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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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For more information, please see full course syllabus of Algebra 1
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Lecture Comments (8)

1 answer

Last reply by: Dr Carleen Eaton
Sun Mar 3, 2013 9:05 PM

Post by Erica Rapetti on March 3, 2013

For example two, do you always start with grouping? What about if you have a GCF for all four what do you do then, do you group or go with the GCF?

1 answer

Last reply by: S A A Mazeed Sumon
Mon May 3, 2010 12:13 AM

Post by Timothy miranda on April 18, 2010

*correction* examples 2 and 3 show that the two binomials are supposed to be multiplied.

3 answers

Last reply by: Dr Carleen Eaton
Sat Jun 29, 2013 11:21 AM

Post by Timothy miranda on April 18, 2010

When showing factoring by grouping (7:40) she puts a + sign in between the 'common binomial factor' and the binomial with the GCF's in it.

Those two binomials are supposed to be multiplied. Examples of this are shown in Examples 1-3 of this video.

Factoring Using Greatest Common Factor

  • You can use the distributive property to factor the greatest common factor out of each polynomial in a sum or difference of polynomials.
  • For a polynomial with 4 terms, factor a GCF out of the first two terms and then factor the GCF out of the second two terms. Then factor the common binomial factor. This is called factoring by grouping.
  • Sometimes in factoring by grouping, you must use the fact that one binomial factor is the additive inverse of another one. Factoring –1 out of one of these binomials produces two identical binomial factors and enables you to complete the factoring by grouping.
  • The Zero Product Property states that if a product of two factors is 0, then one or both of the factors must be equal to 0. This property allows you to solve equations in which a factored polynomial is equal to 0.
  • The solutions of an equation are also called the roots of the equation.

Factoring Using Greatest Common Factor

Factor:
4x2y3z4 − 12x2y2z + 18xy3
  • GCF = 2xy2
2xy2( 2xyz4 − 6xyz + 9y )
Factor:
13a3b4c + 26a2bc3 − 39ab3c5
  • GCF: 13abc
13abc( a2b3 + 2ac2 − 3b2c4 )
Factor:
8r7s5t4 − 16r6s3t2 − 48r8s6t4
  • GCF: 8r6s3t2
8r6s3t2( rs2t2 − 2 − 6r2s3t2 )
Factor:
( 5x2 − 10xy + 4xy − 8y2 )
  • ( 5x2 − 10xy )
    GCF: 5x
  • ( 4xy − 8y2 )
    GCF: 4y
  • 5x( x − 2y ) + 4y( x − 2y )
( 5x + 4y )( x − 2y )
Factor:
( 6x2 + 9xy − 14xy − 21y2 )
  • ( 6x2 + 9xy )
    GCF: 3x
  • ( − 14xy − 21y2 )
    GCF: ( − 7y )
  • 3x( 2x + 3y ) + ( − 7y )( 2x + 3y )
( 3x − 7y )( 2x + 3y )
Factor:
( 13s2 − 39st − 25st + 75t2 )
  • ( 13s2 − 39st )
    GCF: 13s( s − 3t )
  • ( − 25st + 75t2 )
    GCF: 25t( − s + 3t ) = 25t( s − 3t )
  • 13s( s − 3t ) − 25t( s − 3t )
( 13s − 25t )( s − 3t )
Factor:
16x3 − 4x2 − 3 + 12x
  • ( 16x3 − 4x2 ) + ( − 3 + 12x )
  • 4x2( 4x − 1 ) + 3( − 1 + 4x )
  • 4x2( 4x − 1 ) + 3( 4x − 1 )
( 4x2 + 3 )( 4x − 1 )
Factor:
34y4 − 17y2 − 24y + 48y3
  • ( 34y4 − 17y2 ) + ( − 24y + 48y3 )
  • 17y2( 2y2 − 1 ) + 24y( − 1 + 2y2 )
  • 17y2( 2y2 − 1 ) + 24y( 2y2 − 1 )
( 17y2 + 24y )( 2y2 − 1 )
Solve:
( 6m + 4 )( 5m − 10 ) = 0
  • 6m + 4 = 0
  • 6m = 4
  • m = [4/6] = [2/3]
  • 5m − 10 = 0
  • 5m = 10
  • m = 2
{ [2/3],2 }
Solve:
( 2a − 14 )( 6a + 36 ) = 0
  • 2a − 14 = 0
  • a = 7
  • 6a + 36 = 0
  • 6a = − 36
  • a = − 6
{ − 6,7 }

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Factoring Using Greatest Common Factor

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Distributive Property 0:05
    • Example: Binomial
    • Example: Trinomial
  • Factoring by Grouping 4:17
    • Example: Four Terms
  • Zero Product Property 8:21
    • Example
  • Example 1: Factor the Polynomial 10:38
  • Example 2: Factor the Polynomial 13:43
  • Example 3: Factor the Polynomial 19:59
  • Example 4: Solve the Polynomial 22:58