This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1916 edition. Excerpt: ...that gives ua the factors (x--2) (x--7). 146. Model D.--In the product x2-5 x-24, the straight products are x2 and--24;--5 x is the sum of the cross products. The terms giving x2 for a product had like signs and the terms giving--24 for a product had unlike signs; then the two cross products had unlike ...
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1916 edition. Excerpt: ...that gives ua the factors (x--2) (x--7). 146. Model D.--In the product x2-5 x-24, the straight products are x2 and--24;--5 x is the sum of the cross products. The terms giving x2 for a product had like signs and the terms giving--24 for a product had unlike signs; then the two cross products had unlike signs, and the minus cross product was the greater. To get x2 for a straight product, we should have to multiply x and x. To get 24, we could multiply 1 and 24, 2 and 12, 3 and 8, In each case, the larger number would have to or 4 and 6. be negative. Trying each in succession, we find that the pair of factors (x + 3) (x-8) gives the right cross products, so we need not try farther. EXERCISE 47 Name the straight products and the sum of the cross products in the following expressions, and find their factors: 1. x2 + 5z + 6 e. z2+ 5 z+ 4 2. z2-2 x + 1 7. z2 + 7 x + 12 a. x2-3 x + 2 8. z2-7 x + 6 4. x2-4 x + 3 9. z2 + 7 x + 10 5. s? + 4 x + 4 10. x2-8 x + 13 EXERCISE 48 Factor: 1. (x-18)2-49 7. (A + 90)2-36 2. (x-20)2-121 8. (n + 101)2-625 3. (a-32)2-100 9. (25-x)2-25 4. (2 x-5)2-144 10. (x-38)2-225 5. (x-35)2-81 11. (3 x + 51)2-144 e. (p + 28)2-900 12. (m-59)2-400 148. In factoring an expression like x2 + 56 x + 768, an easier method than finding all pairs of factors of 768 and selecting the right pair is to reduce the expression to the form of the difference of two squares. We may then apply Theorem A to it, as in Exercise 48. Taking the first two terms, and remembering Theorem 1 of Section 141, we can say that x2 + 56 x is a part of a perfect square, the square of some binomial. Since x2 is the square of the first term of the binomial, then the first term of the binomial must be x. Again, 56 x is twice the product of both terms, so 28 x is...
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