polynomial review what is a polynomial? an algebraic expression consisting of one or more summed...

Polynomial Review

What is a polynomial? An algebraic expression consisting of one or more

summed terms, each term consisting of a coefficient and one or more variables raised to natural number exponent.

Examples:

X2 + 3x – 3 2x + 1 5y3 + 2y2 - 4y - 8

Terms

What is a term? A term is the product of a coefficient and one or

more variables raised to a natural number exponent Examples:

X 2xy 3y3 x 4

Terms cont.

Identify the coefficient and degree for each example

X Coefficient – 1 Degree – 1

2xy Coefficient – 2 Degree – 2

3y3 x Coefficient – 3 Degree – 4

4 Coefficient – 4 Degree - 0

Types of Polynomials

Monomial Has 1 term

3ab 2y3 xy

Binomial Has 2 terms combined with addition or subtraction

ax2 + bx 2x + 3

Trinomial Has 3 terms combined with addition or subtraction

ax2 + bx + c 2x + 3x +1

Multiplying Polynomials

Binomial x Binomial – FOIL Method First Outer Inner Last

F – x * x O – x * 2

I – 1 * x L – 1 * 2

FOIL cont.

F: x * x = x2

O: x * 2 = 2xI: 1 * x = xL: 1 * 2 = 2

Add the terms:x2 + 2x + x + 2

Combine like terms:x2 + 3x + 2

Chapter 6 – Factoring Polynomials and Solving Equations

6.1 Introduction To Factoring


● Factors – 2 or more numbers that multiply to a new number Factors of 4: 1,4 and 2,2

1 * 4 = 4 and 2 * 2 = 4 Factors of 32: 1, 32 2, 16 4, 8

• 1 * 32 = 32 2 * 16 = 32 and 4 * 8 = 32


• Factors of a polynomial – 2 or more polynomials that multiply to a higher degree polynomial

x2 – x x2 and x have a common factor of x

– x2 = x * x– x = 1 * x

We can use the distributive property to pull the common factor out of each term.

– x(x – 1)


• Find the common factors of the following terms: 8x2 + 6x

• 8x2 = 2x * 4x• 6x = 2x * 3

The common factor is 2x Pull out the common factor using the distributive property.

• 2x(4x + 3)


10x + 6 Step 1: Find the common factors for the terms

• Common factor: 2 Step 2: Pull the common factor out from each

term• 10x = 2 * 5x• 6 = 2 * 3

Step 3: Multiply the common factor by the sum of the other factors

• 2(5x + 3)


6x2 + 15x Step 1: Find the common factors for the terms

• Common factor: 3x Step 2: Pull the common factor out from each

term• 6x2 = 3x * 2x• 15x = 3x * 5

Step 3: Multiply the common factor by the sum of the other factors

• 3x(2x + 5)


• Factor the polynomial: 3z3 + 9z2 – 6z Find the common factor: 3z Pull out the common factor: z2 +3z – 2 Write the polynomial as the product of the factors

• 3z (z2 + 3z – 2)


• Factor the polynomial: 2x2y2 + 4xy3

Find the common factor: 2xy2 Pull out the common factor: x + 2y Write the polynomial as the product of the

factors: • 2xy2(x + 2y)


• Find the GCF and write each polynomial as the product of the factors.

9x2 + 6x

8a2b3 – 16a3b2

66t + 16t2


• Find the GCF and write each polynomial as the product of the factors.

9x2 + 6x• 3x (3x + 2)

8a2b3 – 16a3b2

• 8a2b2(b - 2a) 66t + 16t2

• 2t(33 + 8t)


• Factoring by Grouping What is common between the following terms?

5x(x + 3) + 6(x+ 3)

The common factor is the binomial (x + 3)

We can rewrite the polynomial as the product of the common factor and the sum of the other factors:

(x + 3)(5x + 6)


• Identify the common factors x2(2x – 5) – 4x(2x – 5)

5x(3x – 2) + 2(3x – 2)

(z + 5)z + (z + 5)4


• Identify the common factors x2(2x – 5) – 4x(2x – 5)

• 2x – 5 5x(3x – 2) + 2(3x – 2)

• 3x – 2 (z + 5)z + (z + 5)4

• z + 5


• Factor by grouping: 2x3 – 4x2 + 3x – 6 Step 1: Group the terms that have common

factors (2x3 – 4x2) + (3x – 6)

Step 2: Identify the common factors for each group

• Common factor of 2x3 – 4x2 : 2x2

• Common factor of 3x – 6: 3

2x3 – 4x2 + 3x – 6 cont.

