Precalculus Review Booklet - Topics 8-13

Topics • (8) Solving for x (keeping equations balanced) • (9) Factoring and distributing; difference of two squares • (10) Completing the square; quadratic formula • (11) The Fundamental Theorem of Algebra • (12) Difference of two cubes; long division • (13) One cannot divide a quantity by zero. VIII. Solving for x (keeping equations balanced) An equation is a statement of the equality of [at least] two expressions. To maintain an equation’s equality, one must be certain to treat one side of an equation as one does the other side when manipulating either in a manner that changes its value. Consider the manipulations one might make to one side of an equation, usually in an effort to isolate a variable or to isolate all terms containing that variable – and scalar terms, which contain x0: adding a quantity, subtracting a quantity, multiplying by a quantity, dividing by a quantity, squaring, cubing, finding the square root, finding the cube root, making it the exponent of a quantity, finding its logarithm with respect to a particular positive real base, etc. One tries to isolate a variable (say x), or all terms containing x – including scalar terms, to solve for the value or values of x that satisfy a given equation. For example, assume we are given the equation 4x3 + 15x2 + 3x + 17 = 2x3 + 7x2 + 5x – 10. We have a cubic expression, which may take any real value from negative to positive infinity provided x may take any such value, set equal to another cubic expression. Though we may allow x to be any real number, only three x-values in all of the complex numbers – including at least one real number (see section XXVII) – will make this equation true. Let us manipulate our given equation, carefully holding it in balance, to present this cubic equation in a standard form. 4x3 + 15x2 + 3x + 17 – 2x3 – 7x2 – 5x + 10 2x3 + 8x2 – 2x + 27 = 2x3 + 7x2 + 5x – 10 – 2x3 – 7x2 – 5x + 10 = 0 + 0 + 0 + 0 This standard form is such that one side of the equation equals zero and the leading coefficient on the other side is positive. This standard form (call it standard-form1) does not require that the lead coefficient (the coefficient of the term with highest exponent) equal 1, but one may encounter a case in which one desires such a lead coefficient. To unitize the lead coefficient of the left side of 2x3 + 8x2 – 2x + 27 = 0, we multiply both sides of the equation by ½, the reciprocal of the current lead coefficient. The right side of the equation remains zero, but the left side becomes x3 + 4x2 – x + 27/2. (Let us say that this equation is now in standardform2.) Note that the scalar term is now not an integer. Non-integral coefficients and scalars make mental or manual calculation a bit more difficult. Provided one began working with a polynomial equation expressed with only integral coefficients, if one foresees the need to test xvalues, one may prefer to present an equation in standard-form1 rather than standard-form2. To facilitate the testing of x-values – or the application of a formula to the same end, which is finding a solution or solutions – is a primary reason to standardize the form of a polynomial equation. (Testing integral x-values with standard-form1 of the equation in our example shows that a real number between -4 and -5 satisfies the equation.) One also standardizes to facilitate 2 the distinction of one polynomial equation from others of the same degree (i.e., with the same highest exponent). The latter reason favors standard-form2. IX. Factoring and distributing; difference of two squares Assume we are asked to find the values of x which satisfy the equation x2 – 3x + 2 = 0. Although trial and error likely will lead to the correct solutions quickly, not all equations are as amenable to trial and error. Factoring the expression x2 – 3x + 2 is the quickest way to find the solutions and will allow you to be certain you have found all of the solutions. x2 – 3x + 2 = (x–2)(x–1). Therefore, if x = 2 or x = 1, the expression will equal 0 and the equation will be satisfied. To test that (x–2)(x–1) = x2 –3x + 2, we need to distribute (x–2) over (x–1) [or (x–1) over (x–2)]. That is, we need to multiply every term occurring in (x–2) by every term occurring in (x–1). We have each of two terms to multiply by each of two other terms; we should end with two times two = four products. When distributing a binomial over another binomial, as we are in the case