1.
Complete: \((k-\square)(k-5)=k^2-\square k+135\)
Solution: Since the constant term is \(135\) and one factor is \((k-5)\), the other number is \(135\div 5=27\). Therefore \((k-27)(k-5)=k^2-32k+135\).
2.
Factor: \(-5m^2+20m+60\)
Solution: Factor out \(-5\): \(-5(m^2-4m-12)\). Since \((-6)(2)=-12\) and \((-6)+2=-4\), the factorization is \(-5(m-6)(m+2)\).
3.
Factor: \(-4d^2-28d+240\)
Solution: Factor out \(-4\): \(-4(d^2+7d-60)\). Since \(12(-5)=-60\) and \(12+(-5)=7\), the factorization is \(-4(d-5)(d+12)\).
4.
Factor: \(-3b^2+15b+18\)
Solution: Factor out \(-3\): \(-3(b^2-5b-6)\). Since \((-6)(1)=-6\) and \((-6)+1=-5\), the factorization is \(-3(b-6)(b+1)\).
5.
Factor completely: \(2x^2-16xy+30y^2\)
Solution: Factor out \(2\): \(2(x^2-8xy+15y^2)\). Since \((-5y)(-3y)=15y^2\) and \(-5y-3y=-8y\), the factorization is \(2(x-5y)(x-3y)\).
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