Step 3: write each grouping as the product of the factors

• 2x2(x – 2) + 3(x – 2) Note: the parenthesis are the same Step 4: Distribute the common factor from

each grouping• 2x2(x – 2) + 3(x – 2)

• (x – 2)(2x2 + 3)


• Factor by Grouping: 3x + 3y + ax + ay (3x + 3y) + (ax + ay) 3(x + y) + a(x + y)

(x + y) (3 + a) Check by using FOIL: (x + y) (3 + a)

• 3x + ay + 3y + ay

3x + 3y + ax + ay cont.

• Let's Factor this polynomial again by grouping the x terms and y terms 3x + 3y + ax + ay = 3x + ax + 3y + ay (3x + ax) + (3y + ay) x(3 + a) + y(3 + a)

(3 + a) (x + y)

3x + 3y + ax + ay cont.

• Compare Grouping 1 and Grouping 2

Grouping 1: (x + y) (3 + a) Grouping 2: (3 + a) (x + y)

• Are these the same?

Reflect

• Does it matter how you group the terms? No, you will get the same answer However, you need to group terms that have a common factor other than 1.

Reflect

• Do the parenthesis have to be the same for each term after you have factored the GCF?

Yes

• How many terms do you need to factorby grouping?

At least 4

Practice

• Factor the following Polynomials 6x3 – 12x2 – 3x + 6

2x5 – 8x4 + 6x3 – 24x2

4t3 – 12t2 + 3t – 9

• Check by multiplying the factors

Practice cont.

• Answers: 3(2x2 – 1)(x – 2)

2x2 (x2 + 3)(x – 4)

(4t2 + 3)(t – 3)

???? Questions ????

6.2 Factoring Trinomials – x2 + bx + c


• When factoring a polynomial in the form of x2 + bx + c, we will be reversing the FOIL process

• So, let's review FOIL

FOIL

• F (x + m) (x + n) = x2

FOIL

• F (x + m) (x + n) = x2

• O (x + m) ( x + n) = nx

FOIL

• F (x + m) (x + n) = x2

• O (x + m) ( x + n) = nx• I (x + m) (x + n) = mx

FOIL

• F (x + m) (x + n) = x2

• O (x + m) ( x + n) = nx• I (x + m) (x + n) = mx• L (x + m) (x + n) = nm

FOIL

• F (x + m) (x + n) = x2

• O (x + m) ( x + n) = nx• I (x + m) (x + n) = mx• L (x + m) (x + n) = nm

x2 + nx + mx + nm

FOIL

• Can we combine like terms? Yes, O and I in FOIL give us like terms

• x2 + nx + mx + nm

What is the common factor?• X

Pull out the x:• x2 + (n + m)x + nm

Factoring

Let's

• Standard Form:

x2 + bx + c

Compare

• (x +n)(x + m): x2 + (n + m)x + nm

Factoring

• Based on the Standard Form: x2 + bx + c, what is the b in the polynomial we found:

x2 + (n + m)x +nm

b = (n + m)

Factoring

• Based on the Standard Form: x2 + bx + c, what is the c in the polynomial we found:

x2 + (n + m)x +nm

c = nm


• To factor the trinomial x2 + bx + c, find two numbers m and n that satisfy the following conditions:

m * n = c m + n = b

Such that x2 + bx + c = (x + m)(x + n)


• Remember that (m + n)x was the sum of the O and I in FOIL

and nm was the L in FOIL.

x2 + 7x + 12

• Step 1: Factor x2

• x2 + 7x + 12

• (x )(x )

x2 + 7x + 12

• Step 2: Identify all of the factors of 12

1 * 12 = 12 2 * 6 = 12 3 * 4 = 12

x2 + 7x + 12

• Step 3: Identify the factors of 12 that add to 7

1 * 12 = 12 1 + 12 = 13 2 * 6 = 12 2 + 6 = 8 3 * 4 = 12 3 + 4 = 7

x2 + 7x + 12

• Step 4: Complete the factors

(x + 3)(x + 4)

x2 + 7x + 12

• Step 5: Check by FOIL

(x + 3)(x + 4)

