “x” to the “most
basic unknown” in the list (the thing that, if you
knew its value, the other values could be
found)
What is the most basic unknown in this list?
Length of first piece is most basic, because
problem describes second in terms of the first,
and third in terms of second
Give the name “x” to the length of first
Example Continued
4.
Give all other unknowns in the word list an algebraic
expression name that includes the variable, “x”
How would the length of the second be named?
x+5
How would the length of the third be named?
2(x + 5)
Word List of Unknowns:
Algebra Names:
Length of first
x
Length of second
x+5
Length of third
2(x + 5)
Example Continued
5.
Read the problem one last time to determine what
information has been given, or implied by the problem,
that has not been used in giving an algebra name to
the unknowns and use this information to write an
equation about the unknowns
What other information is given in the problem that has
not been used?
Total length of pipe is 31 inches
How do we say, by using the algebra names, that the
total length of the three pieces is 31?
x + (x + 5) + 2(x + 5) = 31
Example Continued
6.
Solve the equation and answer the original question
This is a linear equation so solve using the appropriate
steps:
x + (x + 5) + 2(x + 5) = 31
x + x + 5 + 2x + 10 = 31
4x + 15 = 31
4x = 16
x=4
Is this the answer to the original question?
No, this is the length of the first piece.
How do we find the length of the third piece?
The length of the third piece is 2(x + 5):
2(4 + 5) = (2)(9) = 18 inches = length of third piece
Example
A rectangular piece of metal is 2 inches longer than it is
wide. Four inch squares are cut from each corner to
make a box with a volume of 32 cubic inches. What
were the original dimensions of the metal?
4
4
x
4
x2
4
Unknowns
L Rec x  2
W Rec x
L Box x  2  8
W Box x  8
Height Box  4
x4
V  LWH
Impossible
32  x  6x  8 4
32  x 2 14x  48 4
x  10
2
32  4 x  56x  192
2
0  4 x  56x  160
W

10
in
.
2
0  x  14x  40
.
0  x  10x  4
L  12 in.
x  10  0 OR x  4  0


Example
The product of two consecutive odd integers is equal to
negative one minus the sum of the two integers. Find all
possible answers for the two integers.
x  4x  3  0
x  1x  3  0
x  1  0 OR x  3  0
x  3
x  1
x2 1
x  2  1 .
2
Unknowns
First Odd Int x
Next Odd Int x  2
xx  2  1  x  x  2
2
x  2x  1  2x  2
2
x  2 x  1  2 x  2
2
x  2 x  2 x  3
Homework Problems
• Section: 5.6
• Page: 375
• Problems: Odd: 7 – 29
• MyMathLab Homework Assignment 5.6 for
practice
• MyMathLab Quiz 5.6 for grade