Transcript Polynomial Equations

Standard: MM1A2e Factor expressions by greatest
common factor, grouping, trial and error, and special
products .
MM1A3a: Solve quadratic equations in
the form ax2 + bx + c = 0 where a = 1, by
using factorization and finding square
roots where applicable.
Today’s Question:
How do we factor polynomials?
Standard: MM1A2e, MM1A3a.
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Plot the following equations:
y=x
y = 2x + 5
y = -0.5x – 2
SLOPE
Y-INTERCEPT
Hint: y = mx + b
Or make a table of x and y (pick an x, calculate a
y). Points plot (x, y),
x is the horizontal direction (think of the
horizon)
y is the vertical direction
6
5
y = 2x + 5
4
3
y=x
2
y = -0.5x - 2
1
-8
-6
-4
-2
2
-1
-2
-3
-4
-5
4
6
8
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Polynomial equations are really products of
linear equations.
Make a table of values and graph: y = (x + 2)
Make a table of values and graph : y = (x – 3)
To distinguish between the two graphs, we can
replace the “y”s with function notation, such as:
f(x) = (x + 2)
 g(x) = (x – 3)

x
f(x) = (x + 2)
g(x) = (x – 3)
-3
-2
-1
0
-1
0
1
2
-6
-5
-4
-3
1
2
3
3
4
5
-2
-1
0
4
6
1
10
8
6
f(x)
4
2
-15
-13
-11
-9
-7
-5
-3
0
-1
-2
-4
-6
-8
-10
gx)
1
3
5
7
9
11
13
15
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Now, multiply f(x) by g(x) to get h(x)
h(x) = (x + 2)(x – 3) = x2 – x – 6
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Remember the pattern?
(x + a)(x + b) = x2 + (a + b)x + ab
Add h(x) to your earlier table of values and
graph h(x) on the same graph
x
f(x) = (x + 2)
g(x) = (x – 3)
-3
-2
-1
0
-6
-5
h(x) = f(x) * g(x)
h(x) = (x + 2)(x - 3)
h(x) = x2 – x – 6
6
0
-1
0
1
1
2
3
-4
-3
-2
-4
-6
-6
2
3
4
4
5
6
-1
0
1
-4
0
6
10
8
6
f(x)
4
gx)
2
-15
-13
-11
-9
-7
-5
-3
0
-1
-2
-4
-6
-8
-10
1
3
5
7
9
11
13
15
h(x)
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What do you notice about the graphs of f(x),
g(x), and h(x)?
f(x) and g(x) are linear
 h(x) is curvy – the curve is called a parabola
 h(x) crosses the x-axis at the same places as f(x) and
g(x)
 y = 0 at every x intercept
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The domain of a function are the set of all
inputs. It is the “independent” variable, the
one you choose.
The range of a function are the set of all
outputs. It is the “dependent” variable, the one
you calculate.
What is the domain and range of the above
functions:
f(x) = (x + 2)
g(x) = (x - 3)
h(x) = (x + 2)(x – 3) = x2 – x – 6
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When we study Linear Equations we
emphasize slope and y-intercept
When we study Polynomials we emphasize
zeros, x-intercepts, solutions, roots – which are
all different words for the basically the same
thing.
We start by moving all terms to one side of the
equation, making them equal zero.
WE ARE THEN LOOKING FOR THE VALUES
OF X THAT MAKE THE EQUATION EQUAL
ZERO!
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Understand value of y is zero at every x
intercept
Understand these factors are equations and the
zeros of these equations are the zeros of the
polynomial.
Understand that if products equal zero, that at
least one of the terms must equal zero (the zero
product rule)
School work talks about factoring by itself, but
in real life you factor to find solutions. For this
reason, we will be emphasizing factoring to
find solutions to problems.
The process of finding the zeros varies based on
the form of the problem.
If the problem is already factored and equals zero,
simply use the zero product rule to find the
zeros.
Example: Solve (x + 3)(x – 4) = 0
Either x + 3 = 0 or x – 4 = 0, so
x = -3 or x = 4
NOTE: The context of the problem may exclude
one of the answers.
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Show the Factoring by Greatest
Common Factor video – on my
Wiki
Always start by factoring out the Greatest
Common Factor (GCF), and use the zero
product rule to find the zeros.
Example: Solve: 5x2 + 15x = 0
5*x*x+3*5*x=0
5x(5 * x * x + 3 * 5 * x) = 0
5x(x + 3) = 0
Therefore, either 5x = 0 or x + 3 = 0
5x = 0 or x + 3 = 0
5 5
-3 -3
So x = 0 or x = -3
1.
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Pg 79, # 5 – 30 by 5’s & 36
and page 80, # 33 (8)
2.
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If the problem is a trinomial of the form
ax2 + bx + c = 0, we must factor the trinomial
by finding the factors of a * c that add to b.
If a = 1, then we are looking for the factors of
c that add to b.