• x2 + 4x + 3x + 12• x2 + 7x +12

x2 + 13x + 30


• x2 + 13x + 30

• (x )(x )

x2 + 13x + 30

• Step 2: Find all factors of 30

• 1 * 30 = 30• 2 * 15 = 30• 3 * 10 = 30• 6 * 5 = 30

x2 + 13x + 30


• 1 * 30 = 30 1 + 30 = 31• 2 * 15 = 30 2 + 15 = 17• 3 * 10 = 30 3 + 10 = 13• 6 * 5 = 30 6 + 5 = 11

x2 + 13x + 30


(x + 3)(x + 10)

x2 + 13x + 30


(x + 3)(x + 10)

• x2 + 10x + 3x + 30• x2 + 13x + 30

z2 – 8z + 15

• Step 1: Factor z2

• z2 – 8z + 20

• (z )(z )

z2 – 8z + 15


• 1 * 15 = 15• -1 * -15 = 15• 3 * 5 = 15• -3 * -5 = 15

z2 – 8z + 15

• Step 3: Identify the factors of 15 that add to -8

• 1 * 15 = 15 1 + 15 = 16• -1 * -15 = 15 -1 + -15 = -16• 3 * 5 = 15 3 + 5 = 8• -3 * -5 = 15 -3 + -5 = -8

z2 – 8z + 15


(z – 3)(z – 5)

z2 – 8z + 15


(z – 3)(z – 5)

• z2 – 3z – 5z + 15• z2 – 8z + 15

y2 – 3y – 4

• Step 1: Factor y2

• y2 – 3y – 4

• (y )(y )

y2 – 3y – 4

• Step 2: Find all factors of -4

• 1 * -4 = -4• -1 * 4 = -4• -2 * 2 = -4

y2 – 3y – 4

• Step 3: Identify the factors of -4 that add to -3

• 1 * -4 = -4 1 + -4 = -3• -1 * 4 = -4 -1 + 4 = 3• -2 * 2 = -4 -2 + 2 = 0

y2 – 3y – 4


(y + 1)(y – 4)

y2 – 3y – 4


(y + 1)(y – 4)

• y2 – 4y + 1y – 4 • y2 – 3y – 4

x2 + 9x + 12


• x2 + 9x + 12

• (x )(x )

x2 + 9x + 12


• 1 * 12 = 12• 2 * 6 = 12• 3 * 4 = 12

x2 + 9x + 12


• 1 * 12 = 12 1 + 12 = 13• 2 * 6 = 12 2 + 6 = 8• 3 * 4 = 12 3 + 4 = 7

x2 + 9x + 12

• No Factors of 12 will add to 9

• What do you think this means? The Trinomial, x2 + 9x + 12, is prime

• The only factors are itself and 1

7x2 + 35x + 42

• Step 1: Factor 7 from all 3 terms• 7*x2 + 7*5x + 7 *6

• 7(x2 + 5x + 6)

7x2 + 35x + 42


• 7(x2 + 5x + 6)

• 7(x )(x )

7x2 + 35x + 42


• 1 * 6 = 6• 2 * 3 = 6

7x2 + 35x + 42


• 1 * 6 = 6 1 + 6 = 7• 2 * 3 = 6 2 + 3 = 5

7x2 + 35x + 42

• Step 5: Complete the factors 7(x + 2)(x + 3)

7x2 + 35x + 42

• Step 6: Check by FOIL 7(x + 2)(x + 3)

• 7(x2 + 3x + 2x + 6)• 7(x2 + 5x + 6)• 7x2 + 35x + 42

Reflect

• Is it possible to factor every trinomial? No, if a trinomial is prime it cannot be factored

• How do you know if a trinomial is prime? If you cannot find 2 numbers that multiply to nm and add to (m + n)

Practice

• z2 + 9z + 20

• t2 – 2t – 24

• x2 + 5x – 4

• 2x4 – 4x3 – 6x2

Practice Answers

• z2 + 9z + 20 (z + 4)(z + 5)• t2 – 2t – 24 (t – 6)(t + 4)• x2 + 5x – 4 No factors – Trinomial is prime• 2x4 – 4x3 – 6x2

2x2(x – 3)(x + 1)

???? Questions ????

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