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Chapter 13
Factoring
Polynomials
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13.1
The Greatest CommonFactor
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Martin-Gay,Developmental Mathematics 4
Factors
Factors(either numbers or polynomials)
When an integer is written as a product of
integers, each of the integers in the product is a
factorof the original number.When a polynomial is written as a product of
polynomials, each of the polynomials in the
product is a factorof the original polynomial.
Factoringwriting a polynomial as a product of
polynomials.
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Martin-Gay,Developmental Mathematics 5
Greatest common factorlargest quantity that is afactor of all the integers or polynomials involved.
F inding the GCF of a List of I ntegers or Terms
1) Prime factor the numbers.
2) Identify common prime factors.
3) Take the product of all common prime factors. If there are no common prime factors, GCF is 1.
Greatest Common Factor
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Martin-Gay,Developmental Mathematics 6
Find the GCF of each list of numbers.
1) 12 and 8
12 = 2 238 = 222
So the GCF is 22= 4.
2) 7 and 20
7 = 1 7
20 = 2 2 5
There are no common prime factors so the
GCF is 1.
Greatest Common Factor
Example
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Find the GCF of each list of numbers.
1) 6, 8 and 46
6 = 2 3
8 = 2 2 2
46 = 2 23
So the GCF is 2.
2) 144, 256 and 300
144 = 2 2 2 3 3
256 = 2 2 2 2 2 2 2 2
300 = 2 2 3 5 5
So the GCF is 2 2= 4.
Greatest Common Factor
Example
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1) x3andx7
x3
= x x xx7 = x x xxxxx
So the GCF is x x x=x3
2) 6x5
and 4x3
6x5= 2 3 x x x
4x3= 2 2 x x x
So the GCF is 2 x x x= 2x3
Find the GCF of each list of terms.
Greatest Common Factor
Example
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Find the GCF of the following list of terms.
a3b2, a2b5and a4b7
a3b2= a a a b b
a2b5= a a b b b b b
a4b7= a a a a b b b b b b b
So the GCF is a a b b = a2b2
Notice that the GCF of terms containing variables will use the
smallest exponent found amongst the individual terms for each
variable.
Greatest Common Factor
Example
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The first step in factoring a polynomial is tofind the GCF of all its terms.
Then we write the polynomial as a product by
factor ing outthe GCF from all the terms.
The remaining factors in each term will form a
polynomial.
Factoring Polynomials
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Factor out the GCF in each of the following
polynomials.
1) 6x39x2+ 12x=
3 x 2 x23 x 3 x+ 3 x 4 =
3x(2x23x+ 4)
2) 14x3y+ 7x2y7xy =
7 x y 2 x2+ 7 x y x7 x y 1 =
7xy(2x2+x1)
Factoring out the GCF
Example
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Factor out the GCF in each of the following
polynomials.
1) 6(x+ 2)y(x+ 2) =
6 (x+ 2)y(x+ 2) =
(x+ 2)(6y)
2)xy(y+ 1)(y+ 1) =
xy(y+ 1)1 (y+ 1)=
(y+ 1)(xy1)
Factoring out the GCF
Example
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Remember that factoring out the GCF from the terms of
a polynomial should always be the first step in factoring
a polynomial.
This will usually be followed by additional steps in the
process.
Factor 90 + 15y218x3xy2.
90 + 15y2
18x3xy2
= 3(30 + 5y2
6xxy2
) =3(5 6 + 5y26 xxy2) =
3(5(6 + y2)x(6 + y2)) =
3(6 + y2
)(5x)
Factoring
Example
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13.2
Factoring Trinomials of theForm x2+ bx+ c
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Factoring Trinomials
Recall by using the FOIL method thatF O I L
(x+ 2)(x+ 4) =x2+ 4x+ 2x+ 8
=x2+ 6x+ 8
To factorx2+ bx+ cinto (x+ one #)(x+ another #),note that bis the sum of the two numbers and cis the
product of the two numbers.
So well be looking for 2 numbers whose product iscand whose sum is b.
Note: there are fewer choices for the product, sothats why we start there first.
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Factor the polynomialx2+ 13x+ 30.
Since our two numbers must have a product of 30 and a
sum of 13, the two numbers must both be positive.
