By Bernard Kolman; Arnold Shapiro

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**Example text**

The term containing x can be found by adding the product of the "inners" and the product of the "outers" : (2x 3)(5x - 2) I �x I + + (x - 1)(2x + 3) - l)(3x + 3) -2x - p(2-3x + 2x2 r oox3x +r +x (x + 1) ( 2 x + 3) 2x2 x -3. (2x 2)(2x -2) (�x +4x22)(2r -2) (2x m4xx - (2x + t)(2-4x -4x +Ox (2x + 2)(2x -2) 4x2 - 4. Thus, (2x 3)(5x - 2) Sum = 10x2 = -4x l lx I Ix - 6. EXAMPLE 4 Multiply. (a) We diagram the process so that you can learn to do these mentally. ) Sum = = Thus, (b) + Once more, we have + ) Sum = Thus, ((2xx+-3)2)(x(3x+ - = PROGRESS CHECK 4 Multiply mentally.

Let's work through this problem: (2x + 3)(5x - 2) = 2x(5x - 2) + 3(5x - 2) (2x)(5x) + 2x( - 2) + 3(5x) + 3( - 2) = l 0x 2 - 4x + 15x 6 = - 53 54 POLYNOMIALS We have stopped just short of the last step because we want to show the relationships between the factors and the products. If we take the product of the first term of each expression (2x + 3)(5 x - 2) 10x2 + we have the term containing x2• Similarly, taking the product of the last term of each expression (2x + 3)(5x - 2) -6 we have the constant term.

A) x < 0 Answers (a) (c) • • (b) x � - I I I + I + I I I -2 - 1 -2 - 1 0 0 2 2 (c) x < - 2 (b) I + I -2 - 1 0 I 2 • 31 32 THE REAL NUMBER SYSTEM We also write double inequalities such as - l ::; x < 2 The solution set to this inequality consists of all real numbers which satisfy - 1 ::; x and x < 2 that is, all numbers between - 1 and 2 and including - 1 itself. We can easily graph the solution set on a real number line. -5 I -4 I -3 + -2 I -1 0 -1 EXAMPLE 6 Graph -3 < x < - 1, 5 x 2 I 3 4 5 I 3 4 I <2 a real number.