check all of the possible first steps in factoring a polynomial with four terms. factor out a gcf factor the difference of cubes factor a sum of cubes factor a difference of squares factor a perfect-square trinomial factor by grouping

get check all of the possible first steps in factoring a polynomial with four terms. factor out a gcf factor the difference of cubes factor a sum of cubes factor a difference of squares factor a perfect-square trinomial factor by grouping from EN Bilgi.

Factoring polynomials by taking a common factor (article)

Learn how to factor a common factor out of a polynomial expression. For example, factor 6x²+10x as 2x(3x+5).

Taking common factors

Factoring polynomials by taking a common factor

Learn how to factor a common factor out of a polynomial expression. For example, factor 6x²+10x as 2x(3x+5).

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What you should be familiar with before this lesson

The GCF (greatest common factor) of two or more monomials is the product of all their common prime factors. For example, the GCF of

6x 6x 6, x and 4x^2 4x 2 4, x, squared is 2x 2x 2, x .

If this is new to you, you'll want to check out our greatest common factors of monomials article.

What you will learn in this lesson

In this lesson, you will learn how to factor out common factors from polynomials.

The distributive property:

a(b+c)=ab+ac a(b+c)=ab+ac

a, left parenthesis, b, plus, c, right parenthesis, equals, a, b, plus, a, c

To understand how to factor out common factors, we must understand the distributive property.

For example, we can use the distributive property to find the product of

3x^2 3x 2 3, x, squared and 4x+3 4x+3 4, x, plus, 3 as shown below:

\blueE 3\overgroup{\overgroup{\blueE{x^2}(4x+3)=\blueE3}\blueE{x^2}(4x)+\blueE 3}\blueE{x^2}(3)

3 x 2 (4x+3)=3 ​ x 2 (4x)+3 ​ x 2 (3)

start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, plus, 3, right parenthesis, equals, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, right parenthesis, plus, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis

Notice how each term in the binomial was multiplied by a common factor of

\blueE{3x^2} 3x 2

start color #0c7f99, 3, x, squared, end color #0c7f99

.

However, because the distributive property is an equality, the reverse of this process is also true!

\blueE 3\overgroup{\blueE{x^2}(4x)+\blueE3\overgroup{\blueE{x^2}(3)=\blueE 3}}\blueE{x^2}(4x+3)

3 x 2 (4x)+3 x 2 (3)=3 ​ ​ x 2 (4x+3)

start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, right parenthesis, plus, start color #0c7f99, 3, end color #0c7f99, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 3, right parenthesis, equals, start color #0c7f99, 3, end color #0c7f99, with, \overgroup, on top, with, \overgroup, on top, start color #0c7f99, x, squared, end color #0c7f99, left parenthesis, 4, x, plus, 3, right parenthesis

If we start with 3x^2(4x)+3x^2(3) 3x 2 (4x)+3x 2 (3)

3, x, squared, left parenthesis, 4, x, right parenthesis, plus, 3, x, squared, left parenthesis, 3, right parenthesis

, we can use the distributive property to factor out

\blueE{3x^2} 3x 2

start color #0c7f99, 3, x, squared, end color #0c7f99

and obtain 3x^2(4x+3) 3x 2 (4x+3)

3, x, squared, left parenthesis, 4, x, plus, 3, right parenthesis

.

The resulting expression is in factored form because it is written as a product of two polynomials, whereas the original expression is a two-termed sum.

Check your understanding

Write

PROBLEM 1 2x(3x)+2x(5) 2x(3x)+2x(5)

2, x, left parenthesis, 3, x, right parenthesis, plus, 2, x, left parenthesis, 5, right parenthesis

in factored form. Choose 1 answer: Choose 1 answer:

Factoring out the greatest common factor (GCF)

To factor the GCF out of a polynomial, we do the following:

Find the GCF of all the terms in the polynomial.

Express each term as a product of the GCF and another factor.

Use the distributive property to factor out the GCF.

Let's factor the GCF out of

2x^3-6x^2 2x 3 −6x 2

2, x, cubed, minus, 6, x, squared

.

Step 1: Find the GCF

2x^3=\maroonD2\cdot \goldD{x}\cdot \goldD{x}\cdot x

2x 3 =2⋅x⋅x⋅x

2, x, cubed, equals, start color #ca337c, 2, end color #ca337c, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10, dot, x

6x^2=\maroonD2\cdot 3\cdot \goldD{x}\cdot \goldD{x}

6x 2 =2⋅3⋅x⋅x

6, x, squared, equals, start color #ca337c, 2, end color #ca337c, dot, 3, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10

So the GCF of 2x^3-6x^2 2x 3 −6x 2

2, x, cubed, minus, 6, x, squared

is

\maroonD2 \cdot \goldD x \cdot \goldD x=\blueE{2x^2}

2⋅x⋅x=2x 2

start color #ca337c, 2, end color #ca337c, dot, start color #e07d10, x, end color #e07d10, dot, start color #e07d10, x, end color #e07d10, equals, start color #0c7f99, 2, x, squared, end color #0c7f99

Step 2: Express each term as a product of

. \blueE{2x^2} 2x 2

start color #0c7f99, 2, x, squared, end color #0c7f99

Source : www.khanacademy.org

Practice determining Factor Techniques Flashcards

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Check all of the possible first steps in factoring a polynomial with four terms.