Positive factors of 30 Sum of Factors
1, 30 31
2, 15 17
3, 10 13
Note, there are other factors, but once we find a pairthat works, we do not have to continue searching.
Sox
2
+ 13x+ 30 = (x+ 3)(x+ 10).
Factoring Polynomials
Example
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Factor the polynomialx211x+ 24.
Since our two numbers must have a product of 24 and a
sum of -11, the two numbers must both be negative.Negative factors of 24 Sum of Factors
1,24 25
2,
12
14
3,8 11
Sox211x+ 24 = (x3)(x8).
Factoring Polynomials
Example
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Factor the polynomialx22x35.
Since our two numbers must have a product of35 and a
sum of2, the two numbers will have to have different signs.Factors of35 Sum of Factors
1, 35 34
1,35 34
5, 7 2
5,7 2
Sox22x35 = (x+ 5)(x7).
Factoring Polynomials
Example
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Factor the polynomialx26x+ 10.
Since our two numbers must have a product of 10 and a
sum of6, the two numbers will have to both be negative.
Negative factors of 10 Sum of Factors
1,10 11
2,5 7Since there is not a factor pair whose sum is6,
x26x+10 is not factorable and we call it a primepolynomial.
Prime Polynomials
Example
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You should always check your factoring
results by multiplying the factored polynomial
to verify that it is equal to the original
polynomial.
Many times you can detect computational
errors or errors in the signs of your numbers
by checking your results.
Check Your Result!
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13.3
Factoring Trinomials ofthe Form ax2+ bx+ c
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Factoring Trinomials
Returning to the FOIL method,F O I L
(3x + 2)(x + 4) = 3x2+ 12x + 2x + 8
= 3x2
+ 14x + 8To factor ax2+ bx+ cinto (#1x+ #2)(#3x+ #4),note that ais the product of the two first coefficients,cis the product of the two last coefficients and bis
the sum of the products of the outside coefficientsand inside coefficients.
Note that bis the sum of 2 products, not just 2numbers, as in the last section.
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Factor the polynomial 25x2+ 20x+ 4.
Possible factors of 25x2are {x, 25x} or {5x, 5x}.
Possible factors of 4 are {1, 4} or {2, 2}.We need to methodically try each pair of factors until we find
a combination that works, or exhaust all of our possible pairs
of factors.
Keep in mind that, because some of our pairs are not identicalfactors, we may have to exchange some pairs of factors and
make 2 attempts before we can definitely decide a particular
pair of factors will not work.
Factoring Polynomials
Example
Continued.
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Martin-Gay,Developmental Mathematics 24
We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 20x.
{x, 25x} {1, 4} (x+ 1)(25x+ 4) 4x 25x 29x
(x+ 4)(25x+ 1) x 100x 101x
{x, 25x} {2, 2} (x+ 2)(25x+ 2) 2x 50x 52x
Factorsof 25x2 ResultingBinomials Product ofOutside Terms Product ofInside Terms Sum ofProductsFactorsof 4
{5x, 5x} {2, 2} (5x+ 2)(5x+ 2) 10x 10x 20x
Factoring Polynomials
Example Continued
Continued.
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Martin-Gay,Developmental Mathematics 26
Factor the polynomial 21x241x+ 10.
Possible factors of 21x2are {x, 21x} or {3x, 7x}.
Since the middle term is negative, possible factors of 10
must both be negative: {-1, -10} or {-2, -5}.
We need to methodically try each pair of factors untilwe find a combination that works, or exhaust all of our
possible pairs of factors.
Factoring Polynomials
Example
Continued.
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Martin-Gay,Developmental Mathematics 27
We will be looking for a combination that gives the sum of
the products of the outside terms and the inside terms equal
to 41x.
Factors
of 21x2Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Factors
of 10
{x, 21x}{1, 10}(x1)(21x10) 10x 21x 31x
(x
10)(21x1) x 210x 211x
{x, 21x} {2, 5} (x2)(21x5) 5x 42x 47x
(x5)(21x2) 2x 105x 107x
Factoring Polynomials
Example Continued
Continued.
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Factors
of 21x2Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
Factors
of 10
(3x
5)(7x2) 6x 35x 41x
{3x, 7x}{1, 10}(3x
1)(7x10) 30x 7x 37x
(3x10)(7x1) 3x 70x 73x
{3x, 7x} {2, 5} (3x2)(7x5) 15x 14x 29x
Factoring Polynomials
Example Continued
Continued.
