(Section 0.7: Factoring Polynomials) 0.7.6 § Solutiona) ax2+bx+c=x2+9x+20=x+5()x+4We want 5 and 4, because they have product = c= 20 and (since this is the a=1 case) sum = b= 9. We can rearrange the factors: x+4x+5. b) x2x2±±20x+100102±Guess that this is aPST for now.²³´´µ´´=x±10Check:2x±10=±20x²³´µ´2,orx±10x±10=x±102c) x2±4x±12=x±6x+2How do we know we need ±6 and +2? The constant term, c, is negative, so use opposite signs: one "+" and one "±." The middle coefficient, b, is negative, so the negative number must be higher in absolute value than the positive number; it “carries more weight.” 2-factorizations of –12 (which is c) Think: What? • What?? = –12. Sum = b= ±4? (a=1 case ) ±12, +1 No ±6, +2 Yes±Can stop ±4, +3 No d) F+O+I+L=3x2±20x±7=3x+1x±7F= First product (product of the First terms)
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