Let EGC J
(Fig. 281) be a rectangle of equal sides, and within it draw the two lines, A H and F D, parallel with the lines of the rectangle, and at such a distance from them that the sides, A B and B D, of the rectangle, A B CD, shall be of equal length. We then have in this figure the three squares, E G C J, A B CD, and FGBH, also the two equal rectangles, EFAB and BHD J.
Let E Fbe represented by a and F G by b, then the area of A B C D will be a x a = a2; the area of FG B H will be bxb = b2; the area of E FA B will be a x b - a b, and that of B HD J will be the same. Putting these areas together thus -
a2+2ab + b2,
the sum equals the area of the whole figure - equals the product of EG x E C - equals the product -
(a + b) x (a + b).
So, therefore, we have -
(a + b) (a + b) r= a2 + 2ab + b2; (112.)
or, in general, the square of a binomial equals the square of the first, plus twice the first by the second, plus the square of the second. This result is obtained graphically. The same result may be obtained by algebraic multiplication, combining each factor of the multiplier with each factor of the multiplicand and adding the products, thus -
a + b a + b
a2 + a b
ab + b2
a2 + 2 a b + b2.
The same result as above shown by graphical representation.
413. - Graphical Squaring of the Difference of Two
Factors. - Let the line E C (Fig. 281) be represented by c,
and the line A E and A C as before respectively by b and
a, then -
EC-AE = AC.
c - b = a. From this, squaring both sides, we have -
(c-b)2 = a2.
The area of the square A B C D may be obtained thus: From the square E G C J take the rectangle E G x E A and the rectangle F G x D J, minus the square F G B H, or from c2 take the rectangle c b, and the rectangle c b, minus the square, b2, and the remainder will be the square, a2; or, in proper form -
c2 - c b - c b + b2 as a2
In deducting from c2 the rectangle cb twice, we have taken away the small square twice; therefore, to correct this error, we have to add the small square, or b2 Then, when reduced, the expression becomes -
c2 -2cb+b2-a2 = (c-b)2 This result is obtained graphically. The result by algebraic process will now be sought. The square of a quantity may be obtained by multiplying the quantity by itself, or -
C - b
c - b
c2 - bc -bc+b2
(c - b)2 = c2 - 2bc + b2 (113.)
In this process, as before, each factor of the multiplier is combined with each factor of the multiplicand and the several products annexed with their proper signs (Art. 415), and thus, by algebraic process, a result is obtained precisely like that obtained graphically. This result is the square of the difference of c and b; and since c and b may represent any quantities whatever, we have this general -
Rule. - The square of the difference of two quantities is equal to the sum of the squares of the two quantities, minus twice their product.
414. - Graphical Product of the Sum and Difference
of Two Quantities. - Let the rectangle A B C D (Fig. 282) have its sides each equal to a. Let the line E F be parallel with A B and at the distance b from it, also, the line F G made parallel with B D, and at the distance b from it. Then the line E F equals a + b, and the line E C equals a - b. Therefore the area of the rectangle E F C G equals a + b,
multiplied by a - b. From the figure, for the area of this rectangle, we have -
ABCD-ABEH+HFDG = EFCG
or, by substitution of the symbols,
a2 - a b + b (a - b).
Multiply the last quantity thus -
ab-b2 = b(a- b).
Substituting this in the above we have -
a2 - a b + a b - b2 = ( a + b) x (a - b).
Two of these like quantities, having contrary signs, cancel each other and disappear, reducing the expression to this -
a2 - b2 = (a + b) x (a - b).
The correctness of this result is made manifest by an inspection of the figure, in which it is seen that the rectangle EFCG is equal to the square A BCD minus the square BJHF. For ABEH equals BJDG. Now, if from the square A B CD we take away ABEH, and place it so as to cover BJDG, we shall have the rectangle EFCG plus the square BJHF-, showing that the square A BCD is equal to the rectangle EFCG plus the square BJHF; or -
a2 =(a + b) x (a - b) + b2
The last quantity may be transferred to the first member of the equation by changing its sign (Art. 403). Therefore -
a2-b2 = (a + b) x (a - b) as was before shown.
The result here obtained is derived from the geometrical figure, or graphically. " Precisely the same result may be obtained algebraically; thus -
a + b a - b
a2 +ab -ab-b2
(a + b)x (a-b) = a2... -b2
Here the two like quantities, having unlike signs, cancel each other and disappear, leaving as the result only the difference of the squares.
The result here obtained is general; hence we have this -
Rule. - The product of the sum and difference of two quantities equals the difference of their squares.