Math I, Sections 2.5 – 2.9Factoring Polynomials

Standard: MM1A2e Factor expressions by greatest common factor, grouping, trial and error, and special products .

MM1A3a: Solve quadratic equations inthe 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.

Warm-Up 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

Warm-Up6

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

Polynomial Equations 1A3e, Section 2.5

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)

Your table should look like this:

x f(x) = (x + 2)

g(x) = (x – 3)

-3 -1 -6

-2 0 -5

-1 1 -4

0 2 -3

1 3 -2

2 4 -1

3 5 0

4 6 1

Your graphs should look like this:

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

-10

-8

-6

-4

-2

0

2

4

6

8

10

f(x)

gx)

Polynomial Equations

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 and graph h(x) on the same graph

Your table should look like this:

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 -6

2 4 -1 -4

3 5 0 0

4 6 1 6

Your graphs should look like this:

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

-10

-8

-6

-4

-2

0

2

4

6

8

10

f(x)

gx)

h(x)

Polynomial Equations

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

Domain and Range1A1d

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

Polynomial Equations 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!

Polynomial Equations 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.

Solving Polynomial Equations 1A3a, Section 2.5

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) = 0Either x + 3 = 0 or x – 4 = 0, so

x = -3 or x = 4NOTE: The context of the problem may

exclude one of the answers.

Video

Show the Factoring by Greatest Common Factor video – on my Wiki

Solving Polynomial Equations Factoring out GCF 1A2e, (2.5)

1. Always start by factoring out the Greatest Common Factor (GCF), and use the zero product rule to find the zeros.

Example: Solve: 5x2 + 15x = 05 * x * x + 3 * 5 * x = 0

5x(5 * x * x + 3 * 5 * x) = 05x(x + 3) = 0

Therefore, either 5x = 0 or x + 3 = 05x = 0 or x + 3 = 0

5 5 -3 -3So x = 0 or x = -3

GCF Problems

Pg 79, # 5 – 30 by 5’s & 36and page 80, # 33 (8)

Solving Polynomial Equations 1A2e, 1A3a, Section 2.6

2. If the problem is a trinomial of the formax2 + 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.

Video

Show the Trinomial Factoring a = 1 video – on my Wiki

Solving Polynomial Equations 1A2e, 1A3a, Section 2.6

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 findThe factors of c that add to b:

Since “b” is positive, we are only looking at positive numbers4 + 5 = 9, so the factors are:

(x + 4)(x + 5) = 0x = -4 or x = -5

 

Solving Polynomial Equations 1A2e, 1A3a, Section 2.6

-11 -10 1 -11 -10-1 11 10

Your Turn Solve:x2 - 10x - 11= 0Make a factor tree to findThe 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 

Practice

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

Factor Completely 1A2e (2.9)

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.

Video

Show the Factor by Grouping video – on my Wiki

Factor Completely 1A2e (2.9)

Factor completely the following expressions:

Example 1: 6x2 -10x -12x +202x(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)

Factor Completely 1A2e (2.9)

Your turn: Factor completely the following expression:

2x3 + 14x2 + 3x + 212x2(x + 7) + 3(x + 7)

(x + 7)(2x2 + 3)

Factor Completely 1A3a (2.9)

4. 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 = 0Factor by grouping: 4(4x2 – 1) + x(4x2 – 1) =

0The Greatest Common Factor is (4x2 – 1), so

(4x2 – 1)(4 + x) = 0x = ±0.5 or x = -4

Practice

Page 95, # 5 – 30 by 5’s and 31 & 32

Skip 2.7 since a ≠ 1

Solving Special Polynomial Equations 1A2e (2.8)

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

Video 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) = 0x = 7/3 or x = -7/3

Solving Special Polynomial Equations 1A2e, (2.8)

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 = 0Factor out the GCF first:

3(x2 + 2x +1) = 03(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.

Solving Special Polynomial Equations 1A2e, (2.8)

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.

Solving Special Polynomial Equations 1A2e, (2.8)

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 if you see the pattern.

Practice

Page 91, # 16 – 21 all and 28 & 29 (8)

Solving f(x) = g(x) 1A1i

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

Solving f(x) = g(x) 1A1i

Steps to solve for the intersections:1. Set the equations equal to each other2. Move everything to one side of the

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

3. Factor4. Use Zero Product Rule to find the roots.5. Substitute the roots into the original

equations to find the corresponding values of y

Solving f(x) = g(x) 1A1i

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 + 3Move everything to one side: x2 - 9 = 0

Can be solved as special function or by square root – show both

(x + 3)(x – 3) = 0x = -3 or x = 3

Substitute to find the values of y: y = 9 or -3Intersections: (-3, -3) and (3, 9)

Solving f(x) = g(x) 1A1i

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 -5x2 + 7x = 0

Solve be GCF: x(x + 7) = 0

x = 0 or x = -7y = -5 or 9

Points of Intersection: (0, -5) and (-7, 9)

Solving f(x) = g(x) 1A1i

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 - 4x2 – 2x – 8 = 0

Solve be finding factors of -8 that add to -2(x + 2)(x - 4) = 0x = -2 or x = 4

y = 0 or 12Points of intersection are: (-2, 0) and (4, 12)

Solving f(x) = g(x) 1A1i

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 = 4ss2 - 4s = 0s(s – 4) = 0

s = 0 or s - 4 = 0s = 0 or s = 4

Solving f(x) = g(x) 1A1i

Your turn: find the radius of a circle such that the area equals the perimeter.

f(x) = A = r2

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

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

Solving Quadratics by Square Roots (2.13)

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)

Warm-up

Solve: x2 – 2x - 80 = 0(x - 10)(x + 8) = 0x = 10 or x = -8

Solve: 2(2x- 3)2 + 4 = 122(2x- 3)2 = 8(2x- 3)2 = 4

2x- 3 = 2 or 2x – 3 = -22x = 5 or 2x = 1

x = 2.5 or x = 0.5

Quick Review of QuadraticFunction

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

-10

-8

-6

-4

-2

0

2

4

6

8

10

f(x)

gx)

h(x)

Cubics (3.1)

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

y = ax3 + bx2 + cx + d

Cubics (3.1) 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:1. f(x) = x+ 12. g(x) = 0.5x + 23. 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

Cubics

-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

Graph of Cubic Function

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 maximumTalk about end conditions when a > 0 and a

< 0What is the domain and range of the cubic?

Factoring Cubics 1A2e (3.2)

They are not easy to factor. We have two patterns we can use:

y = x3 + 3x2y + 3xy2 + y3 factors intoy = (x + y)3

y = x3 - 3x2y + 3xy2 - y3 factors intoy = (x – y)3

Factoring Cubics 1A2e (3.2)

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) = 0“x” = x, “y” = 2

Substitute: x3 - 3x2(2) + 3x(2)2 - 23

Clean it up: x3 - 6x2 + 12x - 8, which matchesFactor via patterns:

-3(x – 2)3 = 0x = 2 with duplicity of three

Factoring Cubics 1A2e (3.2)

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) = 0 “x” = x, “y” = 5

Substitute: x3 + 3x2(5) + 3x(5)2 + 53

Clean it up: x3 + 15x2 + 75x + 125, which matchesFactor via patterns:

x(x + 5)3 = 0x = 0 or x = -5 with duplicity of three

Practice

Page 132, # 4 – 6 all, 11, 12, 16 (6)