Algebra I : Semester II : Polynomials

Sections:

Introduction  |   Section 1  |   Section 2  |   Section 3  |  Section 4  |  Section 5  |  Section 6

Section Five

Part 1  |  Part 2  |  Part 3

Algebra 1: Section 5: Polynomials Solving Equations by Factoring Trinomials Here is a puzzle problem: The sum of a negative number and its square is 72. Find the number. You have solved these kinds of problems before by making lists. This time, let’s learn how to solve these using equations. Start by learning how to solve equations by factoring trinomials, and then do the puzzle problem. You will use the zero product property again in these problems. Factor and then set each factor equal to zero.   Steps for Solving in Factored Form Set the equation equal to zero.  Factor the polynomial if it is not already in factored form. Set each factor equal to zero. Solve each of the equations and write the solution set. Check. Example 1: Solving a Problem by Factoring Solve x2 – 13x + 36 = 0 Factor     (x – 9) (x – 4) = 0 Set = 0     x – 9 = 0      or     x – 4 = 0 Solve       x = 0 + 9              x = 0 + 4                 x = 9                    x = 4 Check: x2 – 13x + 36 = 0 (9)2 – 13(9) + 36 = 0 81 – 117 + 36 = 0 0 = 0  √ x2 – 13x + 36 = 0 (4)2 – 13(4) + 36 = 0 16 – 52  + 36 = 0 0 = 0  √ Therefore the solution is {9, 4}. Example 2: Set = 0 and then Solve Solve n2 – 5n = 24. To use the technique, the trinomial must be equal to zero. Subtract 24         n2 – 5n – 24 = 0 Factor                 (n + 3) (n – 8) = 0 Set = 0                n + 3 = 0      or     n – 8 = 0 Solve                   n = 0 – 3              n = 0 + 8                             n = -3                   n = 8 Check: n2 – 5n = 24 (-3)2 – 5(-3) = 24 9 + 15 = 24 24 = 24  √ n2 – 5n = 24 (8)2 – 5(8) = 24 64 – 40 = 24 24 = 24  √  Therefore the solution is {-3, 8}. Example 3: Solving a Puzzle Problem The sum of a negative number and its square is 72. Find the number. Use these steps: Pick a variable for the unknown number. Let n = the negative number. Write the sentence as an equation. Subtract 72 Write in decreasing order n + n2 – 72 = 72 – 72 n2 + n – 72 = 0 Factor (n – 8) (n + 9) = 0 Set = 0 Solve n – 8 = 0   or  n + 9 = 0 n = 8        or    n = -9 {8, -9} Check by using the word problem. The sum of a number and its square is 72. Is 8 + (8)2 = 8 + 64 = 72?                 yes Is (-9) + (-9)2 = -9 + 81 = 72?         yes Write the solution. I am only looking for a negative number, so the solution is -9.   Practice Solve the following equations and word problems. Equation Factors Solution x2 + 2y – 63 = 0 (x – 7)(x + 9) = 0 {7, -9} z2 + 2z – 8 = 0 (z – 2)(z + 4) = 0 {2, -4} x2 + 25 = 10x First subtract 10x and then put the terms in order. x2 – 10x + 25 = 0 (x – 5)(x – 5) = 0 {5, 5} k2 – 10k = -9 k2 – 10k + 9 = 0 (k – 9)(k – 1) = 0 {1, 9} m2 – m – 6 = 0 (m + 2)(m – 3) = 0 {-2, 3} x2 + 12x + 32 = 0 (x + 4)(x + 8) = 0 {-4, -8} The sum of a positive number and its square is 90. What is the number? x + x2 = 90 x2 + x – 90 = 0 (x – 9)(x + 10) = 0 {9, -10} Since we only want the positive number, the solution is 9. The length of a postcard is 9 centimeters (cm) more than its width. Its area is 112 cm2.  Find its dimensions. Sketch the rectangle. Let width = w and length = (w + 9).         Since A = LW   112 = w(w + 9)   112 = w2 + 9w   112 – 112 = w2 + 9w – 112    0 = w2 + 9w – 112    0 = (w – 7)(w + 16)    {7, -16}    Since the width of a postcard cannot measure -16, the solution for the width is 7.    If w = 7 cm, the length is 7 + 9 = 16 cm and the area is 7(16) = 112 cm2.    The dimensions of the postcard are 7cm by 16 cm. A poster in the shape of a triangle has an area of 40 cm2. If the base is (2h + 6) cm and the height is h cm, what is the measure of h? Sketch the triangle. You know the area is 40 cm2. A = ½bh b = 2h + 6 and h = h   A = ½ (2h + 6) (h) 40 = (h + 3)(h) 40  = h2 + 3h 40 – 40 = h2 + 3h – 40 0 = h2 + 3h – 40 0 = (h – 5)(h + 8) {5, -8} Since the height cannot be negative, the height must be 5 cm   The square of a number is 20 more than eight times the number.  Find the number. x2 = 20 + 8x x2 – 8x – 20 = 0 (x + 2)(x – 10) = 0 {-2, 10} Since you can use either number, the solutions are -2 and 10. Now go on to the next part

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