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What is the correct factorised form of (k^2-15k+56)?

Answer and explanation

Correct answer: (k-7)(k-8)

The direct answer is A: \((k-7)(k-8)\). For a monic quadratic \(k^2+pk+q\), find two numbers whose sum is \(p\) and whose product is \(q\). Here the required sum is \(-15\) and the required product is \(56\). The numbers \(-7\) and \(-8\) work because \((-7)+(-8)=-15\) and \((-7)(-8)=56\). Therefore \(k^2-15k+56=(k-7)(k-8)\). Multiplying A confirms it: \(k^2-8k-7k+56=k^2-15k+56\). B gives \(k^2+15k+56\), so its middle sign is wrong. C gives \(k^2-18k+56\), so the middle coefficient is -18 rather than -15. D gives \(k^2-57k+56\), so its middle coefficient is also wrong. The positive constant 56 means the two numbers have the same sign; because their sum is negative, both must be negative. Memory cue: for a monic quadratic, check the pair by both sum and product, not by product alone.

Related tags

Quadratic FactorisationInteger PairsSignsFactorisationExploring Algebraic IdentitiesMathematicsClass 9 Mcq

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