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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

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    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

    Cengage Learning, 2 Oca 2012 - 912 sayfa

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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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    263 Polynomials 317 Factoring 371

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    A-1 Glossary 61

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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

    Kaynakça bilgileri

    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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    Do you want to see answer or more ?
    James 16 day ago
    4

    Guys, does anyone know the answer?

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