Summary 391 Chapter 5 SUMMARY Section 5.1 A polynomial in x is defined by a finite sum of terms of the form ax n, where a is a real number and n is a whole number. a is the coefficient of the term. n is the degree of the term. The degree of a polynomial is the largest degree of its terms. The term of a polynomial with the largest degree is the leading term. Its coefficient is the leading coefficient. A one-term polynomial is a monomial. A two-term polynomial is a binomial. A three-term polynomial is a trinomial. Addition and Subtraction of Polynomials and Polynomial Functions 7y 4 2y 2 3y 8 is a polynomial with leading coefficient 7 and degree 4. f1x2 4x 3 6x 11 f is a polynomial function with leading term 4x 3 and leading coefficient 4. The degree of f is 3. For f 1x2 4x 3 6x 11, find f 1 12. f 1 12 41 12 3 61 12 11 9 To add or subtract polynomials, add or subtract like terms. Example 4 1 4x 3 y 3x 2 y 2 2 17x 3 y 5x 2 y 2 2 4x 3 y 3x 2 y 2 7x 3 y 5x 2 y 2 11x 3 y 8x 2 y 2 Section 5.2 Multiplication of Polynomials To multiply polynomials, multiply each term in the first polynomial by each term in the second polynomial. Special Products 1. Multiplication of conjugates 1x y21x y2 x 2 y 2 The product is called a difference of squares. 2. Square of a binomial 1x y2 2 x 2 2xy y 2 1x y2 2 x 2 2xy y 2 The product is called a perfect square trinomial. 1x 2213x 2 4x 112 3x 3 4x 2 11x 6x 2 8x 22 3x 3 10x 2 19x 22 13x 5213x 52 13x2 2 152 2 9x 2 25 14y 32 2 14y2 2 12214y2132 132 2 16y 2 24y 9
392 Chapter 5 Polynomials Section 5.3 Division of Polynomials Division of polynomials: 1. For division by a monomial, use the properties a b c for c 0. a c b c and a b c a c b c 12a 2 6a 9 12a2 6a 9 4a 2 3 a 2. If the divisor has more than one term, use long division. 3. Synthetic division may be used to divide a polynomial by a binomial in the form x r, where r is a constant. 13x 2 5x 12 1x 22 3x 11 x 2 3x 2 5x 1 Answer: 13x 2 6x2 11x 1 1 11x 222 2 3 5 1 6 22 3 11 23 23 3x 11 23 x 2 13x 2 5x 12 1x 22 Answer: 3x 11 23 x 2
Summary 393 Section 5.4 The greatest common factor (GCF) is the largest factor common to all terms of a polynomial. To factor out the GCF from a polynomial, use the distributive property. A four-term polynomial may be factored by grouping. Steps to Factor by Grouping 1. Identify and factor out the GCF from all four terms. 2. Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. (Sometimes it is necessary to factor out the opposite of the GCF.) 3. If the two pairs of terms share a common binomial factor, factor out the binomial factor. Greatest Common Factor and Factoring by Grouping 3x 2 1a b2 6x1a b2 3x1a b2x 3x1a b2122 3x1a b21x 22 60xa 30xb 80ya 40yb 1036xa 3xb 8ya 4yb4 1033x12a b2 4y12a b24 1012a b213x 4y2 Section 5.5 Factoring Trinomials AC-Method To factor trinomials of the form ax 2 bx c: 1. Factor out the GCF. 2. Find the product ac. 3. Find two integers whose product is ac and whose sum is b. (If no pair of numbers can be found, then the trinomial is prime.) 4. Rewrite the middle term bx as the sum of two terms whose coefficients are the numbers found in step 3. 5. Factor the polynomial by grouping. 10y 2 35y 20 512y 2 7y 42 ac 1221 42 8 Find two integers whose product is 8 sum is 7. The numbers are 8 and 1. 532y 2 8y 1y 44 532y1y 42 11 y 424 51y 4212y 12 and whose
