1 / 9

“Just the factors Ma’am!”

“Just the factors Ma’am!”. A guide to factoring. Greatest Common Factor. List the prime factors of each term. Identify the factors common to all terms. Multiply the common factors together. This is the Greatest Common Factor. For all real numbers a and b: a 2 – b 2 = (a+b)(a-b)

harlow
Télécharger la présentation

“Just the factors Ma’am!”

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. “Just the factors Ma’am!” A guide to factoring.

  2. Greatest Common Factor • List the prime factors of each term. • Identify the factors common to all terms. • Multiply the common factors together. • This is the Greatest Common Factor.

  3. For all real numbers a and b: a2 – b2 = (a+b)(a-b) Where a is the square root of a2 and b is the square root of b2 for example: X2 – 64 = (x+8)(x-8) Difference of Two Squares

  4. Perfect Square TrinomialsA perfect square trinomial is one where the first and last terms are perfect squares.The middle term is twice the product of one factor from the first term and one factor from the last term. For every real number a and b: a2 + 2ab + b2 = (a+b)(a+b) = (a+b)2 a2 – 2ab + b2 = (a-b)(a-b ) = (a-b)2 For example: x2 + 8x + 16 = (x+4)(x+4) = (x+4)2 4n2 – 12n + 9 = (2n-3)(2n-3) = (2n-3)2

  5. Factoring by GroupingGroup terms in pairs.Factor out the common factor in each pair.The remaining two parenthesis should contain the same binominal.Combine common factors into a parenthesis. 3n3 – 12n2 + 2n – 8 (3n3 – 12n2) (2n – 8) 3n2(n-4) 2(n-4) (3n2 +2)(n-4)

  6. Multiply together the coefficients of the first and last term. Find two numbers that have this number as their product and the middle term as their sum. Substitute these two terms in for the middle term. Solve using grouping. (15)(4) = 60 (12)(5) = 60 12 + 5 = 17 (20)(3) = 60 20 + 3 = 23 √ 4w2 + 20w + 3w + 15 (4w2 + 20w) (3w + 15) 4w(w + 5) 3(w + 5) (4w + 3) (w + 5) Factoring by GroupingTrinomial4w2 + 23w + 15

  7. • Factor out the GCF • Factor the remaining trinomial using the methods of grouping. •Place the GCF in front of the factored trinomial. 4(3p2 + 5p – 2) (3)(-2) = -6 (3)(-2) = -6 3 + -2 = 1 (6)(-1) = -6 6 + -1 = 5 √ 3p2 + 6p – p – 2 (3p2 + 6p) (-p – 2) 3p(p + 2) -1(p + 2) (3p - 1)(p + 2) 4(3p – 1)(p + 2) Factoring out a GCF12p2 + 20p - 8

  8. Sum and Difference of Cubes For all real numbers a and b: a3 + b3 = (a+b)(a2 – ab + b2) a3 – b3 = (a-b)(a2 + ab + b2) Where a is the cube root of a3 and b is the cube root of b3 for example: x3 + 64 = (x+4)(x2 - 4x + 16) n3 – 125 = (n-5)(n2 + 5n + 25)

  9. Summary

More Related