of (x–2)(x–1), we can use the F.O.I.L. technique. F.O.I.L. is an acronym composed of the four initial letters of first, outer, inner, and last. In this case, first = x*x = x2, outer = x*(-1) = -x, inner = (-2)*x = -2x, and last = (-2)(-1) = 2. First + outer + inner + last = x2 –x –2x + 2 = x2 – 3x + 2. When trying to factor a quadratic expression; i.e., an expression in which only positive integral exponents occur and in which the greatest exponent is 2; think about unF.O.I.L.ing the expression. Factoring a quadratic expression whose lead coefficient is 1 requires one to seek two numbers which multiply to equal the scalar – in our example, 2 – and which sum to equal the coefficient of the x-term – in this case, -3. The integers -2 and -1 satisfy these criteria. For many quadratics, though, one will not be able to find two integers or two rational numbers which satisfy these criteria. In these cases, the equation has either irrational or complex roots. (Roots are the x-values at which an expression in x equals zero or an equation in x is satisfied.) The quadratic formula, which we will derive in section III, will help one to find complex roots. Now assume we are asked to find both of the roots of x2 – 121 = 0. The left side of this equation is an example of a difference of two squares. (121 is the square of 11.) Therefore, we can employ the following factorization process: x2 – y2 = (x–y)(x+y). (F.O.I.L. (x–y)(x+y) to test whether this is true or not.) In this case, we have that y2 = 112 =121. Then the factorization of the expression x2 – 121 is (x–11)(x+11). Now we see easily that the roots of (x–11)(x+11) = 0 are 11 and -11. Consider the equation x2 – 43 = 0. Despite that 43 is not the square of an integer, we still may factor x2 – 43 as (x–43½)(x+43½). 43 is the square of the irrational numbers 43½ and -43½. X. Completing the square; the quadratic formula Let’s consider the equation x2 + 10x + 24 = 0. If we add 1 to each side of the equation x2 + 10x + 24 = 0, we will have x2 + 10x + 25 = 1. The left side of the equation is now equal to (x+5)(x+5) = (x+5)2. We have completed the square on the left side of the equation. Our equation now reads (x+5)2 = 1. Subtract 1 from each side to arrive at (x+5)2 – 1 = 0. Now we may employ the formula for a difference of two squares. Employing that formula, we get (x+5–1)(x+5+1) = 0; that is, (x+4)(x+6) = 0. We could have factored x2 + 10x + 24 by looking at numbers whose product is 24 and whose sum is 10. 6 and 4 match these criteria. Solving the 3 equation by completing the square, though, offered us an opportunity to begin to work with a powerful technique. The power of completing the square is captured in the quadratic formula. Let’s derive the quadratic formula. The general form of a quadratic equation is ax2 + bx + c = 0 in which a0, b, c are real numbers. We must stipulate that a0 because if a = 0, the equation would not be quadratic. (The equation would be bx + c = 0, which is a line.) We begin by dividing both sides of the equation by a. We get x2 + (b/a)x + (c/a) = 0. To complete the square, we must add a number to each side of the equation. What is this number? We know from having used the F.O.I.L. technique that the completed square’s general appearance is (x+y)2 in which y is ½ of the coefficient of the x-term. In x2 + (b/a)x + (c/a), the x-term’s coefficient is b/a. Therefore, y = ½(b/a) = b/2a. Let’s F.O.I.L. (x+(b/2a))2. We get x2 + (b/a)x + (b2/4a2). Therefore, we know that to complete the square we must add [(-c/a) + (b2/4a2)] to each side of the equation x2 + (b/a)x + (c/a) = 0. Adding [(-c/a) + (b2/4a2)] to each side, we get x2 + (b/a)x + (b2/4a2) = (-c/a) + (b2/4a2). The left side of the equation is now a square. We have (x+(b/2a))2 = [(-c/a) + (b2/4a2)]. On the right side of the equation, we have the sum of two fractions. Let’s find a common denominator and express this sum as one fraction (with a bipartite numerator). One of the fractions has a as its denominator and the other has 4a2 as its denominator. If we multiply both the numerator and denominator of (-c/a) by 4a, we will have a common denominator of 4a2. Expressed with common denominators, [(-c/a) + (b2/4a2)] = [(-4ac/4a2) + (b2/4a2)] = (b2 – 4ac)/(4a2). Now subtract (b2 – 4ac)/(4a2) from each side of the equation. We get (x+(b/2a))2 – [(b2 – 4ac)/(4a2)] = 0. If we regard (b2 – 4ac)/(4a2) as a square (any real number can be regarded as a square) we have a difference of