Transcript Section 1

Chapter 4
Polynomials
TSWBAT determine if an
expression is a monomial,
binomial, or polynomial;
determine the degree of a
monomial or polynomial; add
and subtract two polynomials.
Section 1 – Polynomials cont.
• Monomial – a constant, a variable, or
•
•
the product of a constant and one or
more variables.
Ex. -7, u, 1/3m 2, -s 2t 3, 6xy 3
Binomial – a polynomial that has two
terms. Ex. 2x +3
Trinomial – a polynomial that has
three terms. Ex. X2 + 4x – 5
Section 1 – Polynomials cont.
• Degree of a variable – the number
of times the variable occurs as a
factor in the monomial or the
exponent of the variable.
Ex. 6xy 3 has variables x and y.
X has degree 1 and y has
degree 3.
Section 1 – Polynomials cont.
• Degree of a monomial – the sum of
the degrees of the variables in
the monomial. A nonzero constant
has degree 0. The constant 0 has
no degree.
Ex. 6xy 3 has degree 0+1+3=4
-s 2t 3 has degree 0+2+3=5
u has degree 1
-7 has degree 0
Section 1 – Polynomials cont.
• Similar (or like) Monomials – monomials
that are identical or that differ only in
their coefficients.
Ex. -s 2t 3 and 2s 2t 3 are similar
6xy 3 and 6x 3y are not similar
Section 1 – Polynomials cont.
• Polynomial – a monomial or a sum of
monomials. The monomials in a
polynomial are called the terms of
the polynomial.
Ex. x2 + (-4)x + 5 or x2 – 4x + 5
The terms are x2, -4x, and 5
Section 1 – Polynomials cont.
• Simplified polynomial – a polynomial
in which no two terms are similar.
The terms are usually arranged in
order of decreasing degree of one
of the variables, usually x.
Ex. 2x3 – 5 + 4x + x3
is not simplified but
3x3 + 4x – 5 is simplified.
Section 1 – Polynomials cont.
• Degree of a polynomial – the
greatest of the degrees of its
terms after it has been
simplified.
Ex. x4 – 2x2y3 + 6y – 11
has terms with degree 4, 5,
1, and 0. Thus the
polynomial has degree 5 the
largest degree of its terms.
Section 1 – Polynomials cont.
• Addition of Polynomials – to add two or
more polynomials, write their sum and
then simplify by combining similar terms.
Ex.
Add 2x2 – 3x + 5 and x3 – 5x2 + 2x – 5
2x2 – 3x + 5
+ x3 – 5x2 + 2x – 5
x3 – 3x2 - x
Section 1 – Polynomials cont.
• Subtracting Polynomials – to subtract one
polynomial from another, add the opposite of
each term of the polynomial you are subtracting.
Ex. Subtract
2x2 – 3x + 5 from x3 – 5x2 + 2x – 5
x3 – 5x2 + 2x – 5
x3 – 5x2 + 2x – 5
- (2x2 – 3x + 5)
- 2x2 + 3x – 5
x3 – 7x2 + 5x – 10
Multiplying Polynomials
TSWBAT multiply Polynomials using
the FOIL method, the Rules of
Special Products, and the
distributive Property.
Section 3 – Multiplying Polynomials
If you are multiplying two binomials
together it is best to use the FOIL
method of multiplication.
• FOIL Method –
F - Multiply the two first terms
O – Multiply the two outer terms
I – Multiply the two inner terms
L – Multiply the two last terms
Section 3 – Multiplying
Polynomials cont.
Example of FOIL
(2a – b)(3a + 5b)
F – (2a)(3a) = 6a2
O – (2a)(5b) = 10ab
I – (-b)(3a) = -3ab
L – (-b)(5b) = -5b2
Combining the four
terms into one
polynomial and
combining the like
terms of O and I
we get the
following as a
result:
6a2 + 7ab – 5b2
Section 3 – Multiplying
Polynomials cont.
