Review of Algebra 2 REVIEW of ALGEBRA
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Review of Algebra 2 I REVIEW OF ALGEBRA Review of Algebra GGGGGGGGGGGGGGG Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real numbers have the following properties: a ϩ b b ϩ a ab ba (Commutative Law) ͑a ϩ b͒ ϩ c a ϩ ͑b ϩ c͒ ͑ab͒c a͑bc͒ (Associative Law) a͑b ϩ c͒ ab ϩ ac (Distributive law) In particular, putting a Ϫ1 in the Distributive Law, we get Ϫ͑b ϩ c͒ ͑Ϫ1͒͑b ϩ c͒ ͑Ϫ1͒b ϩ ͑Ϫ1͒c and so Ϫ͑b ϩ c͒ Ϫb Ϫ c EXAMPLE 1 (a) ͑3xy͒͑Ϫ4x͒ 3͑Ϫ4͒x 2y Ϫ12x 2y (b) 2t͑7x ϩ 2tx Ϫ 11͒ 14tx ϩ 4t 2x Ϫ 22t (c) 4 Ϫ 3͑x Ϫ 2͒ 4 Ϫ 3x ϩ 6 10 Ϫ 3x If we use the Distributive Law three times, we get ͑a ϩ b͒͑c ϩ d͒ ͑a ϩ b͒c ϩ ͑a ϩ b͒d ac ϩ bc ϩ ad ϩ bd This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have ͑a ϩ b͒͑c ϩ d͒ In the case where c a and d b, we have ͑a ϩ b͒2 a2 ϩ ba ϩ ab ϩ b2 or 1 ͑a ϩ b͒2 a2 ϩ 2ab ϩ b2 Similarly, we obtain 2 ͑a Ϫ b͒2 a2 Ϫ 2ab ϩ b2 REVIEW OF ALGEBRA N 3 EXAMPLE 2 (a) ͑2x ϩ 1͒͑3x Ϫ 5͒ 6x 2 ϩ 3x Ϫ 10x Ϫ 5 6x 2 Ϫ 7x Ϫ 5 (b) ͑x ϩ 6͒2 x 2 ϩ 12x ϩ 36 (c) 3͑x Ϫ 1͒͑4x ϩ 3͒ Ϫ 2͑x ϩ 6͒ 3͑4x 2 Ϫ x Ϫ 3͒ Ϫ 2x Ϫ 12 12x2 Ϫ 3x Ϫ 9 Ϫ 2x Ϫ 12 12x 2 Ϫ 5x Ϫ 21 Fractions To add two fractions with the same denominator, we use the Distributive Law: a c 1 1 1 a ϩ c ϩ ϫ a ϩ ϫ c ͑a ϩ c͒ b b b b b b Thus, it is true that ϩ a c a ϩ c b b b But remember to avoid the following common error: | a a ϩ a b ϩ c b c (For instance, take a b c 1 to see the error.) To add two fractions with different denominators, we use a common denominator: ϩ a ϩ c ad bc b d bd We multiply such fractions as follows: a ؒ c ac b d bd In particular, it is true that Ϫ a Ϫ a a b b Ϫb To divide two fractions, we invert and multiply: a b a ϫ d ad c b c bc d 4 I REVIEW OF ALGEBRA EXAMPLE 3 x ϩ 3 x 3 3 (a) ϩ 1 ϩ x x x x 3 x 3͑x ϩ 2͒ ϩ x͑x Ϫ 1͒ 3x ϩ 6 ϩ x 2 Ϫ x (b) ϩ x Ϫ 1 x ϩ 2 ͑x Ϫ 1͒͑x ϩ 2͒ x 2 ϩ x Ϫ 2 2 ϩ ϩ x 2x 6 x 2 ϩ x Ϫ 2 s2t ut s2t 2u s2t 2 c) ؒ Ϫ) u Ϫ2 Ϫ2u 2 x x ϩ y ϩ 1 y y x ϩ y x x͑x ϩ y͒ x 2 ϩ xy (d) ϫ y x Ϫ y y x Ϫ y y͑x Ϫ y͒ xy Ϫ y 2 1 Ϫ x x Factoring We have used the Distributive Law to expand certain algebraic expressions. We some- times need to reverse this process (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest situation occurs when the expres- sion has a common factor as follows: Expanding 3x(x-2)=3x@-6x Factoring To factor a quadratic of the form x2 ϩ bx ϩ c we note that ͑x ϩ r͒͑x ϩ s͒ x 2 ϩ ͑r ϩ s͒x ϩ rs so we need to choose numbers r and s so that r ϩ s b and rs c. EXAMPLE 4 Factor x 2 ϩ 5x Ϫ 24. SOLUTION The two integers that add to give 5 and multiply to give Ϫ24 are Ϫ3 and 8. Therefore x 2 ϩ 5x Ϫ 24 ͑x Ϫ 3͒͑x ϩ 8͒ EXAMPLE 5 Factor 2x 2 Ϫ 7x Ϫ 4. SOLUTION Even though the coefﬁcient of x2 is not 1, we can still look for factors of the form 2x ϩ r and x ϩ s, where rs Ϫ4. Experimentation reveals that 2x 2 Ϫ 7x Ϫ 4 ͑2x ϩ 1͒͑x Ϫ 4͒ Some special quadratics can be factored by using Equations 1 or 2 (from right to left) or by using the formula for a difference of squares: 3 a2 Ϫ b2 ͑a Ϫ b͒͑a ϩ b͒ REVIEW OF ALGEBRA N 5 The analogous formula for a difference of cubes is 4 a3 Ϫ b3 ͑a Ϫ b͒͑a2 ϩ ab ϩ b2͒ which you can verify by expanding the right side. For a sum of cubes we have 5 a3 ϩ b3 ͑a ϩ b͒͑a2 Ϫ ab ϩ b2͒ EXAMPLE 6 (a) x2 Ϫ 6x ϩ 9 ͑x Ϫ 3͒2 (Equation 2; a x, b 3) (b) 4x2 Ϫ 25 ͑2x Ϫ 5͒͑2x ϩ 5͒ (Equation 3; a 2x, b 5) (c) x3 ϩ 8 ͑x ϩ 2͒͑x2 Ϫ 2x ϩ 4͒ (Equation 5; a x, b 2) x 2 Ϫ 16 EXAMPLE 7 Simplify . x 2 Ϫ 2x Ϫ 8 SOLUTION Factoring numerator and denominator, we have 2 Ϫ ͑ Ϫ ͒͑ ϩ ͒ ϩ x 16 x 4 x 4 x 4 x 2 Ϫ 2x Ϫ 8 ͑x Ϫ 4͒͑x ϩ 2͒ x ϩ 2 To factor polynomials of degree 3 or more, we sometimes use the following fact. 6 The Factor Theorem If P is a polynomial and P͑b͒ 0, then x Ϫ b is a factor of P͑x͒. EXAMPLE 8 Factor x 3 Ϫ 3x 2 Ϫ 10x ϩ 24. SOLUTION Let P͑x͒ x 3 Ϫ 3x 2 Ϫ 10x ϩ 24. If P͑b͒ 0, where b is an integer, then b is a factor of 24. Thus, the possibilities for b are Ϯ1, Ϯ2, Ϯ3, Ϯ4, Ϯ6, Ϯ8, Ϯ12, and Ϯ24. We ﬁnd that P͑1͒ 12,P͑Ϫ1͒ 30 ,P͑2͒ 0 . By the Factor Theorem, x Ϫ 2 is a factor. Instead of substituting further, we use long division as follows: x 2 Ϫ x Ϫ 12 x Ϫ 2 ͒x 3 Ϫ 3x 2 Ϫ 10x ϩ 24 x 3 Ϫ 2x2 Ϫx2 Ϫ 10x Ϫx 2 ϩ 2x Ϫ 12x ϩ 24 Ϫ 12x ϩ 24 Therefore x 3 Ϫ 3x 2 Ϫ 10x ϩ 24 ͑x Ϫ 2͒͑x 2 Ϫ x Ϫ 12͒ ͑x Ϫ 2͒͑x ϩ 3͒͑x Ϫ 4͒ Completing the Square Completing the square is a useful technique for graphing parabolas or integrating rational functions. Completing the square means rewriting a quadratic ax2 ϩ bx ϩ c 6 I REVIEW OF ALGEBRA in the form a͑x ϩ p͒2 ϩ q and can be accomplished by: 1. Factoring the number a from the terms involving x. 2. Adding and subtracting the square of half the coefﬁcient of x. In general, we have b ax 2 ϩ bx ϩ c aͫx2 ϩ xͬ ϩ c a b b 2 b 2 aͫx 2 ϩ x ϩ ͩ ͪ Ϫ ͩ ͪͬ ϩ c a 2a 2a b 2 b2 aͩx ϩ ͪ ϩ ͩc Ϫ ͪ 2a 4a EXAMPLE 9 Rewrite x 2 ϩ x ϩ 1 by completing the square. 1 SOLUTION The square of half the coefﬁcient of x is 4. Thus 2 ϩ ϩ 2 ϩ ϩ 1 Ϫ 1 ϩ ϩ 1 2 ϩ 3 x x 1 x x 4 4 1 (x 2) 4 EXAMPLE 10 2x 2 Ϫ 12x ϩ 11 2͓x 2 Ϫ 6x͔ ϩ 11 2͓x 2 Ϫ 6x ϩ 9 Ϫ 9͔ ϩ 11 2͓͑x Ϫ 3͒2 Ϫ 9͔ ϩ 11 2͑x Ϫ 3͒2 Ϫ 7 Quadratic Formula By completing the square as above we can obtain the following formula for the roots of a quadratic equation. 7 The Quadratic Formula The roots of the quadratic equation ax 2 ϩ bx ϩ c 0 are Ϫb Ϯ sb2 Ϫ 4ac x 2a EXAMPLE 11 Solve the equation 5x 2 ϩ 3x Ϫ 3 0. SOLUTION With a 5,b 3 ,c Ϫ3 , the quadratic formula gives the solutions Ϫ3 Ϯ s32 Ϫ 4͑5͒͑Ϫ3͒ Ϫ3 Ϯ s69 x 2͑5͒ 10 The quantity b2 Ϫ 4ac that appears in the quadratic formula is called the discriminant. There are three possibilities: 1. If b2 Ϫ 4ac Ͼ 0, the equation has two real roots. 2. If b2 Ϫ 4ac 0, the roots are equal. 3. If b2 Ϫ 4ac Ͻ 0, the equation has no real root. (The roots are complex.) REVIEW OF ALGEBRA N 7 These three cases correspond to the fact that the number of times the parabola y ax 2 ϩ bx ϩ c crosses the x-axis is 2, 1, or 0 (see Figure 1). In case (3) the quad- ratic ax 2 ϩ bx ϩ c can’t be factored and is called irreducible. y y y 0 x 0 x 0 x FIGURE 1 Possible graphs of y=ax@+bx+c (a) b@-4ac>0 (b) b@-4ac=0 (c) b@-4ac<0 EXAMPLE 12 The quadratic x 2 ϩ x ϩ 2 is irreducible because its discriminant is negative: b2 Ϫ 4ac 12 Ϫ 4͑1͒͑2͒ Ϫ7 Ͻ 0 Therefore, it is impossible to factor x 2 ϩ x ϩ 2.