Adição e Fatoração de Frações e Polinômios

To add two fractions with the same denominator we use the Distributive Law a c b a b c b But remember to avoid the following common error a b c a b a c EXAMPLE 7 Simplify x² 16 x² 2x 8 SOLUTION Factoring numerator and denominator we have x² 16 x² 2x 8 x 4x 4 x 4x 2 x 4 x 2 To factor polynomials of degree 3 or more we sometimes use the following fact 6 The Factor Theorem If Px is a polynomial and Pb 0 then x b is a factor of Px EXAMPLE 8 Factor x³ 3x² 10x 24 SOLUTION Let Px x³ 3x² 10x 24 If Pb 0 where b is an integer then b is a factor of 24 Thus the possibilities for b are 1 2 3 4 6 8 12 and 24 We find that P1 12 P1 30 P2 0 By the Factor Theorem x 2 is a factor Instead of substituting further we use long division as follows x² x 12 x 2 x³ 3x² 10x 24 x³ 2x² x² 10x x² 2x 12 12 x 2x 3x 4 Therefore x³ 3x² 10x 24 x 2x 3x 4 EXAMPLE 11 Solve the equation 5x² 3x 3 0 SOLUTION With a 5 b 3 c 3 the quadratic formula gives the solutions x 3 3² 453 25 3 69 10 The quantity b² 4ac that appears in the quadratic formula is called the discriminant There are three possibilities 1 If b² 4ac 0 the equation has two real roots 2 If b² 4ac 0 the roots are equal 3 If b² 4ac 0 the equation has no real root The roots are complex EXAMPLE 14 a 18 2 18 2 9 3 b x²y x²y xy Notice that x² x because indicates the positive square root See Absolute Value In general if n is a positive integer x a means xⁿ a If n is even then a 0 and x 0 Thus 8 2 because 2² 8 but 8 and 8 are not defined The following rules are valid ab ab EXAMPLE 15 x⁴ x³x x²x xx To rationalize a numerator or denominator that contains an expression such as a b we multiply both the numerator and the denominator by the conjugate radical a b Then we can take advantage of the formula for a difference of squares a ba b a² b² a b EXAMPLE 16 Rationalize the numerator in the expression x 4 2 x SOLUTION We multiply the numerator and the denominator by the conjugate radical x 4 2 x 4 2 x x 4 2 xx 4 2 x 4 2 x 4 4 xx 4 2 x xx 4 2 1 x 4 2 Notice that we have included the endpoints 2 and 3 because we are looking for values of x such that the product is either negative or zero The solution is illustrated in Figure 3 Recall that the symbol means the positive square root of Thus f s means s² r and s 0 Therefore the equation a² a is not always true It is true only when a 0 If a 0 then a 0 so we have a² a In view of 12 we then have the equation a² a which is true for all values of a