• Besides the FOIL Method there
are also three special products to
know, that will help you multiply
polynomials and their terms faster.
Special Products
(a+b)2 = a2 + 2ab + b2
(a-b)2 = a2 -2ab + b2
(a+b)(a-b) = a2 + b2
Section 3 – Multiplying
Polynomials cont.
• Besides these methods of FOIL and
Special Products there is also the old
fashioned Distribution Method.
Ex. (2x+3)(x2+4x-5)
2x(x2+4x-5)+3(x2+4x-5)
2x(x2)+2x(4x)-2x(5)+3(x2)+3(4x)-3(5)
2x3+8x2-10x+3x2+12x-15
Combine Similar Terms
2x3+11x2+2x-15
Exponents
TSWBAT Use the laws of
exponents in polynomial
multiplication and simplification.
Section 2 – Laws of Exponents
• There are five laws of exponents
–
Let a and b be real numbers
and m and n be positive integers.
Then:
1. am * an = am+n
2. (ab)m = ambm
3. (am)n = am*n
4. (a/b)n = an/bn
5. am/an = am-n
Section 2 – Laws of Exponents
• Examples
Factoring
TSWBAT find the prime
factorization of a number and the
greatest common factor or least
common multiple of two or more
numbers or monomials.
Section 4 – Prime Factorization
• Factor Set – The set of all numbers
•
•
•
that are a factor of a given number.
Factor – One of two numbers that
multiplied together form a given
number.
Prime Number or Prime – a number
whose only factors are 1 and itself.
Composite Number – any number that
has more than one set of factors.
Activity – Primes Chart
Section 4 – Prime Factorization
• Greatest Common
Factor (GCF) – of two
or more integers is the
greatest integer that
is a factor of each.
Ex. GCF of 15 and
30
1. Find all the
factors of each
number.
2. Find the
greatest number in
both sets. This is your
GCF.
Factors Factors
of 15
of 30
1 15
1 30
2 15
3 5
3 10
5 6
Section 4 – Prime Factorization
• Least Common Multiple
12 24 36 48
60
(LCM) – of two or more
30 60 90 120 150
integers is the least
positive integer having
each as a factor.
1. Find the multiples
of the two numbers.
Ex. Find the LCM
of 12 and 30.
2. Find the least
common number
among both
numbers. That is
your LCM.
Section 4 – Prime Factorization
• The GCF and LCM also hold true for
Monomials with a slight variation as you
now need to be aware of the variables.
• GCF of Monomials – the common factor
with the greatest degree and greatest
numerical coefficient.
• LCM of Monomials – the common multiple
that has the least degree and least
positive numerical coefficient.
Section 4 – Prime Factorization
Ex. of GCF of Monomials 48u2v2 and 60uv3w.
1. Find GCF of Coefficients
48 – 1,48, 2,24, 3,16, 4,12, 6,8
60 – 1,60, 2,30, 3,20, 4,15, 5,12, 6,10
2. Compare the variables. Only variables in both
can be used in GCF. Then use the smaller
exponent variable.
u2 or u – use u
v2 or v3 – use v2
3. Therefore our GCF is 12uv2
Section 4 – Prime Factorization
Ex. of LCM of Monomials 48u2v2 and 60uv3w
1. Find LCM of coefficients.
48,96,144,192,240,288
60,120,180,240,300,360
2. Compare the variables. All variables in
both must be used in LCM. Then use the
greater exponent variable.
u2 or u – use u2 v2 or v3 – use v3 use w
3. Therefore the LCM is 240u2v3w
Section 4 – Prime Factorization
• Prime Factorization –
is the factorization
of a number down to
the product of a set
of prime numbers.
The most common way
to find the prime
factorization is to use
a factor tree working
down to all prime
factors.
• Ex. of Prime
Factorization
936=2*468
2*2*234
2*2*2*117
2*2*2*3*39
2*2*2*3*3*13
Therefore the
Prime Factorization
is 23*32*13
Factoring Quadratic Polynomials
• Quadratic Polynomials – or second-degree
polynomials are polynomials of the form
ax2+bx+c where (a≠0).
