Henry solved the equation –x² + x + 6 = 4 as follows:
(3 – x)(x + 2) = 4
x + 2 = 4, or 3 – x = 4
x = 2, or x = –1
Next he tackled the equation –x² – 3x + 10 = 6. He proceeded as follows:
(5 + x)(2 – x) = 6
5 + x = 6, or 2 – x = 6
x = 1, or x = –4
Again both solutions checked out.
Then he multiplied the equation by –1 to obtain a more “standard” form,
x² + 3x – 10 = –6. He applied his method again:
(x + 5)(x – 2) = –6
x + 5 = –6, or x – 2 = –6
x = –11, -4

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