factor out a GCF

factor the difference of cubes

factor a sum of cubes

factor a difference of squares

factor a perfect-square trinomial

factor by grouping

Click card to see definition 👆

A and F

Click again to see term 👆

A polynomial has two terms. Check all of the factoring methods that should be considered.

common factor difference of cubes sum of cubes

difference of squares

perfect-square trinomial

factoring by grouping

Click card to see definition 👆

A, B, C, and D

Click again to see term 👆

1/11 Created by Izzy_464

This is from edge 2020

Terms in this set (11)

Check all of the possible first steps in factoring a polynomial with four terms.

factor out a GCF

factor the difference of cubes

factor a sum of cubes

factor a difference of squares

factor a perfect-square trinomial

factor by grouping A and F

A polynomial has two terms. Check all of the factoring methods that should be considered.

common factor difference of cubes sum of cubes

difference of squares

perfect-square trinomial

factoring by grouping

A, B, C, and D

Which polynomials are prime? Check all of the boxes that apply.

x^2 + 9 x^2 - 9 x^2 + 3x + 9 -2x^2 + 8 A and C

Which constant term would mean that the expression is completely factored?x2 - 3x + ____

10

Which expression does not factor?

m^3 + 1 m^3 - 1 m^2 + 1 m^2 - 1 C

What is the completely factored form of 25x^4 - 16y^2?

(5x^4 + 4y)(5x - 4y)

(5x^3 + 4y)(5x^2 - 4y)

(5x^2 + 4y)(5x^2 - 4y)

25x^4 - 16y^2 C

What is the completely factored form of 2x^2 - 32?

(2x^2 + 16)(x - 16) 2(x + 4)(x - 4) 2(x + 8)(x - 4) 2(x - 8)(x - 4) B

What is the completely factored form of x^4y - 4x^2y - 5y?

y(x^2 - 5)(x^2 + 1) y(x^2 + 5)(x^2 - 1) (x^2y - 5)(x^2 + 1) (x^2y + 5)(x^2 - 1) A

Which of the following statements about 42xy - 49x + 30y - 35 are true?Check all of the boxes that apply.

One of the factors is (6y + 7).

One of the factors is (7x + 5).

One of the factors is (6y - 7).

One of the factors is (7x - 5).

This expression is prime.

B and C

A student factors a^6 - 64 to (a^2 - 4)(a^4 + 4a^2 + 16).

Which statement about (a^2 − 4)(a^4 + 4a^2 + 16) is correct?

The expression is equivalent and is completely factored.

The expression is equivalent, but the (a^2 - 4) term is not completely factored.

The expression is equivalent, but the (a^4 + 4a^2 + 16) term is not completely factored.

The expression is not equivalent.

B

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Algebra: Beginning and Intermediate

Intended for combined introductory and intermediate algebra courses, this text retains the hallmark features that have made the Aufmann texts market leaders: an interactive approach in an objective-based framework: a clear writing style, and an emphasis on problem-solving strategies. The acclaimed Aufmann Interactive Method, allows students to try a skill as it is introduced with matched-pair examples, offering students immediate feedback, reinforcing the concept, identifying problem areas, and, overall, promoting student success.Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.

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Algebra: Beginning and Intermediate

Richard N. Aufmann, Joanne Lockwood

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Intended for combined introductory and intermediate algebra courses, this text retains the hallmark features that have made the Aufmann texts market leaders: an interactive approach in an objective-based framework: a clear writing style, and an emphasis on problem-solving strategies. The acclaimed Aufmann Interactive Method, allows students to try a skill as it is introduced with matched-pair examples, offering students immediate feedback, reinforcing the concept, identifying problem areas, and, overall, promoting student success.

Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.

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Dizin

İçindekiler

Real Numbers and Variable Expressions

1

Solving Equations and Inequalities

71

Linear Functions and Inequalities in Two Variables

173

Chapters 1 Through 3

251

Systems of Equations and Inequalities

263 Polynomials 317 Factoring 371

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419

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605

Functions and Relations

621

Exponential and Logarithmic Functions

653 Conic Sections 705 Appendix 747

Solutions to Chapter Problems

501

Answers to Selected Exercises

A-1 Glossary 61

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Rational Exponents and Radicals

487

Quadratic Equations and Inequalities

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2012 Cengage Learning absolute value angles annual simple interest axis of symmetry binomial coefficient complex number contains the point coordinates copied Copyright 2012 Cengage cost decimal denominator distance Distributive Property domain Due to electronic duplicated eB ook and/or electronic rights eleetronie eopied Evaluate Example exponent exponential expression Factor find Find the equation first flmction formula fraction front the eB function GETTING READY given Graph the solution inequality integers intersect inverse invested line that contains linear nearest tenth ook and/or eChapter(s ordered pairs parabola party Due percent plane point-slope formula polynomial Problem quadratic equation radical expression real numbers rectangle rewrite Rights Reserved scanned seanned set-builder notation Simplify slope solution set Subtract suppressed front system of equations third party content triangle trinomial Try Exercise variable expression vertex vertical whole width x-coordinate x-intercepts zero

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Başlık Algebra: Beginning and Intermediate

Yazarlar Richard N. Aufmann, Joanne Lockwood

Baskı 3

Yayıncı Cengage Learning, 2012

ISBN 1133709397, 9781133709398

Uzunluk 912 sayfa

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