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Martin-Gay,Developmental Mathematics 29
Check the resulting factorization using the FOIL method.
(3x
5)(7x2)=
= 21x26x35x+ 10
3x(7x)F
+ 3x(-2)
O
- 5(7x)
I
- 5(-2)
L
= 21x241x+ 10
So our final answer when asked to factor 21x241x+ 10
will be (3x5)(7x2).
Factoring Polynomials
Example Continued
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Factor the polynomial 3x27x+ 6.
The only possible factors for 3 are 1 and 3, so we know that, if
factorable, the polynomial will have to look like (3x )(x )in factored form, so that the product of the first two terms in the
binomials will be 3x2.
Since the middle term is negative, possible factors of 6 must both
be negative: {1, 6} or {2, 3}.
We need to methodically try each pair of factors until we find a
combination that works, or exhaust all of our possible pairs of
factors.
Factoring Polynomials
Example
Continued.
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We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 7x.
{1, 6} (3x1)(x6) 18x x 19x
(3x
6)(x1) Common factor so no need to test.{2, 3} (3x2)(x3) 9x 2x 11x
(3x3)(x2) Common factor so no need to test.
Factorsof 6
ResultingBinomials
Product ofOutside Terms
Product ofInside Terms
Sum ofProducts
Factoring Polynomials
Example Continued
Continued.
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Martin-Gay,Developmental Mathematics 34
Since the product of the last two terms of the binomials
will have to be30, we know that they must be
different signs.
Possible factors of30 are {1, 30}, {1,30}, {2, 15},
{2,15}, {3, 10}, {3,10}, {5, 6} or {5,6}.
We will be looking for a combination that gives the sum
of the products of the outside terms and the inside terms
equal tox.
Factoring Polynomials
Example Continued
Continued.
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Factors
of -30
Resulting
Binomials
Product of
Outside Terms
Product of
Inside Terms
Sum of
Products
{-1, 30} (3x1)(x+ 30) 90x -x 89x
(3x + 30)(x1) Common factor so no need to test.
{1, -30} (3x+ 1)(x30) -90x x -89x
(3x30)(x+ 1) Common factor so no need to test.
{-2, 15} (3x2)(x+ 15) 45x -2x 43x
(3x + 15)(x2) Common factor so no need to test.
{2, -15} (3x+ 2)(x15) -45x 2x -43x
(3x15)(x+ 2) Common factor so no need to test.
Factoring Polynomials
Example Continued
Continued.
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Martin-Gay,Developmental Mathematics 37
Check the resulting factorization using the FOIL method.
(3x10)(x+ 3) =
= 3x2+ 9x10x30
3x(x)
F
+ 3x(3)
O
10(x)
I
10(3)
L
= 3x2x30
So our final answer when asked to factor the polynomial
6x2y22xy260y2will be 2y2(3x10)(x+ 3).
Factoring Polynomials
Example Continued
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13.4
Factoring Trinomials of
the Form x2+ bx+ cby Grouping
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Factoring polynomials often involves additionaltechniques after initially factoring out the GCF.
One technique is factor ing by grouping.
Factorxy+y+ 2x+ 2 by grouping.
Notice that, although 1 is the GCF for all four
terms of the polynomial, the first 2 terms have a
GCF ofyand the last 2 terms have a GCF of 2.
xy+y+ 2x+ 2 =xy+ 1 y+ 2x+ 21 =
y(x+ 1)+ 2(x+ 1)= (x+ 1)(y+ 2)
Factoring by Grouping
Example
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1) x3+ 4x+x2+ 4 = xx2+ x 4 + 1x2+ 1 4 =
x(x2+ 4)+ 1(x2+ 4)=
(x2+ 4)(x+ 1)
2) 2x3x210x+ 5 = x2 2xx2 15 2x5 (1) =
x2
(2x
1)5(2x
1)=(2x1)(x25)
Factor each of the following polynomials by grouping.
