Short Answer TypeFactorise the following expressions.
p2 + 6p +8
We have   p2+ 6p + 8 = p2 + 6p + 9 - 1
                              = [(p)2 + 2(p)(3) + 32] - 1
                              = (p + 3)2 - 12  [Using a2 + 2ab + b2 = (a + b)2]
                              = [(p + 3) + 1] [(p + 3) - 1]
                              = (p + 4)(p + 2)
∴       p2 + 6p + 8 = (p + 4)(p + 2)
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