How do you find all values of k so that 2x^2+kx+122x2+kx+12 can be factored?

1 Answer
Steve M
Dec 5, 2016

k in {+-10,+-11,+-12,+-24} k∈{±10,±11,±12,±24}

Explanation:

The rule to factorise any quadratic is to find two numbers such that

"product" = x^2 " coefficient "xx" constant coefficient"product=x2 coefficient × constant coefficient
"sum" \ \ \ \ \ \ = x " coefficient"

So for 2x^2+kx+12 we seek two numbers such that

"product" = 1*12 = 24
"sum" \ \ \ \ \ \ = k

So if we looks at the factors of 24 and compute their sum we get

{: ("factor1", "factor2", "sum"),(24,1,25),(12,2,12),(6,4,10),(3,8,11),(-24,-1,-25),(-12,-2,-12),(-6,-4,-10),(-3,-8,-11) :}

Hence k in {+-10,+-11,+-12,+-24}

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