Factoring by Grouping
Example
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Martin-Gay,Developmental Mathematics 42
Factor 2x9y+ 18xyby grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2x+ 189yxy= 2x+ 2 99 yx y=
2(x+ 9)y(9 +x) =
2(x+ 9)y(x+ 9)= (make sure the factors are identical)(x + 9)(2y)
Factoring by Grouping
Example
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13.5
Factoring Perfect Square
Trinomials and theDifference of Two Squares
P f S T i i l
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Recall that in our very first example in Section4.3 we attempted to factor the polynomial
25x2+ 20x+ 4.
The result was (5x+ 2)2, an example of a
binomial squared.
Any trinomial that factors into a singlebinomial squared is called a perfect square
trinomial.
Perfect Square Trinomials
P f S T i i l
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In the last chapter we learned a shortcut for squaring a
binomial
(a+ b)2= a2+ 2ab+ b2
(ab)2= a22ab+ b2
So if the first and last terms of our polynomial to be
factored are can be written as expressions squared, and
the middle term of our polynomial is twice the product
of those two expressions, then we can use these twoprevious equations to easily factor the polynomial.
a2+ 2ab+ b2 =(a+ b)2
a22ab+ b2 = (ab)2
Perfect Square Trinomials
P f t S T i i l
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Factor the polynomial 16x28xy+y2.
Since the first term, 16x2, can be written as (4x)2, and
the last term,y2is obviously a square, we check the
middle term.
8xy= 2(4x)(y) (twice the product of the expressions
that are squared to get the first and last terms of the
polynomial)Therefore 16x28xy+y2= (4xy)2.
Note: You can use FOIL method to verify that the
factorization for the polynomial is accurate.
Perfect Square Trinomials
Example
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Difference of Two Squares
Another shortcut for factoring a trinomial is when we
want to factor the difference of two squares.
a2b2= (a + b)(ab)
A binomial is the difference of two square if
1.both terms are squares and
2.the signs of the terms are different.
9x225y2
c4+ d4
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Difference of Two Squares
Example
Factor the polynomialx29.
The first term is a square and the last term, 9, can be
written as 32. The signs of each term are different, so
we have the difference of two squares
Thereforex29 = (x3)(x+ 3).
Note: You can use FOIL method to verify that thefactorization for the polynomial is accurate.
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13.6
Solving Quadratic
Equations by Factoring
Z F t Th
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Zero Factor Theorem
Quadratic Equations Can be written in the form ax2+ bx+ c= 0.
a, band care real numbers and a0.
This is referred to as standard form.Zero Factor Theorem
If aand bare real numbers and ab= 0, then a= 0
or b= 0. This theorem is very useful in solving quadratic
equations.
S l i Q d ti E ti
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Steps for Solving a Quadratic Equation byFactoring
1) Write the equation in standard form.
2) Factor the quadratic completely.3) Set each factor containing a variable equal to 0.
4) Solve the resulting equations.
5) Check each solution in the original equation.
Solving Quadratic Equations
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Solvex25x= 24.
First write the quadratic equation in standard form.
x2
5x24 = 0 Now we factor the quadratic using techniques from
the previous sections.
x25x24 = (x8)(x+ 3) = 0
We set each factor equal to 0.
x8 = 0 orx+ 3 = 0, which will simplify to
x= 8 orx=3
Solving Quadratic Equations
Example
Continued.
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Check both possible answers in the original
equation.
825(8) = 6440 = 24 true
(3)25(3) = 9(15) = 24 true
So our solutions forxare 8 or3.
Example Continued
Solving Quadratic Equations
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Martin-Gay,Developmental Mathematics 54
Solve 4x(8x+ 9) = 5
First write the quadratic equation in standard form.
32x2+ 36x= 5
32x2+ 36x5 = 0
Now we factor the quadratic using techniques from the
previous sections.
32x2+ 36x5 = (8x1)(4x+ 5) = 0
We set each factor equal to 0.
8x1 = 0 or 4x+ 5 = 0
Solving Quadratic Equations
Example
Continued.
8x= 1 or 4x =5, which simplifies tox = or5.
4
1
8
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Martin-Gay,Developmental Mathematics 55
Check both possible answers in the original equation.
1 1 1
4 8 9 4 1 9 4 (10) (10) 58
1
8
1
8 8 2
true
5 54 8 9 4 10 9 4 ( 1) ( 5)( 1) 545 54 44 true
So our solutions for x are or .8
1
4
5
Example Continued
Solving Quadratic Equations
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Finding x intercepts
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Find thex-intercepts of the graph ofy= 4x2+ 11x+ 6.
