Step i find the gcf of the coefficients gcf8 6 2 step

This preview shows page 3 - 6 out of 6 pages.

Step I: Find the GCF of the coefficients. GCF(8, 6) = 2. Step II: Find the GCF of the variable parts. GCF( x 3 , x 2 ) = x 2 . Step III: Divide all the monomials by the GCFs and rewrite with the GCFs out front: 8 x 3 + 6 x 2 = + 2 2 2 3 2 2 6 2 8 2 x x x x x = 2 x 2 (4 x + 3) 3. Factor 18 y 3 ( x – 1) + 21 y ( x – 1) Step I: Find the GCF of the coefficients: GCF(18, 21) = 3. Step II: Find the GCF of the variable parts: GCF( y 3 ( x – 1), y ( x – 1)) = y ( x – 1) Step III: Divide all the monomials by the GCFs and rewrite with the GCFs out front: 18 y 3 ( x – 1) + 21 y ( x – 1) = - - + - - - ) 1 ( 3 ) 1 ( 21 ) 1 ( 3 ) 1 ( 18 ) 1 ( 3 3 x y x y x y x y x y = 3 y ( x – 1)(6 y 2 + 7) 6. Factor 3 ab + 7 b .
Image of page 3

Subscribe to view the full document.

Math 1300 Section 4.1 Notes 4 7. Factor -28 a 3 b 7 – 36 a 2 b 5 . 8. Factor 8 a 2 bc – 12 ab 2 c . 9. Factor 9 xy 3 + 27 xy x 5 . (Hint: What is the leading coefficient?) 10. Factor (9 – 3 x )(27 y 2 ) – (9 – 3 x )(15 y 3 )
Image of page 4
Math 1300 Section 4.1 Notes 5 Factoring by Grouping If a polynomial contains four or more terms, it may be helpful to put the terms into groups of two and factor out a common factor from each of these groups. This is called grouping . Examples: 1. Factor x 2 y + 6 x + 3 xy 2 + 18 y Step I: Group the terms so that each group shares a common factor: x 2 y + 6 x + 3 xy 2 + 18 y = ( x 2 y + 6 x ) + (3 xy 2 + 18 y ) Step II: Factor out the common terms from each group: ( x 2 y + 6 x ) = x ( xy + 6) (3 xy 2 + 18 y ) = 3 y ( xy + 6) Step III: Rewrite the polynomial as the sum of the factored groups: x 2 y + 6 x + 3 xy 2 + 18 y = x ( xy + 6) + 3 y ( xy + 6) Step IV: Factor the resulting polynomial from Step III: x ( xy + 6) + 3 y ( xy + 6) = ( xy + 6)( x + 3 y ) 2. Factor 2 x 3 + 3 x 2 + 2 x + 3 Step I: Group the terms so that each group shares a common factor: 2 x 3 + 3 x 2 + 2 x + 3 = (2 x 3 + 3 x 2 ) + (2 x + 3) Step II:
Image of page 5

Subscribe to view the full document.

Image of page 6

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern