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22. (ab+cd)? 26. (r+2). 30. (m- n). 23. (2 ab- 3 cd)? 27. (r->)2. 31. (2 p+ 9). 24. (x+?). 28. (a + b)2 32. (3 m-in)? 25. (x – 3)

29. (x - 1 y)2 For further exercises on this topic see the review exercises, p. 114, and Appendix, p. 299.

62. Geometric Meaning of Formulas V and VI. The expression (a+b)2 means the area of the square whose side is (a+b), and this area may be broken up into four parts (shaded in Fig. 27). Now, one of these parts (lower right-hand corner) is seen to be a square whose side is a, and hence this part has an area equal to a?. Similarly another part has an area of 62, while the remaining two parts are rectangles each having the area ab. The sum of the four parts is therefore, a2+62+ab+ab, which may be written a2+2ab+62. Thus, the figure illustrates how (a + b)2 = a2+2 ab + b2; that is, it illustrates Formula V.

While it is not so easy to illustrate the meaning of Formula VI, a careful examination of Fig. 28 will show that it does so. Note that the part (ab)2 is equal to the area of the whole square, or a, minus the area of the shaded border. But the shaded border is equal to ab+ab-b2, or 2 ab-62. Thus (ab)2 = 92—(2 ab-62), which reduces to a2–2 ab+b?, or Formula VI.

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63. Trinomials That Are Perfect Squares. We have seen in $8 59, 60 that the square of a binomial is always a trinomial. Thus, (2 x+y)?= (2x)2+2(2 x • y)+ya=4x2+4 xy+y.

(Formula V) (2—3 y)2=X2 — 2(x • 3 y)+(3 y)2=X2 – 6 xy+9 ya.

(Formula VI) A trinomial obtained in this way is called a trinomial square, or a perfect square. It must be carefully observed that not all trinomials are perfect squares. The essentials for this are that two of the terms must be squares and positive, while the remaining term (usually the middle one) must be equal to twice the product of the square roots of the square terms. This will now be illustrated by means of several examples.

EXAMPLE 1. x2 +6 x+9 is a perfect square because its terms x2 and 9 are squares and positive, while its remaining term, 6 x, is equal to 2 : Vx2.79. Moreover, since the sign of the term 6 x in the given trinomial is +, we know by Formula V that we here have the square of the sum of two numbers. In fact, what we have is (x+3)2.

EXAMPLE 2. m2—4 mn +4 na is a perfect square because its terms ma and 4 na are squares and positive, while its remaining term, 4 mn, is equal to 2. V m2 . V4 n?. Moreover, since the sign of 4 mn in the given trinomial is –, we know from Formula VI that we here have the square of the difference of two numbers. It is, in fact, (m-2 n)a.

EXAMPLE 3. 9x2 – 10 x+4 is not a perfect square since the term 10 x does not equal 2 .V9 x2 . V4.

EXAMPLE 4. 4 x2 – 12 x– 9 is not a perfect square since its last term is negative instead of positive.

ORAL EXERCISES In each of the following exercises tell whether the given trinomial is a perfect square, and if so whether it is the square of the sum or of the difference of two numbers.

1. x2+2 xy+y?. 3. 62— 146+49. 5. ma+8-4m. 2. a?–6 a+9. 4. p2+10 rs-52. 6. 2— 16t+64. 7. 4x2 – 8 x +4.

14. a2b2 2 ab+4. 8. x2–2 xyy?.

15. 121 a x2 – 22 ax+1. 9. 16 m2+12 mn +9 n?.' 16. 4 r2+1+44 r. 10. a?+2 a+1.

17. 8 p2 – 16 r+4. 11. 962 +66+3.

18. 144 x2 – 24 x+1. 12. 25 22 – 10 2-1. . 19. 81 x2+36 x+4. 13. 36 a2+12 ab+b2. 20. 14 a2+28 ab+962.

