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nydus/The First Steps in AlgebraPublic
Page 76 of 226
Table of Contents

Case III.

  1. 49a2b24.
  1. 25a4b49.
  1. 9a8b616x10.
  1. 144x2y21.
  1. 100x6y2z41.
  1. 1121a4b8c12.
  1. 25a264x6y6.
  1. 16x1625y18.

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)216a2.

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+3y4a.

The difference of these roots is x+3y4a.

Therefore (x+3y)216a2=(x+3y+4a)(x+3y4a).

  1. Resolve into factors a2(3b5c)2.

The square roots of the terms are a and (3b5c).

The sum of these roots is a+(3b5c), or a+3b5c.

The difference of these roots is a(3b5c), or a3b+5c.

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