two squares. (b2 – 4ac)/(4a2) is the square of (b2 – 4ac)½/2a. Our formula for the difference of two squares gives us (x + (b/2a) – [(b2 – 4ac)½/2a])(x + (b/2a) + [(b2 – 4ac)½/2a]) = 0. Observing that (b/2a) and [(b2 – 4ac)½/2a] share a denominator, we rewrite the equation at the end of the last sentence as (x + ([b – (b2 – 4ac)½]/2a))(x + ([b + (b2 – 4ac)½]/2a)) = 0. The roots of this equation are therefore x = -[b – (b2 – 4ac)½]/2a and x = -[b + (b2 – 4ac)½]/2a. If we bring the outer negative signs within the expressions, we get x = [-b + (b2 – 4ac)½]/2a and x = [-b – (b2 – 4ac)½]/2a. Employing a bit of shorthand, we may rewrite the roots as x = [-b  (b2 – 4ac)½]/2a. This is the quadratic formula. It can find the roots of any quadratic equation. (Note that if (b2 – 4ac) < 0, both roots will be non-real, complex numbers.) XI. The Fundamental Theorem of Algebra We established the quadratic formula by finding the roots of the general quadratic equation: ax2 + bx +c = 0 (a0, b, c are real numbers). The general cubic equation is a3x3 + a2x2 + a1x + a0 = 0 (an  and a30) and the general quartic equation is a4x4+ a3x3 + a2x2 + a1x + a0 = 0 (an  and a40). Every term of these equations is the product of a real coefficient and xn for some positive integer n. The standard presentation of a general equation in which n is the greatest exponent is anxn + an-1xn-1 + an-2xn-2 + … + a2x2 + a1x + a0 = 0 (all coefficients , all exponents , an0). The Fundamental Theorem of Algebra states that this equation has n roots among the set of all complex numbers. Note that the set of real numbers is a proper subset of the complex numbers [proper indicates that the subset excludes some member or members of the parent set]. Specifically, the real numbers are the numbers a+0i. Quadratic equations have two roots, cubic equations have three roots, quartic equations have four roots, quintic equations have five roots, etc. 4 XII. Difference of two cubes; long division Let’s factor x3 – 8. Since 8 is the cube of 2, x3 – 8 is an example of a difference of two cubes. A difference of two cubes, like a difference of two squares, has a nice factorization. Let’s find it using long division. Long division may be employed with polynomial expressions; i.e., expressions such as quadratics, cubics, quartics, etc.; as it is employed with numbers. Since x = 2 is a root of x3 – 8, we know that x – 2 is factor of x3 – 8. Let’s divide x3 – 8 by x – 2. We begin by looking at the lead terms of the divisor and the expression to be divided. We ask: what multiplied by x equals x3? The answer is x2. Write x2 above x3 – 8. Multiply x – 2 by x2 to arrive at the expression x3 – 2x2. Now write x3 – 2x2 below x3 – 8 and subtract the former from the latter to arrive at 2x2 – 8. Now we ask: what multiplied by x equals 2x2? The answer is 2x. Write +2x above x3 – 8 after x2. Multiply x – 2 by 2x to arrive at 2x2 – 4x. Write 2x2 – 4x beneath 2x2 – 8 and subtract the former from the latter to arrive at 4x – 8. What times x equals 4x? The answer is 4. Write +4 above x3 – 8 after 2x. Multiply x – 2 by 4 to arrive at 4x – 8. Writing 4x – 8 beneath 4x – 8 and subtracting former from latter yields zero. Our long division is done. x2 x - 2 + 2x + 4 x3 - 8 bring down - (x3 - 2x2) 2x2 - 8 bring down - (2x2 - 4x) 4x - 8 (4x - 8) 0 We have factored x3 – 8 into (x – 2)(x2 + 2x + 4). Notice that in the expression x2 + 2x + 4, the x-term’s coefficient is 2 and the scalar term is 4 = 22. (The general formula for the difference of two cubes is x3 – y3 = (x – y)(x2 + yx + y2).) Then, to the expression x2 + 2x + 4 we may apply the quadratic formula. Note first that (b2 – 4ac) [the determinant] = 22 – (4*1*4) = 4 – 16 = -12. Therefore, the roots of x2 + 2x + 4 = 0 are non-real, complex numbers. The quadratic formula gives us that the roots are x = (-2  (-12)½)/2 = (-2  12½i)/2 = (-2  (22*3)½i)/2 = (-2  2*3½i)/2 = -1  3½i. Now we have the three roots of x3 – 8 = 0: x = 2, (-1 + 3½i), and (-1 – 3½i). Factor the following equations over the real numbers (i.e., do not factor them into expressions containing imaginary numbers): 1. x3 + 64 = 0 2. x5 + 512x2 = 0 3. x3 + x2 + x – 3 = 0 4. x6 – y6 = 0. 5 XIII. One cannot divide a quantity by zero. What is 1 divided by 0? This question cannot be answered simply. To begin to see this, divide 1 by very small numbers. For example, 1 divided by .0001 equals 10,000. 