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" The product of two binomials having a common term equals the square of the common term, plus the algebraic sum of the unlike terms multiplied by the common term, plus the algebraic product of the unlike terms. "
Complete School Algebra - Page 103
by Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton - 1919 - 507 pages
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A First Course in Algebra

Frederick Charles Kent - Algebra - 1913 - 292 pages
...product of x and the sum of 3 and 5. In general, IX (x + a)(x + b) = x" + (a + b)x + ab. This means that the product of two binomials having a common term equals the square of the common term plus the product of the common term and the algebraic sum of the other two terms, plus also the product of the...
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Advanced Algebra

Joseph Victor Collins - Algebra - 1913 - 360 pages
...Ans. (5 x3 + 1) (5 z 1 - 1) = ? (3 z" - 2 yp) (3 z" + 2 yp) = ? IV. O + a(jr + 6=jr 2 + ( THEOREM. The product of two binomials having a common term equals the square of the common term, and the algebraic sum of the other terms times the common term, and the algebraic product of the other...
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First Course in Algebra

William Benjamin Fite - Algebra - 1913 - 358 pages
...The product of two binomials having a common term is equal to the square of the common term, plus the sum of the unlike terms multiplied by the common term, plus the product of the unlike terms. • EXERCISES Expand the following by this formula: 1. (a; + 1)(* + 2)....
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First Course in Algebra

William Benjamin Fite - Algebra - 1913 - 304 pages
...The product of two binomials having a common term is equal to the square of the common term, plus the sum of the unlike terms multiplied by the common term, plus the product of the unlike terms. EXERCISES Expand the following by this formula : 1. (x + 1)(x + 2). 17....
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Elementary Algebra: First Course

John Charles Stone - 1914 - 380 pages
...each obtained ? In general, That is, To find the product of two binomials having a common term, take the square of the common term, plus the algebraic sum of the unlike terms times the common term, plus the algebraic product of the unlike terms. EXAMPLE 1. — (n + 2)(n +3)=...
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School Algebra: First Course

Henry Lewis Rietz, Arthur Robert Crathorne, Edson Homer Taylor - Algebra - 1915 - 304 pages
...bx + ab x2 + (a + b)x + ab от (ж + a) (ж + Ь) = хг + (a + Ь)ж + ab. In words this reads : The product of two binomials having a common term equals the square of the common term plus the product of the common term by the sum of the other terms, plus the product of the other terms. EXERCISES...
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School Algebra: First-[second], Book 1

Henry Lewis Rietz, Arthur Robert Orathorne, Edson Homer Taylor - Algebra - 1915 - 300 pages
...ax bx + ab _ x2 + (a + b)x + ab or (x + ) (x + ft) = ж 2 + (a + b)x + ab. In words this reads : The product of two binomials having a common term equals the square of the common term plus the product of the common term by the sum of the other terms, plus the product of the other terms. EXERCISES...
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Intermediate Algebra

Herbert Ellsworth Slaught, Nels Johann Lennes - Algebra - 1916 - 280 pages
...twice their product. Example 3. (x + a) (x + 6) = jr2 + (a -f- 6) x + ab. The product of two Ыnomials having a common term equals the square of the common term plus the sum of the unlike terms multiplied by the common term, plus the product of the unlike terms. Example...
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First Course in Algebra

Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton - Algebra - 1917 - 344 pages
...multiplication x +b + bx + ab gives the formula (x+ d)(x + V) = This may be expressed in words as follows : IV. The product of two binomials having a common term...the application of IV in the following : EXAMPLES l.'(x + 1) (ж + 2) = a? +(1+ 2) ж + 2 = ж2 + Зж + 2. 2. (о - 2)0 - 3) = <? + (- 2 - S) с + 6...
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Elementary Algebra

Elmer Adelbert Lyman, Albertus Darnell - Algebra - 1917 - 520 pages
...5 x + 6. ж2 +(а + &)ж + ao. This gives (дг + )(x + b) = or2 + (a + ft)дг + aft. In words : The product of two binomials having a common term equals the square of the common term, plus the product of the common term by the sum of the other terms, plus the product of the other terms. The...
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