Ex. X2+0x+25
• Quadratic Polynomials have three terms each
with its own name.
– Quadratic Term – is the ax2 term
– Linear Term – is the bx term
– Constant Term – is the c term
Factoring Quadratic Polynomials
• Quadratic Trinomial – a quadratic
polynomial in which a, b, and c are all
nonzero integers.
Ex. x2+2x-15
• In this section we will factor a special
type of quadratic polynomial the perfect
square trinomial. In the next section we
will look at how to factor any Quadratic
Polynomial.
Section 5 – Factoring Polynomials
• 5 Special Factorings – In section 3 we had
3 special products. Those products play a
role here in factoring as well. Besides
those three there are also two other
special cases in factoring. On the
following slide you will find these 5 cases.
(Hint – Take an index card and write
these 5 cases on it to help you when
doing Problems!!!)
Section 5 – Factoring Polynomials
Special Factorings
Perfect Square Trinomials
Positive a2 + 2ab + b2 = (a + b)2
Negative a2 - 2ab + b2 = (a - b)2
Difference of Squares
a2 – b2 = (a + b)(a – b)
Sum and Difference of Cubes
Sum a3 + b3 = (a + b)(a2 – ab + b2)
Difference a3 - b3 = (a - b)(a2 + ab + b2)
Section 5 – Factoring Polynomials
• Example of a Perfect Square Trinomial
•
•
Positive –> z2+6z+9= (z+3)2
Negative -> 4s2-4st+t2 = (2s-t)2
Example of a Difference of Squares
25x2-16a2 = (5x+4a)(5x-4a)
Example of Cubes
Sum ->8u3+v3 = (2u+v)(4u2-2uv+v2)
Difference -> y3-1 = (y-1)(y2+y+1)
Section 5 – Factoring Polynomials
Common Numerical Powers Activity
• You can also factor a polynomial by factoring
out the Greatest Common Monomial Factor.
Ex. 2x4-4x3+8x2
Step 1 – Find the GCF, which here it is 2x2
Step 2 – Divide out 2x2 from each term.
Step 3 – Write out your answer.
2x2(x2-2x+4)
Section 5 – Factoring Polynomials
• One last way to factor a polynomial is to
rearrange and group terms that you might be
able to find a GCF from.
Ex. 3xy-4-6x+2y
If we group the 1st and 3rd terms together
they have a common factor of 3x and then
grouping the 2nd and 4th terms they have a
common factor of 2.
(3xy-6x)+(2y-4) -> 3x(y-2)+2(y-2)
Now we have a common factor of (y-2) that we
can pull out and we are left with (y-2)(3x+2).
Section 6 – Factoring Quadratic
Polynomials
• Factoring a Quadratic Polynomial –
ax2+bx+c can be factored into the
product (px+q)(rx+s) where p, q, r,
and s are integers.
Then ax2+bx+c = prx2+(ps+qr)x+qs.
So a = pr, b=(ps+qr), and c = qs
Section 6 – Factoring Quadratic
Polynomials
• Lets look at an example of how to factor a
Quadratic Polynomial.
15t2-16t+4 -> 1. Lets begin by looking at
the pr term which is 15 here. There are
two ways in which to factor 15t2. They
are (t___)(15t____) or (3t____)(5t____).
Section 6 – Factoring Quadratic
Polynomials
• 2. We need to look at the qs term of 4.
Since 4 is positive, both factors are either
positive or negative. To determine this we
look at the middle term or -16t. Since this
is negative both of our factors of 4 will be
negative. Thus there are 3 negative
combinations that factor 4. They are
(___-1)(_-4), (_-2)(_-2), or (_-4)(__-1).
Section 6 – Factoring Quadratic
Polynomials
• If we take our two
possible answers
for the first terms
and our three for
our second term,
we find we have 6
possible factoring
possibilities. These
are shown in the
table.