The equation is already written in standard form, so
we lety= 0, then factor the quadratic inx.
0 = 4x2+ 11x+ 6 = (4x+ 3)(x+ 2)
We set each factor equal to 0 and solve for x.
4x+ 3 = 0 orx+ 2 = 0
4x=3 orx=2
x= orx=2
So thex-intercepts are the points (, 0) and (2, 0).
Finding x-intercepts
Example
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13.7
Quadratic Equations
and Problem Solving
Strategy for Problem Solving
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Strategy for Problem Solving
General Strategy for Problem Solving1) Understand the problem
Read and reread the problem
Choose a variable to represent the unknown
Construct a drawing, whenever possible Propose a solution and check
2) Translate the problem into an equation
3) Solve the equation
4) Interpret the result Check proposed solution in problem
State your conclusion
Finding an Unknown Number
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The product of two consecutive positive integers is 132. Find the
two integers.
1.) Understand
Read and reread the problem. If we let
x= one of the unknown positive integers, then
x+ 1 = the next consecutive positive integer.
Finding an Unknown Number
Example
Continued
Finding an Unknown Number
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Martin-Gay,Developmental Mathematics 61
Finding an Unknown Number
Example continued
2.) Translate
Continued
two consecutive positive integers
x (x+ 1)
is
=
132
132
The product of
Finding an Unknown Number
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Martin-Gay,Developmental Mathematics 62
Finding an Unknown Number
Example continued3.) Solve
Continued
x(x+ 1) = 132
x2+x= 132 (Distributive property)
x2+x132 = 0 (Write quadratic in standard form)
(x+ 12)(x11) = 0 (Factor quadratic polynomial)
x+ 12 = 0 orx11 = 0 (Set factors equal to 0)
x=12 orx= 11 (Solve each factor forx)
Finding an Unknown Number
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Martin-Gay,Developmental Mathematics 63
Finding an Unknown Number
Example continued
4.) Interpret
Check: Remember thatxis suppose to represent a positive
integer. So, althoughx= -12 satisfies our equation, it cannot be asolution for the problem we were presented.
If we letx= 11, thenx+ 1 = 12. The product of the two numbers
is 11 12 = 132, our desired result.
State: The two positive integers are 11 and 12.
The Pythagorean Theorem
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Pythagorean Theorem
In a right triangle, the sum of the squares of the
lengths of the two legs is equal to the square of the
length of the hypotenuse.
(leg a)2+ (leg b)2= (hypotenuse)2
leg a
hypotenuse
leg b
The Pythagorean Theorem
The Pythagorean Theorem
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Find the length of the shorter leg of a right triangle if the longer leg
is 10 miles more than the shorter leg and the hypotenuse is 10 miles
less than twice the shorter leg.
The Pythagorean Theorem
Example
Continued
1.) Understand
Read and reread the problem. If we let
x= the length of the shorter leg, then
x+ 10 = the length of the longer leg and
2x10 = the length of the hypotenuse.
+ 10
2 - 10x
x
The Pythagorean Theorem
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The Pythagorean Theorem
Example continued2.) Translate
Continued
By the Pythagorean Theorem,
(leg a)2+ (leg b)2= (hypotenuse)2
x2+ (x+ 10)2= (2x10)23.) Solve
x2+ (x+ 10)2= (2x10)2
x2+x2+ 20x+ 100 = 4x240x+ 100 (multiply the binomials)
2x2
+ 20x+ 100 = 4x2
40x+ 100 (simplify left side)
x= 0 orx= 30 (set each factor = 0 and solve)
0 = 2x(x30) (factor right side)
0 = 2x260x (subtract 2x2+ 20x+ 100 from both sides)
The Pythagorean Theorem
8/13/2019 I_F. Factoring Polynomials
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The Pythagorean Theorem
Example continued4.) Interpret
Check: Remember thatxis suppose to represent the length of
the shorter side. So, althoughx= 0 satisfies our equation, it
cannot be a solution for the problem we were presented.
If we letx= 30, thenx+ 10 = 40 and 2x10 = 50. Since 302 +
402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem
checks out.
State: The length of the shorter leg is 30 miles. (Remember that
is all we were asked for in this problem.)