(Section 0.7: Factoring Polynomials) 0.7.7 e) 4x2+11x+6=4x+3()x+2Method 1: Trial-and-Error ("Guess") Method4xx2x2x±²³F=4x2+1+6+6+1+2+3+3+2±²´´³´´L= 6; need both "+" because of +11xMethod 2: Factoring by Grouping4 and 6 are neither prime nor “1,” so we may prefer this method. We want two integers whose product is ac= (4)(6) = 24 and whose sum is b= 11. We want 8 and 3; split the middle term accordingly. 4x2+11x+6=4x2+8x+3xOK to switch±²³´³+6=4x2+8x+3x+6±Group terms=4+2+3x+2±"Local factoring"=4x+3x+2±"Global factoring"f) 2x2+10x+5 is prime or irreducible over the integers (i.e., it cannot be broken down further using integer coefficients). None of these combinations work:
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