Integers whose product is 9 1 9 1 9 3 3 3 3 Determine the sum of the integers

Integers whose product is 9 1 9 1 9 3 3 3 3 determine

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Integers whose product is 9 1, 9 1, 9 3, 3 3, 3 Determine the sum of the integers. Integers whose product is 9 1, 9 1, 9 3, 3 3, 3 Sum 10 10 6 6 Select 1, 9 because the product is 9 (last term of the trinomial) and the sum is 10 (middle term coefficient of the trinomial). x 2 10 x 9 ( x 9)( x 1) Check: ( x 9)( x 1) x 2 1 x 9 x 9 x 2 10 x 9 YOUR TURN Factor x 2 9 x 20.
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0.4 Factoring Polynomials 43 Answer: (2 t 3)( t 1) Factors of a a x 2 b x c ( n x n )( n x n ) Factors of c EXAMPLE 8 Factoring a Trinomial Whose Leading Coefficient Is Not 1 Factor 5 x 2 9 x 2. Solution: S TEP 1 Start with the fi r st term. Note that 5 x x 5 x 2 . (5 x ; n )( x ; n ) S TEP 2 The product of the last terms should yield 2. 1, 2 or 1, 2 S TEP 3 Consider all possible factors based on Steps 1 and 2. (5 x 1)( x 2) (5 x 1)( x 2) (5 x 2)( x 1) (5 x 2)( x 1) Since the outer and inner products must sum to 9 x , the factored form must be: 5 x 2 9 x 2 (5 x 1)( x 2) Check: (5 x 1)( x 2) 5 x 2 10 x 1 x 2 5 x 2 9 x 2 YOUR TURN Factor 2 t 2 t 3. When the leading coefficient of the trinomial is not equal to 1, then we consider all possible factors using the following procedure, which is based on the FOIL method in reverse. EXAMPLE 9 Factoring a Trinomial Whose Leading Coefficient Is Not 1 Factor 15 x 2 x 6. Solution: S TEP 1 Start with the fi r st term. (5 x ; n )(3 x ; n ) or (15 x ; n )( x ; n ) S TEP 2 The product of the last terms should yield 6. 1, 6 or 1, 6 or 2, 3 or 2, 3 Step 1: Find two F irst terms whose product is the first term of the trinomial. Step 2: Find two L ast terms whose product is the last term of the trinomial. Step 3: Consider all possible combinations found in Steps 1 and 2 until the sum of the O uter and I nner products are equal to the middle term of the trinomial. F ACTORING A TRINOMIAL WHOSE LEADING COEFFICIENT IS NOT 1
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44 CHAPTER 0 Prerequisites and Review S TEP 3 Consider all possible factors (5 x 1)(3 x 6) (15 x 1)( x 6) based on Steps 1 and 2. (5 x 6)(3 x 1) (15 x 6)( x 1) (5 x 1)(3 x 6) (15 x 1)( x 6) (5 x 6)(3 x 1) (15 x 6)( x 1) (5 x 2)(3 x 3) (15 x 2)( x 3) (5 x 3)(3 x 2) (15 x 3)( x 2) (5 x 2)(3 x 3) (15 x 2)( x 3) (5 x 3)(3 x 2) (15 x 3)( x 2) Since the outer and inner products must sum to x , the factored form must be: 15 x 2 x 6 (5 x 3)(3 x 2) Check: (5 x 3)(3 x 2) 15 x 2 10 x 9 x 6 15 x 2 x 6 YOUR TURN Factor 6 x 2 x 12. Answer: (3 x 4)(2 x 3) EXAMPLE 11 Factoring a Polynomial by Grouping Factor x 3 x 2 2 x 2. Solution: Group the terms that have a common factor. ( x 3 x 2 ) (2 x 2) Factor out the common factor in each pair of parentheses. x 2 ( x 1) 2( x 1) Use the distributive property. ( x 2 2)( x 1) EXAMPLE 10 Identifying Prime (Irreducible) Polynomials Factor x 2 x 8. Solution: Write the trinomial as a product of two binomials in general form. x 2 x 8 ( x n )( x n ) Write all of the integers whose product is 8. Factoring by Grouping Much of our attention in this section has been on factoring trinomials. For polynomials with more than three terms we first look for a common factor to all terms. If there is no common factor to all terms of the polynomial, we look for a group of terms that have a common factor. This strategy is called factoring by grouping .
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  • Summer '17
  • juan alberto
  • Fractions, Elementary arithmetic, Greatest common divisor

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