(t-1)(15t-4) (3t-1)(5t-4)
(t-2)(15t-2) (3t-2)(5t-2)
(t-4)(15t-1) (3t-4)(5t-1)
Section 6 – Factoring Quadratic
Polynomials
• To find out which
of these
possibilities work,
we check each of
them by doing the
O and I part of
FOIL on each pair
to find the
grouping that gives
us -16t.
(t-1)(15t-4) -15t-4t=-19t
(t-2)(15t-2) -30t-2t=-32t
(t-4)(15t-1) -60t-t=-61t
(3t-1)(5t-4) -5t-12t=-17t
(3t-2)(5t-2) -10t-6t=-16t
(3t-4)(5t-1) -20t-3t=-23t
Section 6 – Factoring Quadratic
Polynomials
• Therefore our quadratic polynomial
15t2-16t+4 factors down into (3t-2)(5t-2).
• This polynomial is now factored
completely as it is written as the product
of factors that are prime polynomials. It
could also be factored completely if we
had monomials 2x(3y), or a power of a
prime polynomial like (x-2)2 .
Section 6 – Factoring Quadratic
Polynomials
• Prime Polynomial – a polynomial that is not
reducible and the greatest common factor of its
coefficients is 1.
Ex. X2+4x-3 is a Prime Polynomial, since it can
not be factored into two lower degree
polynomials with integers and the GCF of its
coefficients is 1.
2x2+8x-6 is not Prime because the GCF of its
coefficients is 2. Thus a 2 can be factored out.
Section 6 – Factoring Quadratic
Polynomials
• Irreducible polynomial – a polynomial that
contains more than one term and cannot be
expressed as a product of lower degree
polynomials taken from its factor set.
Ex. X2+4x-3
Why? Factor sets are (x-1)(x+3) and
(x+1)(x-3) and neither of these sets gives a
middle term of +4. (They give +2x and -2x.)
Section 7 – Solving Polynomial
Equations
• Polynomial Equation – an equation that is
•
•
equivalent to an equation with a polynomial as
one side and 0 as the other.
Ex. x2-5x-24=0 and x2=5x+24
Root – is a solution to the polynomial equation.
Solution – a value of the variable that satisfies
the equation.
Ex. Roots and solutions for the above
problem are -3 and 8.
Section 7 – Solving Polynomial
Equations
• To solve a polynomial equation we use the
Zero-Product Property.
• Step 1 – Write the equation in Zero
Equals Form.
Ex. If we take the problem x2=5x+24,
we need to rewrite it as x2-5x-24=0
Section 7 – Solving Polynomial
Equations
• Step 2 – Factor the polynomial side of the
•
equation.
Ex. From our equation x2-5x-24=0 we factor
the left side into (x-8)(x+3)=0
Step 3 – Solve the equation by setting each
factor equal to zero.
Ex. Taking our factor of (x-8)(x+3) we set
each equal to zero and solve for x.
x-8=0 and x+3=0 -> x=8 and x=-3
Section 7 – Solving Polynomial
Equations
• Double Root or Double Zero – of an
equation is a solution of the equation that
appears twice.
Ex. If we have (x-2)(x-2) as factors we
have the solution of 2 twice. This means
that 2 is a double root of that equation
and can be written x=2d.r.
Section 7 – Solving Polynomial
Equations
• Multiple Root or Multiple Zero - of
an equation is a solution of the
equation that appears more than twice.
Ex. If we have (t-1)(t-1)(t-1)(t-1) as
factors we have the solution of 1 four
times. This means that 1 is a multiple
root of that equation and can be
written t=1m.r.4.
Section 7 – Solving Polynomial
Equations
• Using the Zero-Product Property we can
also find the zeros of a polynomial
function. That means find the values for
when f(x)=0
Ex. f(x)=x2-9 following the steps we
factor the polynomial side to (x-3)(x+3)
(special products) and then solve to x=3
and x=-3.
