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Compound Integral Expressions.

  1. 2x3−3x2+x and 2x2.
  1. a2−5ab−b2 and 5ab.
  1. a3+2a2b+2ab2 and a2.
  1. a3+2a2b+2ab2 and b3.
  1. 4x2−6xy−9y2 and 2x.
  1. −x2−2xy+y2 and −y.
  1. −a3−a2b2−b3 and −a2.
  1. −x2+2xy−y2 and −y2.
  1. 3a2b2−4ab3+a3b and 5a2b2.
  1. −ax2+3axy2−ay4 and −3ay2.
  1. x12−x10y3−x3y10 and x3y2.
  1. −2x3+3x2y2−2xy3 and −2x2y3.
  1. a3x2y5−a2xy4−ay3 and a7x3y5.
  1. 3a2b2−2ab3+5a3b and 5a2b3.

87. Multiplication. Polynomials by Polynomials.

If we have m+n+p to be multiplied by a+b+c, we may substitute M for the multiplier a+b+c. Then M(m+n+p)=Mm+Mn+Mp.

If now we substitute a+b+c for M, we shall have

find the product of two polynomials, therefore,

Multiply every term of the multiplicand by each term of the multiplier, and add the partial products.

In multiplying polynomials, it is a convenient arrangement to write the multiplier under the multiplicand, and place like terms of the partial products in columns.

  1. Multiply 2x−3y by 5x−4y. ccrrcr2x&−&3&y&&5x&−&4&y&&\cline1−410x2&−&15&xy&&&−&8&xy&+&12y2\hline10x2&−&23&xy&+&12y2

We multiply 2x, the first term of the multiplicand, by 5x, the first term of the multiplier, and obtain 10x2, then −3y, the second term of the multiplicand, by 5x, and obtain −15xy. The first line of partial products is 10x2−15xy. In multiplying by −4y, we obtain for a second line of partial products −8xy+12y2, which is put one place to the right, so that the like terms −15xy and −8xy may stand in the same column. We then add the coefficients of the like terms, and obtain the complete product in its simplest form.

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