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Show the Trinomial Factoring
a = 1 video – on my Wiki
Example: Solve:
x2 + 9x + 20 = 0
Make a factor tree to find
The factors of c that add to b:
Since “b” is positive, we are only
looking at positive numbers
4 + 5 = 9, so the factors are:
(x + 4)(x + 5) = 0
x = -4 or x = -5
20
9
1 20 21
2 10 12
4 5
9
-1 -20 -21
-2 -10 -12
-4 -5 -9
Your Turn Solve:
x2 - 10x - 11= 0
Make a factor tree to find
The factors of c that add to b:
“c” = -11
-11 + 1 = - 10 so the factors are
(x + 1)(x - 11) = 0
x = -1 or x = 11
-11
1 -11
-1 11
-10
-10
10
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Page 83, # 3 – 27 by 3’s and 28 29
(11 problems)
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Factoring a common monomial from pairs of
terms, then looking for a common binomial
factor is called factor by grouping.
A polynomial that cannot be written as a
product of polynomials with integer
coefficients is called unfactorable.
A factorable polynomial with integer
coefficients is factored completely if it is
written as a product of unfactorable
polynomials with integer coefficients.
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Show the Factor by Grouping
video – on my Wiki
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Factor completely the following expressions:
Example 1: 6x2 -10x -12x +20
2x(3x-5) – 4(3x – 5)
(3x – 5)(2x – 4)
Your turn: 6x2 + 27x + 4x2 + 18x
3x(2x + 9) + 2x(2x + 9)
(2x + 9)(3x + 2x)
(2x + 9)(5x)
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Your turn: Factor completely the following
expression:
2x3 + 14x2 + 3x + 21
2x2(x + 7) + 3(x + 7)
(x + 7)(2x2 + 3)
If the equation has a common factor in each
term, factor out the largest common factor and
use the zero product rule to find the zeros.
(2.9)
Your turn: Solve 16x2 – 4 + 4x3 – x = 0
Factor by grouping: 4(4x2 – 1) + x(4x2 – 1) = 0
The Greatest Common Factor is (4x2 – 1), so
(4x2 – 1)(4 + x) = 0
x = ±0.5 or x = -4
4.
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Page 95, # 5 – 30 by 5’s and 31 & 32
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If the polynomial has two terms, look to see if
they contain only the difference of squares
(NOTE: THIS ONE IS VERY IMPORTANT !! )
a2 – b2 = (a + b)(a – b)
Show the Factoring Difference of Squares video
– on my Wiki
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Example: Find the zeros of x2 – 9 = 0
(x + 3)(x – 3) = 0
x = -3 or x = 3
Your Turn: Solve: 9x2 - 49 = 0
(3x + 7)(3x - 7) = 0
x = 7/3 or x = -7/3
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If the polynomial has three terms, look to see if
they are a prefect square trinomial:
 a2 + 2ab + b2 = (a + b)2 and
 a2 - 2ab + b2 = (a - b)2
Example #1: Solve: 3x2 + 6x +3 = 0
Factor out the GCF first:
3(x2 + 2x +1) = 0
3(x + 1)2 = 0
x = -1 with duplicity of two
NOTE: You can do it the “long way” but it is faster if
you see the pattern.
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Factoring the prefect square trinomial:
 a2 + 2ab + b2 = (a + b)2 and
 a2 - 2ab + b2 = (a - b)2
Your turn: Solve: x2 - 6x +9 = 0
(x – 3)2= 0
x = 3 with duplicity of two
NOTE: You can do it the “long way” but it is
faster if you see the pattern.
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Factoring the prefect square trinomial:
 a2 + 2ab + b2 = (a + b)2 and
 a2 - 2ab + b2 = (a - b)2
Your turn: Solve: 18x2 + 60x +50 = 0
2(9x2 + 30x +25) = 0
2(3x + 5)2= 0
x = -5/3 with duplicity of two
NOTE: You can do it the “long way” but it is faster if
you see the pattern.
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Page 91, # 16 – 21 all and 28 & 29 (8)
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We can find the points of intersection between
two curves by setting them equal to each other
and solving the equation.
Popular examples are profit/loss curves and
supply/demand curves
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Steps to solve for the intersections:
1.
2.
1.
2.
3.
Set the equations equal to each other
Move everything to one side of the
equation, making that side equal zero.
(HINT: Keep x2 positive)
Factor
Use Zero Product Rule to find the roots.
