Algebra I : Semester II : Polynomials

Sections:

Introduction  |   Section 1  |   Section 2  |   Section 3  |  a href="section4.html">Section 4  |  Section 5  |   Section 6

Section Six

Part 1  |  Part 2

Algebra 1: Section 6: Polynomials Difference of Two Squares The city hired a team to design a space in the city where there would be a concrete walk/sitting area around a central garden. The team came up with the concept above, where the gray region is concrete and the green region is the garden. How many square feet will be covered in concrete? This is not so tough because you are working with two squares. If you find the area of the large square and subtract the garden area, the result will be the area of the concrete region. The formula for the area of a square is A = s2.  For the larger rectangle A = 1002. For the smaller rectangle A = 702. Area of the concrete region = 1002 - 702. 10,000 – 4,900 = 5,100 ft2 You just found the difference of two squares using numbers. In this section, you will be factoring the difference of two squares using binomials. Let’s review one of the special products rules.   Product of a Sum and Difference (a + b) (a – b) = a2 – b2 To factor the difference of two squares, reverse this rule.   To Factor a2 – b2   Find the square root of a2. Find the square root of b2. Write them in the parentheses as(a + b) (a – b). If in doubt, check by using FOIL. Example 1:  Factoring the Difference of Two Squares Factor each of the following. x2 – 25 x2 is the square of x 25 is the square of 5 x2 – 25 = (x + 5)(x – 5) 4x2 – 9 4x2 is the square of 2x 9 is the square of 3 4x2 – 9 = (2x + 3)(2x – 3) 64x2 – 81y2 64x2 is the square of 8x 81y2 is the square of 9y 64x2 – 81y2 = (8x + 9y)(8x – 9y) Example 2: Factoring the GCF First Factor the GCF first and then factor the binomial. If the binomial cannot be factored, write prime. 8c2 – 32 8c2 – 32 = 8(c2 – 4) = 8 (c + 2)(c – 2) 8c2 – 16 8c2 – 16 = 8(c2 – 2)  c2 – 2 is prime, so the factors are 8(c2 – 2) Example 3: Factoring More than Once Factor 2x4 – 32 completely. First, factor out the GCF:    2(x4 – 16) Second, factor the difference of squares:  2(x2 + 4)(x2 – 4) Third, factor the last binomial again:  2(x2 + 4)(x + 2)(x – 2) Note: x2 + 4 cannot be factored:  it is the sum of two squares, not the difference, and is prime. Example 4: Solving Equations by Factoring Solve each of the following by factoring. Check the solutions. y2 – 9 = 0 Factor:     (y + 3)(y – 3) = 0 Set = 0:    y + 3 = 0 or y – 3 = 0 Solve:       y = 0 – 3     y = 0 + 3                 y = -3          y = 3 Check: y2 – 9 = 0 (3)2 – 9 = 0 9 – 9 = 0 0 = 0  √ y2 – 9 = 0 (-3)2 – 9 = 0 9 – 9 = 0 0 = 0  √ Solution:  {-3, 3} 2x2 = 50 Subtract 50:               2x2 – 50 = 50 – 50                                  2x2 – 50 = 0 Factor:                     2(x2 – 25) = 0                          2(x + 5)(x – 5) = 0 Set = 0:               2 = 0   or  x + 5 = 0   or  x – 5 = 0 Solve:                   False     x = 0 – 5        x = 0 + 5                                           x = -5             x = 5 Check: 2x2 = 50 2(-5)2 = 50 2(25) = 50 50 = 50   √ 2x2 = 50 2(5)2 = 50 2(25) = 50 50 = 50   √ Solution: {-5, 5} 4x2 + 2x = 2x + 49 Subtract 2x and 49:   4x2 + 2x – 2x – 49 = 2x + 49 – 2x – 49                                   4x2 – 49 = 0 Factor:                      (2x + 7)(2x – 7) = 0 Set = 0:                    2x + 7 = 0 or 2x – 7 = 0 Solve:                      2x = 0 – 7      2x = 0 + 7                                  x = -7/2          x = 7/2                            or   x = -3.5        or x = 3.5 Check: 4x2 + 2x = 2x + 49 4(-3.5)2 + 2(-3.5) = 2(-3.5) + 49 4(12.25) + 2(-3.5) = 2(-3.5) + 49 49 – 7 = -7 + 49 42 = 42   √ 4x2 + 2x = 2x + 49 4(3.5)2 + 2(3.5) = 2(3.5) + 49 4(12.25) + 2(3.5) = 2(3.5) + 49 49 + 7 = 7 + 49 56 = 56   √ Solution: {-7/2, 7/2}  or  {-3.5, 3.5} Note: Do not use decimals if they are for fractions which are repeating or non-terminating decimals, such as 1/3 = 0.333…Use fractions in those cases.   Practice Click to get a new problem. Factor the polynomial and then click to see the answer. Remember that your answer is still correct if you have the same binomials reversed. Factor the following problems completely, using the GCF or factoring more than once. Problem Factor 1 Factor 2 x4 – y4 (x2 + y2) (x2 – y2) (x2 + y2)(x + y)(x – y) a3 – 4a a(a2 – 4) a(a + 2)(a – 2) 2x2 – 162 2(x2 – 81) 2(x + 9)(x – 9) t3 – t2 t2(t – 1) Done, t – 1 is prime Solve the following equations by factoring. Equation Factored Solutions x2 – 64 = 0 (x + 8)(x – 8) = 0 {-8, 8} 25m2 = 64 25m2 – 64 = 0 (5m + 8)(5m – 8) = 0 {-8/5, 8/5} or {-1.6, 1.6} 12 – 27x2 = 0 3(4 – 9x2) = 0 3(2 + 3x)(2 – 3x) = 0 {-2/3, 2/3} Do not change to decimals. 18x3 – 25x = 25x 18x3 – 50x = 0 2x(9x2 – 25) = 0 2x(3x + 5)(3x – 5) = 0 {0, -5/3, 5/3} Now go on to the next part

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