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I can either multiply the brackets out using the distributive law:
(x + 5)5 = (x + 5)(x + 5)(x + 5)(x + 5)(x + 5)
= (x2 + 5x + 5x + 52)(x + 5)(x + 5)(x + 5)
= (x2 + 10x + 25)(x + 5)(x + 5)(x + 5)
= (x3 + 15x2 + 75x + 125)(x + 5)(x + 5)
= (x4 + 20x3 + 150x2 + 500x + 625)(x + 5)
= x5 + 25x4 + 250x3 + 1250x2 + 3125x + 3125
However, this is liable to error, so I'd use the binomial theorem to go there directly:
(p + q)n = Σnr=0 (nr)pn-rqr = (n0)pn + (n1)pn-1q1 + ... + (nn-1)pqn-1 + (nn)qn
Where (nr) = nCr = n!/(n-r)!r!
Thus:
(x + 5)5 = (50)x5 + (51)x45 + (52)x352 + (53)x253 + (54)x54 + (55)55
= 1.x5 + 5.x4.5 + 10.x3.25 + 10.x2.125 + 5.x.625 + 1.3125
= x5 + 254 + 250x3 + 1250x2 + 3125x + 3125
(The dot is used to represent "multiply" as its symbol would look too much like the x.)
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0
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