Section 8 – Problem Solving Using
Polynomial Equations
• The seven steps of word problems are:
1. Read and Write the problem
2. Draw a picture
3. Define the variables
4. Re-Read the problem and set up the
basic problem
5. Fill-in values from the problem
6. Solve the equation
7. Check your solutions (this is critical now
because not all values will be accurate solutions)
Section 8 – Problem Solving Using
Polynomial Equations
• Example – Step 1 – Read and Write the
problem. - A graphic artist is designing a
poster that consists of a rectangular prism
with a uniform border. The print is to be
twice as tall as it is wide, and the border is
to be 3 inches wide. If the area of the
poster is to be 680 inches squared, find
the dimensions of the print.
Section 8 – Problem Solving Using
Polynomial Equations
• Step 2 – Draw a picture – See overhead.
• Step 3 – Define the variable – w = width
of print
• Step 4 – Label drawing – See Overhead
• Step 5 – Create equation – using area
formula – A = L*W -> 680 =
(w+6)(2w+6)
Section 8 – Problem Solving Using
Polynomial Equations
• Step 6 – Solve for variable.
•
680 = (w+6)(2w+6) -> 680 = 2w2+18w +36
-> 0 = 2w2+18w-644 -> Divide by 2
-> 0 = w2 +9w -322 -> (w-14)(w+23) = 0
-> w-14 = 0 and w+23 = 0
-> w = 14 or w = -23
Step 7 – Check solutions – We cannot have a
negative length thus x = -23 does not work and
our solution is x = 14.
Section 9 – Solving Polynomial
Inequalities
• Polynomial Inequality – an
inequality that is equivalent to an
inequality with a polynomial as one
side and zero as the other side.
Ex. x2-x-6>0 and x2>x+6
Section 9 – Solving Polynomial
Inequalities
• To solve a polynomial inequality we
start by following the first two steps
of that of an equation.
Step 1 - We make one side zero and
the other side the polynomial.
Ex. x2>x+6 becomes x2-x-6>0
Section 9 – Solving Polynomial
Inequalities
Step 2 - We factor the polynomial side of the
inequality.
Ex. x2-x-6>0 becomes (x-3)(x+2)>0.
Step 3 is actually a two part step.
Part A – if the factors of the polynomial
are >0 we will use the same inequality
sign in Step 4.
Part B – if the factors of the polynomial
are <0 we will use opposite inequality
signs in Step 4.
Section 9 – Solving Polynomial
Inequalities
Step 3 cont. – Ex. (x-3)(x+2)>0 our factors are >0
and thus for Step 4 we will use the same
inequality sign.
Step 4 – is another two part process. If you
remember when we first talked about
inequalities we had the terms Conjunction and
Disjunction, well they return here in this step
when we set the inequality factors to zero.
Section 9 – Solving Polynomial
Inequalities
Step 4 cont. – In this step we have two
conjunctions joined together by a
disjunction. So in our example,
(x-3)(x+2)>0 is possible only if both
factors are positive (conjunction) or
(disjunction) both factors are negative
(conjunction).
Section 9 – Solving Polynomial
Inequalities
Step 4 cont. – Ex. we have the following
possibilities…
x-3>0 and x+2>0 OR x-3<0 and x+2<0
Step 5 – solve for x…
x>3 and x>-2 OR x<3 and x<-2 so…
x>3 OR x<-2.
Step 6 – Graph the solution.
Section 9 – Solving Polynomial
Inequalities
Example 2 – 3t<4 - t2
Step 1 – t2 + 3t – 4 <0
Step 2 – (t+4)(t-1)<0
Step 3 – Use Part B as factors are <0.
Step 4 –
(t+4)>0 and (t-1)<0 OR (t+4)<0 and (t-1)>0
Step 5 – t>-4 and t <1 OR t<-4 and t >1
-4<t<1 OR no solution
Step 6 graph the one remaining solution.