WRITTEN EXERCISES Supply the missing term in each of the following expressions so as to make a perfect square of the trinomial.

1. x2+6 x+( ).

Solution. Since we may write 6 x =2. Vx2. V9, we see that the missing term must be 9. Ans.

CHECK. With 9 supplied for the missing term we find (as in $ 63) that the given trinomial is a perfect square. • 2. a2 +4 a + ( ).

3. 1-2 a+( ). .

4. a2b2+1 ) +1. SOLUTION. The missing term must be twice the product of the square roots of the other terms; that is, it must be 2. Va2b2 . Vi which reduces to 2 ab. Ans. 5. ax2 - ( ) +25.

8. 9 72 — 12 r+( ). 6. ( )-6 pq+4 q?.

9. 16 x2 – ( ) +4. 7. ()-2 xy+ya.

10. ( )-28 s+4 s.

11. x2+( )+1.
12. 100 x2+( )+4.
13. 9 c” – 6 c+( ).
14. 4b2+( )+9c2.
15. ma+( ) +121.
16. (a+b)2+2(a+b)+( ).

17. 452 – ( )+}. 18. 36 2–( )+36., 19. 992—( )+81. 20. a? — 10 a+( ). 21. 4x2 – ( ) +25 22. 22. 122+( )+25 2.

64. Factoring a Trinomial Square. We have seen in $ 63 how we may tell a trinomial square when we see it and tell also whether it is the square of the sum or of the difference of two numbers. We may easily go a step farther and tell just what binomial it is the square of; that is, we may find its two equal binomial factors. We will now illustrate this by examples.

EXAMPLE 1. Find the factors of x2 +10 x+25.

SOLUTION. The square roots of x2 and 25 are x and 5 respectively, while the sign before the remaining term, 10 x, is +. Hence we may write

x2+10 x+25= (x+5)(x+5) Ans. EXAMPLE 2. Find the factors of 4 x2 – 16 x+16.

Solution. The square roots of 4 x2 and 16 are 2 x and 4 respectively, while the sign before the remaining term, 16 x, is Hence we may write

4 x2–16 x+16=(2 x—4)(2 x—4) Ans. From these examples we obtain the following general rule.

RULE FOR FACTORING A TRINOMIAL SQUARE. Find the square roots of the square terms and connect them by the sign of the remaining term.

Note. Finding the two equal factors of a trinomial square in this way is the same as finding its square root. Thus, the answer to Example 1 above means that x+5 is the square root of x2+10 x+25. See $ 14.

WRITTEN EXERCISES In the following list, give the factors of those trinomials that are perfect squares; also state what is the square root of each. (See Note in § 64.)

1. x2 — 2 x+1. 5. a?+6 a+9. 9. 12+12r+36. 2. x2+2 x+1. 6. a?—8 a+16. 10. 9–6 ata?. 3. x2 +4 x+4. 7. a? +10 a +25. 11. 452 +4 st1. 4. x2 —4 x+4. 8. 22–12 r+36. 12. 81 – 72 s+16 s. 13. 100—20 y+y?.

19. 225 x2y2 – 30 xy+4. 14. 81 x2+18 x+1.

20. 16–8 x+x2. 15. 144 m2+72 r+9.

21. 22+32 zy+256 ya. 16. 144 m2 — 72 r+1.

22. 49 k2+28 kp+4 p. 17. 100-a2+84 a +25. 23. 25 m2 — 20 mn+4 n?. 18. 121 52–88 8+16. 24. 4x2 — 2 xy-1 y.

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

65. The Product of the Sum and Difference of Two Numbers. Suppose we have any two numbers, a and b, and let us form their sum, a+b, and their difference, a-b. The product of this sum and difference, when multiplied out as in § 53, looks as follows:

a + b
a 6
a2+ab

- ab-62

a2+0 62 or simply

Q? — 62.
Thus we have the following formula :
Formula VII. (a+b)(a-6)=a? — 62.

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