Substitute the roots into the original
equations to find the corresponding values
of y
Example: find where the parabola
f(x) = x2 + 2x -6 intersects with the line
g(x) = 2x + 3
 Set them equal to each other gives:
x2 + 2x - 6 = 2x + 3
Move everything to one side: x2 - 9 = 0
Can be solved as special function or by square root –
show both
(x + 3)(x – 3) = 0
x = -3 or x = 3
Substitute to find the values of y: y = 9 or -3
Intersections: (-3, -3) and (3, 9)
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Example: find where the parabola
f(x) = x2 + 5x - 5 intersects with the line
g(x) = -2x - 5
Set them equal to each other gives:
x2 + 5x - 5 = -2x -5
x2 + 7x = 0
Solve be GCF:
x(x + 7) = 0
x = 0 or x = -7
y = -5 or 9
Points of Intersection: (0, -5) and (-7, 9)
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Your Turn: find where the parabola
f(x) = x2 - 4 intersects with the parabola
g(x) = 2x2 – 2x - 12
Setting them equal to each other gives:
2x2 – 2x - 12 = x2 - 4
x2 – 2x – 8 = 0
Solve be finding factors of -8 that add to -2
(x + 2)(x - 4) = 0
x = -2 or x = 4
y = 0 or 12
Points of intersection are: (-2, 0) and (4, 12)
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Example: find the length of the side of a square
such that the area equals the perimeter.
f(x) = A = s2
g(x) = P = 4s
Set them equal to each other gives:
s2 = 4s
s2 - 4s = 0
s(s – 4) = 0
s = 0 or s - 4 = 0
s = 0 or s = 4
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Your turn: find the radius of a circle such that
the area equals the perimeter.
f(x) = A = r2
g(x) = P = 2r
Set them equal to each other gives:
r2 = 2r
r2 - 2r = 0
r(r – 2) = 0
r = 0 or r = 2
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Solving Quadratic Equations by Square Root
video – on my Wiki
Do some Guided Practice on page 120,
especially some of 7 – 12.
NOTE: LEAVE ANSWSER IN RADICAL
FORM
Solve: 4(2z – 7)2 = 100
Practice: page 121, # 16 – 21 all
and 24 - 26 all (9 problems)
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Solve:
x2 – 2x - 80 = 0
(x - 10)(x + 8) = 0
x = 10 or x = -8
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Solve:
2(2x- 3)2 + 4 = 12
2(2x- 3)2 = 8
(2x- 3)2 = 4
2x- 3 = 2 or 2x – 3 = -2
2x = 5 or 2x = 1
x = 2.5 or x = 0.5
10
8
6
f(x)
4
gx)
2
-15
-13
-11
-9
-7
-5
-3
0
-1
-2
-4
-6
-8
-10
1
3
5
7
9
11
13
15
h(x)
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A cubic function is a nonlinear function that
can be written in the standard form:
y = ax3 + bx2 + cx + d
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Cubic polynomials are simply three linear
equations multiplied together, just like
quadratics are the products of two linear
equations.
Assume we have the following equations:
f(x) = x+ 1
g(x) = 0.5x + 2
h(x) = 0.5x – 1
The cubic is simply the product of these three:
i(x) = f(x) * g(x) * h(x)
i(x) = (-0.2x + 2)(x + 5)(0.3x -1)
i(x) = 0.25x3 + 0.75x2 – 1.5x -2
We would get graphs that looked like:
10
8
f(x) = x + 1
6
4
2
g(x) = 0.5x + 2
-10
-8
-6
h(x) = 0.5x -1
0
-4
-2
-2
-4
-6
-8
-10
Describe the graph
0
2
4
6
8
10
Talk about the x intercept of the linear functions
and the cubic
Talk about values of x where the cubic is
increasing.
Talk about values of x where the cubic is
decreasing
Talk about local minimum and maximum
Talk about end conditions when a > 0 and a < 0
What is the domain and range of the cubic?
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They are not easy to factor. We have two
patterns we can use:
y = x3 + 3x2y + 3xy2 + y3 factors into
y = (x + y)3
y = x3 - 3x2y + 3xy2 - y3 factors into
y = (x – y)3
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y = x3 + 3x2y + 3xy2 + y3 = (x + y)3
y = x3 - 3x2y + 3xy2 – y3 = (x – y)3
Solve:
-3x3 + 18x2 – 36x + 24 = 0
Factor out the GCF:
-3(x3 - 6x2 + 12x – 8) = 0
“x” = x, “y” = 2
Substitute: x3 - 3x2(2) + 3x(2)2 - 23
Clean it up: x3 - 6x2 + 12x - 8, which matches
Factor via patterns:
-3(x – 2)3 = 0
x = 2 with duplicity of three
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y = x3 + 3x2y + 3xy2 + y3 = (x + y)3
y = x3 - 3x2y + 3xy2 – y3 = (x – y)3
Your turn, solve:
x4 + 15x3 + 75x2 + 125x = 0
Factor out the GCF:
x(x3 + 15x2 + 75x + 125) = 0
“x” = x, “y” = 5
Substitute: x3 + 3x2(5) + 3x(5)2 + 53
Clean it up: x3 + 15x2 + 75x + 125, which matches
Factor via patterns:
x(x + 5)3 = 0
x = 0 or x = -5 with duplicity of three
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Page 132, # 4 – 6 all, 11, 12, 16 (6)