2 Polynomial Equations

7/31/2019 2 Polynomial Equations

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Standard: MM1A2e Factor expressions by greatestcommon factor, grouping, trial and error, and specialproducts .

MM1A3a: Solve quadratic equations inthe form ax2 + bx + c = 0 where a = 1, byusing factorization and finding square

roots where applicable.

Todays 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


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6

5

4

3

2

1

-1

-2

-3

-4

-5

-8 -6 -4 -2 2 4 6 8

y = -0.5x - 2

y = 2x + 5

y = x


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Polynomial equations are really products oflinear 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 canreplace the ys with function notation, such as:

f(x) = (x + 2) g(x) = (x 3)


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x f(x) = (x + 2) g(x) = (x 3)

-3 -1 -6

-2 0 -5

-1 1 -40 2 -3

1 3 -2

2 4 -1

3 5 0

4 6 1


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-10

-8

-6

-4

-2

0

2

4

6

8

10

-15 -13 -11 -9 -7 -5 -3 -1 1 3 5 7 9 11 13 15

f(x)

gx)


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Now, multiply f(x) by g(x) to get h(x)

h(x) = (x + 2)(x 3) = x2 x 6

Remember the pattern?

(x + a)(x + b) = x2 + (a + b)x + ab

Add h(x) to your earlier table of values andgraph h(x) on the same graph


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x f(x) = (x + 2) g(x) = (x 3) h(x) = f(x) * g(x)h(x) = (x + 2)(x - 3)

h(x) = x2 x 6

-3 -1 -6 6

-2 0 -5 0

-1 1 -4 -4

0 2 -3 -6

1 3 -2 -62 4 -1 -4

3 5 0 0

4 6 1 6


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-10

-8

-6

-4

-2

0

2

4

6

8

10

-15 -13 -11 -9 -7 -5 -3 -1 1 3 5 7 9 11 13 15

f(x)

gx)

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 allinputs. It is the independent variable, theone you choose.

The range of a function are the set of alloutputs. It is the dependent variable, the oneyou 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 weemphasize slope and y-intercept

When we study Polynomials we emphasizezeros, x-intercepts, solutions, roots which areall different words for the basically the samething.

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 EQUALZERO!


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Understand value of y is zero at every xintercept

Understand these factors are equations and thezeros 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 zeroproduct rule)

School work talks about factoring by itself, butin real life you factor to find solutions. For thisreason, we will be emphasizing factoring to

find solutions to problems.


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The process of finding the zeros varies based onthe form of the problem.

If the problem is already factored and equals zero,

simply use the zero product rule to find thezeros.

Example: Solve (x + 3)(x 4) = 0

Either x + 3 = 0 or x 4 = 0, sox = -3 or x = 4

NOTE: The context of the problem may excludeone of the answers.


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Show the Factoring by GreatestCommon Factor video on my

Wiki


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1. Always start by factoring out the GreatestCommon Factor (GCF), and use the zeroproduct 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


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Pg 79, # 5 30 by 5s & 36

and page 80, # 33 (8)


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2. If the problem is a trinomial of the form

ax2 + bx + c = 0, we must factor the trinomialby finding the factors of a * c that add to b.

If a = 1, then we are looking for the factors ofc that add to b.


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Show the Trinomial Factoring

a = 1 video on my Wiki


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20 9

1 20 21

2 10 12

4 5 9-1 -20 -21

-2 -10 -12

-4 -5 -9

Example: Solve:x2 + 9x + 20 = 0Make a factor tree to find

The factors of c that add to b:

Since b is positive, we are onlylooking at positive numbers

4 + 5 = 9, so the factors are:(x + 4)(x + 5) = 0x = -4 or x = -5


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-11 -10

1 -11 -10

-1 11 10

Your Turn Solve:x2 - 10x - 11= 0Make 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) = 0x = -1 or x = 11


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Page 83, # 3 27 by 3s and 28 29(11 problems)


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Factoring a common monomial from pairs ofterms, then looking for a common binomialfactor is called factor by grouping.

A polynomial that cannot be written as aproduct of polynomials with integercoefficients is called unfactorable.

A factorable polynomial with integercoefficients is factored completely if it iswritten as a product of unfactorablepolynomials with integer coefficients.


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Show the Factor by Groupingvideo 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 followingexpression:

2x3 + 14x2 + 3x + 21

2x2(x + 7) + 3(x + 7)(x + 7)(2x2 + 3)


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4. If the equation has a common factor in eachterm, factor out the largest common factor anduse 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


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Page 95, # 5 30 by 5s and 31 & 32


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If the polynomial has two terms, look to see ifthey 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: 9x

2

- 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 ifthey are a prefect square trinomial:

a2 + 2ab + b2 = (a + b)2 and a2 - 2ab + b2 = (a - b)2

Example #1: Solve: 3x2 + 6x +3 = 0Factor out the GCF first:

3(x2 + 2x +1) = 0

3(x + 1)2 = 0x = -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 isfaster 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 = 02(9x2 + 30x +25) = 0

2(3x + 5)2= 0x = -5/3 with duplicity of two

NOTE: You can do it the long way but it is faster ifyou 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 betweentwo curves by setting them equal to each otherand solving the equation.

Popular examples are profit/loss curves andsupply/demand curves


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Steps to solve for the intersections:1. Set the equations equal to each other

2. Move everything to one side of the

equation, making that side equal zero.(HINT: Keep x2 positive)

1. Factor

2. Use Zero Product Rule to find the roots.

3. Substitute the roots into the originalequations to find the corresponding valuesof y


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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 squaresuch 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 thatthe area equals the perimeter.

f(x) = A = r2

g(x) = P = 2r Set them equal to each other gives:

r2 = 2r

r2 - 2r = 0r(r 2) = 0

r = 0 or r = 2


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Solving Quadratic Equations by Square Rootvideo 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 alland 24 - 26 all (9 problems)


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Solve: x2 2x - 80 = 0

(x - 10)(x + 8) = 0

x = 10 or x = -8

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


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-10

-8

-6

-4

-2

0

2

4

6

8

10

-15 -13 -11 -9 -7 -5 -3 -1 1 3 5 7 9 11 13 15

f(x)gx)

h(x)


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A cubic function is a nonlinear function thatcan be written in the standard form:

y = ax3 + bx2 + cx + d


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Cubic polynomials are simply three linearequations multiplied together, just likequadratics are the products of two linearequations.

Assume we have the following equations:1. f(x) = x+ 1

2. g(x) = 0.5x + 2

3. 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


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-10

-8

-6

-4

-2

0

2

4

6

8

10

-10 -8 -6 -4 -2 0 2 4 6 8 10

g(x) = 0.5x + 2

We would get graphs that looked like:

Describe the graph

f(x) = x + 1

h(x) = 0.5x -1


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Talk about the x intercept of the linear functionsand 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 twopatterns 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 = 0Factor out the GCF:

-3(x3 - 6x2 + 12x 8) = 0x = 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 = 0Factor out the GCF:

x(x3 + 15x2 + 75x + 125) = 0x = 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)

2 Polynomial Equations

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