Is it possible to factor:

b^2 + [ (b(c - k)) / (d - e) ] + [ (c - k) / 2(d - e)^2 ]

I'm not sure if it factors out, but if it does, would you just put it into the quadratic formula? Thanks.

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Is it possible to factor:

b^2 + [ (b(c - k)) / (d - e) ] + [ (c - k) / 2(d - e)^2 ]

I'm not sure if it factors out, but if it does, would you just put it into the quadratic formula? Thanks.

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## Answers (1)

Let s = b^2 + [ (b(c - k)) / (d - e) ] + [ (c - k) / 2(d - e)^2 ]

Simplifying the appearance by letting d - e = p and c - k = q:

s = b^2 + (q / p)b + q / (2p^2)

The discriminant is (q / p)^2 - 2q / p^2, and that's not a perfect square.

Therefore, you use the formula.

I would stay with p and q, substituting back at the end to get factors in terms of the original letters.