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Case III.

  1. 49a2b2−4.
  1. 25a4b4−9.
  1. 9a8b6−16x10.
  1. 144x2y2−1.
  1. 100x6y2z4−1.
  1. 1−121a4b8c12.
  1. 25a2−64x6y6.
  1. 16x16−25y18.

Find, by resolving into factors, the value of:

  1. (375)2−(225)2.
  1. (579)2−(559)2.
  1. (873)2−(173)2.
  1. (101)2−(99)2.
  1. (7244)2−(7242)2.
  1. (3781)2−(219)2.

If the squares are compound expressions, the same method may be employed.

  1. Resolve into factors (x+3y)2−16a2.

The square root of the first term is x+3y.

The square root of the second term is 4a.

The sum of these roots is x+3y−4a.

The difference of these roots is x+3y−4a.

Therefore (x+3y)2−16a2=(x+3y+4a)(x+3y−4a).

  1. Resolve into factors a2−(3b−5c)2.

The square roots of the terms are a and (3b−5c).

The sum of these roots is a+(3b−5c), or a+3b−5c.

The difference of these roots is a−(3b−5c), or a−3b+5c.

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