Factoring Polynomials: From Grouping to Solving Equations

Chapter 13 Factoring Polynomials § 13.4 Factoring Trinomials of 2 the Form x + bx + c by Grouping Factoring by Grouping Factoring polynomials often involves additional techniques after initially factoring out the GCF. One technique is factoring by grouping. Example Factor xy + y + 2x + 2 by grouping. Notice that, although 1 is the GCF for all four terms of the polynomial, the first 2 terms have a GCF of y and the last 2 terms have a GCF of 2. xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 = y(x + 1) + 2(x + 1) = (x + 1)(y + 2) 39 Factoring by Grouping Factoring a Four-Term Polynomial by Grouping 1) Arrange the terms so that the first two terms have a common factor and the last two terms have a common factor. 2) For each pair of terms, use the distributive property to factor out the pair’s greatest common factor. 3) If there is now a common binomial factor, factor it out. 4) If there is no common binomial factor in step 3, begin again, rearranging the terms differently. • If no rearrangement leads to a common binomial factor, the polynomial cannot be factored. 40 Factoring by Grouping Example Factor each of the following polynomials by grouping. 1) x3 + 4x + x2 + 4 = x · x2 + x · 4 + 1 · x2 + 1 · 4 = x(x2 + 4) + 1(x2 + 4) = (x2 + 4)(x + 1) 2) 2x3 – x2 – 10x + 5 = x2 · 2x – x2 · 1 – 5 · 2x – 5 · (– 1) = x2(2x – 1) – 5(2x – 1) = (2x – 1)(x2 – 5) 41 Factoring by Grouping Example Factor 2x – 9y + 18 – xy by grouping. Neither pair has a common factor (other than 1). So, rearrange the order of the factors. 2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y = 2(x + 9) – y(9 + x) = 2(x + 9) – y(x + 9) = (make sure the factors are identical) (x + 9)(2 – y) 42 § 13.5 Factoring Perfect Square Trinomials and the Difference of Two Squares Perfect Square Trinomials Recall that in our very first example in Section 4.3 we attempted to factor the polynomial 25x2 + 20x + 4. The result was (5x + 2)2, an example of a binomial squared. Any trinomial that factors into a single binomial squared is called a perfect square trinomial. 44 Perfect Square Trinomials In the last chapter we learned a shortcut for squaring a binomial (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 So if the first and last terms of our polynomial to be factored are can be written as expressions squared, and the middle term of our polynomial is twice the product of those two expressions, then we can use these two previous equations to easily factor the polynomial. a2 + 2ab + b2 = (a + b)2 a2 – 2ab + b2 = (a – b)2 45 Perfect Square Trinomials Example Factor the polynomial 16x2 – 8xy + y2. Since the first term, 16x2, can be written as (4x)2, and the last term, y2 is obviously a square, we check the middle term. 8xy = 2(4x)(y) (twice the product of the expressions that are squared to get the first and last terms of the polynomial) Therefore 16x2 – 8xy + y2 = (4x – y)2. Note: You can use FOIL method to verify that the factorization for the polynomial is accurate. 46 Difference of Two Squares Another shortcut for factoring a trinomial is when we want to factor the difference of two squares. a2 – b2 = (a + b)(a – b) A binomial is the difference of two square if 1.both terms are squares and 2.the signs of the terms are different. 9x2 – 25y2 – c4 + d4 47 Difference of Two Squares Example Factor the polynomial x2 – 9. The first term is a square and the last term, 9, can be written as 32. The signs of each term are different, so we have the difference of two squares Therefore x2 – 9 = (x – 3)(x + 3). Note: You can use FOIL method to verify that the factorization for the polynomial is accurate. 48 § 13.6 Solving Quadratic Equations by Factoring Zero Factor Theorem Quadratic Equations • Can be written in the form ax2 + bx + c = 0. • a, b and c are real numbers and a ≠ 0. • This is referred to as standard form. Zero Factor Theorem • If a and b are real numbers and ab = 0, then a = 0 or b = 0. • This theorem is very useful in solving quadratic equations. 50 Solving Quadratic Equations Steps for Solving a Quadratic Equation by Factoring 1) 2) 3) 4) 5) Write the equation in standard form. Factor the quadratic completely. Set each factor containing a variable equal to 0. Solve the resulting equations. Check each solution in the original equation. 51 Solving Quadratic Equations Example Solve x2 – 5x = 24. • First write the quadratic equation in standard form. x2 – 5x – 24 = 0 • Now we factor the quadratic using techniques from the previous sections. x2 – 5x – 24 = (x – 8)(x + 3) = 0 • We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = – 3 Continued. 52 Solving Quadratic Equations Example Continued • Check both possible answers in the original equation. 82 – 5(8) = 64 – 40 = 24 true (–3)2 – 5(–3) = 9 – (–15) = 24 true • So our solutions for x are 8 or –3. 53 Solving Quadratic Equations Example Solve 4x(8x + 9) = 5 • First write the quadratic equation in standard form. 32x2 + 36x = 5 32x2 + 36x – 5 = 0 • Now we factor the quadratic using techniques from the previous sections. 32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0 • We set each factor equal to 0. 8x – 1 = 0 or 4x + 5 = 0 1 8x = 1 or 4x = – 5, which simplifies to x = or − 5 . 8 4 Continued. 54 Solving Quadratic Equations Example Continued • Check both possible answers in the original equation. ( ) ( ( ) ) ( ) (1 + 9=) ( ) 1 4 8 1 8 + 9= 8 1 4 8 1 4 (10) = 8 ( )( ( ) ) ( ) 4 − ( ) 1 (10) = 2 5 true 5 5 5 5 8 − + 9 = 4 − ( −10 + 9 ) = 4 − (−1) = (−5)(−1) = 5 4 4 4 4 true • So our solutions for x are 1 8 or 5 − 4 . 55 Finding x-intercepts Recall that in Chapter 3, we found the x-intercept of linear equations by letting y = 0 and solving for x. The same method works for x-intercepts in quadratic equations. Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x2 term. The intercepts will be where the parabola crosses the x-axis. 56 Finding x-intercepts Example Find the x-intercepts of the graph of y = 4x2 + 11x + 6. The equation is already written in standard form, so we let y = 0, then factor the quadratic in x. 0 = 4x2 + 11x + 6 = (4x + 3)(x + 2) We set each factor equal to 0 and solve for x. 4x + 3 = 0 or x + 2 = 0 4x = –3 or x = –2 x = –¾ or x = –2 So the x-intercepts are the points (–¾, 0) and (–2, 0). 57