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Consider the Following Approach to Factoring 20x2+39x+1820 x ^ { 2 } + 39 x + 18

question 253

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Consider the following approach to factoring. 20x2+39x+1820 x ^ { 2 } + 39 x + 18  Consider the following approach to factoring.  20 x ^ { 2 } + 39 x + 18    We need two integers whose sum is 39 and whose product is 360. To help find these integers, let s prime factor 360.  360 = 5 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5  Now by grouping these factors in various ways, we find that  2 \cdot 2 \cdot 2 \cdot 3 = 24,3 \cdot 5 = 15  and  24 + 15 = 39  So the numbers are 15 and 24, and we can express the middle term of the given trinomial,  39 x  , as  15 x + 24 x  . Therefore, we can complete the factoring as follows:  20 x ^ { 2 } + 39 x + 18 = 20 x ^ { 2 } + 15 x + 24 x + 18 = 5 x ( 4 x + 3 )  + 6 ( 4 x + 3 )  = ( 4 x + 3 )  ( 5 x + 6 )   Factor the trinomial.  20 y ^ { 2 } + 23 y + 6  A)  (4 y + 3) (8 y + 2)  B)  (4 y  -  3) (5 y  -  2)  C)  (4 y + 3) (5 y + 2)  D)  (4 y  -  3) (5 y + 2)  E)  (6 y + 3) (5 y + 3)  We need two integers whose sum is 39 and whose product is 360. To help find these integers, let s prime factor 360. 360=5222335360 = 5 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 Now by grouping these factors in various ways, we find that 2223=24,35=152 \cdot 2 \cdot 2 \cdot 3 = 24,3 \cdot 5 = 15 and 24+15=3924 + 15 = 39 So the numbers are 15 and 24, and we can express the middle term of the given trinomial, 39x39 x , as 15x+24x15 x + 24 x . Therefore, we can complete the factoring as follows: 20x2+39x+18=20x2+15x+24x+18=5x(4x+3) +6(4x+3) =(4x+3) (5x+6) 20 x ^ { 2 } + 39 x + 18 = 20 x ^ { 2 } + 15 x + 24 x + 18 = 5 x ( 4 x + 3 ) + 6 ( 4 x + 3 ) = ( 4 x + 3 ) ( 5 x + 6 ) Factor the trinomial. 20y2+23y+620 y ^ { 2 } + 23 y + 6


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