1 divided by .000001 equals 1,000,000. As we approach zero, the result grows large and heads off to infinity. Infinity is not a number and is not a satisfactory answer to 1/0. However, to say 1/0 is infinity is to offer more information than to say one cannot divide by zero. The situation is a bit more complex, though. In our consideration of tiny divisors, we did not try any negative divisors. Approaching zero from its negative side is necessary to provide us with a full picture of the complexity involved in division by zero. The approach from the negative side has a similar result: negative infinity. One might say that 1/0 is both negatively and positively infinite. Now we have two problems: 1/0 has two answers of infinite magnitude and infinity is not a number. That infinity is not a number is not problematic in circumstances in which one expects or is prepared to receive an infinite answer. Such a circumstance occurs in calculus, for example, when one examines the one-sided limit behavior of a quotient of real-valued functions. Onesided approaches might be necessary to limit the number of possible distinct results to one. Also, that there are many kinds of infinity; e.g., the infinity of the integers and the infinity of the real continuum; necessitates that one specify a number system through which to shrink a denominator to zero. In calculus, the specified number system is always the real numbers or the complex numbers (whose structure is that of two-dimensional real space). In a general algebraic setting, one works without a limit-examining context and, therefore, without such careful restrictions. In the general algebraic setting, one cannot divide by zero. The following deductive sequence gives an example of the problems that can follow from dividing a quantity by zero. (I excerpt the following from Steve Hagen’s book How the World Can Be the Way It Is. Steve Hagen excerpted it from William Poundstone’s Labyrinths of Reason.) 1. Let x = 1 2. Then obviously: x = x 3. Square both sides: x2 = x2 4. Subtract x2 from both sides: x2 – x2 = x2 – x2 5. Factor both sides: x(x – x) = (x + x)(x – x) 6. Factor out the (x – x): x = (x + x) [One cannot factor out (x – x) = 0.] 7. Or: x = 2x 8. And since x = 1: 1 = 2 6 Homework Problem Set 2 Topics • (8) Solving for x (keeping equations balanced) • (9) Factoring and distributing; difference of two squares • (10) Completing the square; quadratic formula • (11) The Fundamental Theorem of Algebra • (12) Difference of two cubes; long division • (13) One cannot divide a quantity by zero. 8a-l. For each of the following, find at least one satisfying x-value: a. x – 4 = 15 e. 2x = 64 i. x2 – 5x + 9 = 3 b. x – 15 = 4 f. log10x = 5 j. 3x3 + 99 = 18 c. x2 – 9 = 16 d. 4x3 – 32 = 0 8m. g. 6x4/3 = 96 h. 2x2/9 = 8 k. x4 – 256 = 0 l. a2x2 = 49 Present the following quintic equation in standard-form1 and standard-form2: 5x5 + 7x4 – 12x3 – 9x2 + 3x – 6 = 8x5 – 3x4 + 5x3 + 16x2 – 10x + 23 9a-d. Factor the following quadratic expressions: a. x2 – 81 b. 15x2 – 105 c. x2 – 10x + 21 d. x2 + 8x – 33 10a-d. Complete the square to find the real roots of the following quadratic equations: a. x2 – 4x – 12 = 0 b. x2 – 14x + 24 = 0 c. x2 + 32x + 31 = 0 d. (x+7)(x+3) = 77 10e-h. Employ the quadratic formula to factor the following equations over the complex numbers: e. x2 – 16x + 63 = 0 f. 2x2 + 24x – 506 = 0 g. x2 + 7x + 2 = 0 h. x2 + 2x + 7 = 0 10i. Can a parabola cross the x-axis only once? 11a. How many complex roots does the following equation have? Remember that the real numbers are a subset of the complex numbers. 8x9 + 6x8 – 4x5 + 10x3 + 2x2 – 1 = 0 11b. What is the highest possible number of times the following function can cross the x-axis? f(x) = x7 – 2x2 + 3 12a-d. Factor the following cubic expressions over the real numbers (i.e., maximally deconstruct these expressions into ones which contain only real numbers.) 3 a. x + 125 b. x3 – 7x – 6 = 0 c. x3 – 2x2 – 4x + 3 = d. x3 – 3x2 + 4x – 2 = 0 0 12e. Divide x5 – 4x2 + 2x + 1 by x – 1. Find two consecutive integers between which the resulting quartic expression has a root. 7 12f. Divide x4 + 4x3 + 6x2 + 4x + 1 by x + 1. Note the form of the resulting cubic expression and divide it by x + 1. Note the form of the resulting quadratic expression and divide it by x + 1. 13. Consider the following infinite sequence: an = 1/n2; n goes from 1 toward 0 through the reciprocals of the integers. Write out the eleventh through thirty-second terms of the sequence. What is the one-billionth term?