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Chapter 2

6. Chapter. Chapter 2. Factoring Polynomials. Factoring Trinomials of the Form a x 2 + bx + c. Section. 6.3. Factoring Trinomials of the Form ax 2 + bx + c. Objective. 1. Factoring Trinomials. Returning to the FOIL method, F O I L

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Chapter 2

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  1. 6 Chapter Chapter 2 Factoring Polynomials

  2. Factoring Trinomials of the Form ax2 + bx + c Section 6.3

  3. Factoring Trinomials of the Form ax2 + bx + c Objective 1

  4. Factoring Trinomials • Returning to the FOIL method, • F O I L • (3x + 2)(x + 4) = 3x2 + 12x + 2x + 8 • = 3x2 + 14x + 8 • To factor ax2 + bx + c into ( x + ) ( x + ), note that a is the product of the two first coefficients, c is the product of the two last coefficients and b is the sum of the products of the outside coefficients and inside coefficients. • Note that b is the sum of 2 products, not just 2 numbers, as in the last section.

  5. Example Factor: 21x2 – 41x + 10 Possible factors of 21x2: 21x2 = x • 21x, 21x2 = 3x • 7x. Since the middle term is negative, possible factors of 10 must both be negative: 10 = –1 • –10, 10 = –2 • –5. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Continued

  6. Factors of 21x2 Factors of 10 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products Example (cont) We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to – 41x. x, 21x 1, 10 (x – 1)(21x – 10) –10x–21x – 31x (x – 10)(21x – 1) –x–210x – 211x x, 21x 2, 5 (x – 2)(21x– 5) –5x–42x – 47x (x – 5)(21x – 2) –2x–105x – 107x Continued

  7. Factors of 21x2 Factors of 10 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products (3x – 5)(7x – 2) –6x–35x –41x Example (cont) 3x, 7x 1, 10 (3x – 1)(7x – 10) –30x –7x–37x (3x – 10)(7x – 1) –3x–70x–73x 3x, 7x 2, 5 (3x – 2)(7x – 5) –15x–14x–29x Continued

  8. F O L I 3x(7x) + 3x(–2) – 5(–2) – 5(7x) Example (cont) Check the resulting factorization using the FOIL method. (3x – 5)(7x – 2)= = 21x2 – 6x – 35x + 10 = 21x2 – 41x + 10 A factored form of 21x2 – 41x + 10 is (3x – 5)(7x – 2).

  9. Example Factor: 3x2 – 7x + 6. The only possible factors for 3 are 1 and 3, so we know that, if factorable, the polynomial will have to look like (3x + )(x + ) in factored form, so that the product of the first two terms in the binomials will be 3x2. Since the middle term is negative, possible factors of 6 must both be negative: {–1, –6} or {–2, –3}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Continued

  10. Factors of 6 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products Example (cont) We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to –7x. –1, –6 (3x – 1)(x – 6) –18x–x –19x (3x – 6)(x – 1) Common factor so no need to test. –2, –3 (3x – 2)(x – 3) –9x –2x –11x (3x – 3)(x – 2) Common factor so no need to test. Continued

  11. Example (cont) Now we have a problem, because we have exhausted all possible choices for the factors, but have not found a pair where the sum of the products of the outside terms and the inside terms is –7. So 3x2 – 7x + 6 is a prime polynomial and will not factor.

  12. Factoring Out the Greatest Common Factor Objective 2

  13. Example • Factor: 3x3 + 17x2 + 10x • Factor out the greatest common factor x. • x(3x2 + 17x + 10) • Factors of 3x2: 3x and x • Factors of 10: 1, 10 or 2 and 5 • Try some combinations of the factors. • x(3x + 1)(x + 10) = x(3x2 + 31x + 10) • x(3x + 2)(x + 5) = x(3x2 + 17x + 10) TRUE

  14. Factoring Perfect Square Trinomials Objective 3

  15. Perfect Square Trinomials A trinomial that is the square of a binomial is called a perfectsquaretrinomial.

  16. Factoring Perfect Square Trinomials a2 + 2ab + b2 =(a + b)2 a2 – 2ab + b2 = (a – b)2

  17. Example Factor: 25x2 + 20x + 4 Possible factors of 25x2: 25x2 = x • 25x, 25x2 = 5x • 5x. Possible factors of 4: 4= 1 • 4, 4 = 2 • 2. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Keep in mind that, because some of our pairs are not identical factors, we may have to exchange some pairs of factors and make 2 attempts before we can definitely decide a particular pair of factors will not work. Continued

  18. Factors of 25x2 Factors of 4 Resulting Binomials Product of Outside Terms Product of Inside Terms Sum of Products 5, 5x 2, 2 (5x + 2)(5x + 2) 10x10x 20x Example (cont) We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 20x. x, 25x 1, 4 (x + 1)(25x + 4) 4x25x 29x (x + 4)(25x + 1) x 100x 101x x, 25x 2, 2 (x + 2)(25x + 2) 2x50x 52x Continued

  19. F O L I 5x(5x) + 5x(2) + 2(2) + 2(5x) Example (cont) Check the resulting factorization using the FOIL method. (5x + 2)(5x + 2) = = 25x2 + 10x + 10x + 4 = 25x2 + 20x + 4 Thus a factored form of 25x2 + 20x + 4 is (5x + 2)(5x + 2) or (5x + 2)2.

  20. Example Factor:

  21. Example Factor: 16x2 – 8xy + y Since we have determined that this trinomial is a perfect square, we can factor it using a special product formula. 16x2 – 8xy + y = (4x)2 – 2 • 4x • y + (y)2 a2 – 2 • a •b + b2 =(4x – y)2 (a – b)2

  22. Example Factor:

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