Expressed in words, this says that the sum of any two numbers multiplied by their difference equals the difference of their squares.

The examples below illustrate this:

EXAMPLE 1.

(x+8)(x-8)= x2-82=x2-64. Ans.

EXAMPLE 2. (2 xy+6) (2 xy−6) = (2 xy)2 — 62

=4x2y2-36. Ans.

EXAMPLE 3. (9x+3y) (9x-3 y) = (9 x)2 - (3 y)2

NOTE. Formula VII is a special case of Formula IV, for if in that formula we use a for x, +b for m, and -b for n, we obtain

(a+b)(a−b)=a2+(−b+b)a+(−b · b) = a2—0 · a—b2= a2—b2.

ORAL EXERCISES

Perform each of the following multiplications by inspection.

1. (a+2)(a-2). 4. (r-5)(r+5).

7. (s−7)(s+7).

2. (s+3)(s-3). 5. (k-10) (k+10). 8. (z+21)(z−21).

31. Find by Formula VII the value of 32 28. SOLUTION. 32 28=(30+2) (30-2)=302-22-900-4-896.

Ans.

32. Find by Formula VII the value of the following (see

For further exercises on this topic, see the review exercises, p. 114, and Appendix, p. 300.

67. Factoring the Difference of Two Squares. Formula VII gives a quick method of factoring the difference of two squares, as illustrated below.

EXAMPLE 1. Factor 4 x2-25 y2.

SOLUTION. 4 x2-25 y2= (2 x)2- (5 y)2

=(2x+5 y) (2 x-5 y). Ans.

EXAMPLE 2. Factor 64 m2n2-9 p2q2.

SOLUTION. 64 m2n2-9 p2q2= (8 mn)2- (3 pq)2

EXAMPLE 3. Factor 16 x4-1.

SOLUTION. 16x4 - 1 = (4 x2)2-1=(4 x2+1)(4 x2-1)

= (4x2+1) (2x+1)(2x-1). Ans.

For further exercises on this topic, see the review exercises, p. 114, and Appendix, p. 300.

EXERCISES - APPLIED PROBLEMS

1. The figure represents a large square, whose side a, and within it lies (in any

=

=

b.

manner) a smaller square whose side Show that the area between the two (shaded in the figure) is represented by the expression (a+b)(a-b).

[HINT. The shaded area = a2-b2.]

2. The result in Ex. 1 gives a quick way of determining the area between any two squares. State what it is.

3. By means of Exs. 1 and 2 answer the following question: What is the area of pavement in the street surrounding a city block one half a mile on a side, the street being 4 rods wide?

[HINT. 1 mile=320 rods. Express your answer in square rods.]

1/2 Mile

FIG. 31.

4. Show that the area between a circle of radius R and

FIG. 32.

a smaller circle of radius r lying (in any manner) within it is equal to (R+r) (R−r).

5. The diameter of a certain sunflower was 10 inches, and the black center containing the seeds (called the disk) measured 4 inches across. What was the area of the outside yellow part (called the rays)? Neglect the slight overlapping of the rays and work by Ex. 4.

6. The figure shows a right triangle whose sides are a and b and whose hypotenuse (long side) is h. Show that a2= (h+b) (h—b); also that b2 (h+a)(ha).

=

[HINT. See Ex. 20, p. 21.]

h

a

FIG. 33.

7. By means of the result in Ex. 6 answer the following: A 20-foot ladder rests against a building, the bottom of the ladder being 12 feet from the cellar wall. How far is the top from the ground?

the square of Suppose, for We first take

8. Formula VII is frequently used to find a number quickly by mental arithmetic. example, we wish to know the value of 162. 6 away from 16, giving 10, then we add 6 to the 16, giving 22. We multiply the 10 and the 22 (as is done easily mentally) giving us 220. Now, all we have to do is to add 62, or 36, to the 220 to get the value desired. This gives 256 as the answer. The reason for this appears from the following. By Formula VII, we have

Hence,

(16-6)(16+6)=162-62.

(16-6)(16+6)+62=162.

Find (mentally) in this way the values of the following: (a) 152.

[HINT. Subtract 5 from the 15, getting 10; then add 5 to it, getting 20. Multiply these and add 52.]

68. Further Study of the Equation. The following principle is useful, as we shall show presently, in the solving of equations:

PRINCIPLE. If either of two numbers is equal to zero, the product of the numbers is itself zero. For example, 2×0=0; (-4)X0=0, etc.