394 Chapter 5 Polynomials Trial-and-Error Method To factor trinomials in the form ax 2 bx c: 1. Factor out the GCF. 2. List the pairs of factors of a and the pairs of factors of c. Consider the reverse order in either list. 3. Construct two binomials of the form Factors of a Factors of c 4. Test each combination of factors until the product of the outer terms and the product of inner terms add to the middle term. 5. If no combination of factors works, the polynomial is prime. The factored form of a perfect square trinomial is the square of a binomial: a 2 2ab b 2 1a b2 2 a 2 2ab b 2 1a b2 2 1 x 21 x 2 10y 2 35y 20 512y 2 7y 42 The pairs of factors of 2 are 2 1. The pairs of factors of 4 are 1 4 1 1 42 2 2 2 1 22 4 1 4 1 12 12y 221y 22 2y 2 2y 4 No 12y 421y 12 2y 2 2y 4 No 12y 121y 42 2y 2 7y 4 No 12y 221y 22 2y 2 2y 4 No 12y 421y 12 2y 2 2y 4 No 12y 121y 42 2y 2 7y 4 Yes Therefore, 10y 2 35y 20 factors as 512y 121y 42. 9w 2 30wz 25z 2 13w2 2 213w215z2 15z2 2 13w 5z2 2 Section 5.6 Factoring Binomials Factoring Binomials: Summary Difference of squares: a 2 b 2 1a b21a b2 Difference of cubes: a 3 b 3 1a b21a 2 ab b 2 2 Sum of cubes: a 3 b 3 1a b21a 2 ab b 2 2 25u 2 9v 4 15u 3v 2 215u 3v 2 2 8c 3 d 6 12c d 2 214c 2 2cd 2 d 4 2 27w 9 64x 3 13w 3 4x219w 6 12w 3 x 16x 2 2
Summary 395 Section 5.7 Additional Factoring Strategies 1. Factor out the GCF (Section 5.4). 2. Identify whether the polynomial has two terms, three terms, or more than three terms. 3. If the polynomial has more than three terms, try factoring by grouping (Section 5.4). 4. If the polynomial has three terms, check first for a perfect square trinomial. Otherwise, factor by using the ac-method or trial-and-error method (Section 5.5). 5. If the polynomial has two terms, determine if it fits the pattern for a difference of squares, difference of cubes, or sum of cubes. Remember, a sum of squares is not factorable over the real numbers (Section 5.6). 6. Be sure to factor the polynomial completely. 7. Check by multiplying. 9x 2 4x 9x 3 x19x 4 9x 2 2 x19x 2 9x 42 x13x 4213x 12 4a 2 12ab 9b 2 c 2 4a 2 12ab 9b 2 c 2 12a 3b2 2 c 2 12a 3b c212a 3b c2 Factor out the GCF. Descending order. Factor the trinomial. Group 3 by 1. Perfect square trinomial. Difference of squares. Section 5.8 Solving Equations by Using the Zero Product Rule An equation of the form ax 2 bx c 0, where a 0, is a quadratic equation. The zero product rule states that if a b 0, then a 0 or b 0. The zero product rule can be used to solve a quadratic equation or higher-degree polynomial equation that is factored and equal to zero. 0 x12x 321x 42 x 0 or 2x 3 0 or x 4 0 x 3 2 or x 4 f1x2 ax 2 bx c 1a 02 defines a quadratic function. The x-intercepts of a function defined by y f1x2 are determined by finding the real solutions to the equation f1x2 0. The y-intercept of a function y f1x2 is at f(0). Find the x-intercepts. f1x2 3x 2 8x 5 0 3x 2 8x 5 0 13x 521x 12 3x 5 0 or x 1 0 x 5 3 or x 1 The x-intercepts are 1 5 3, 02 and (1, 0). Find the y-intercept. f1x2 3x 2 8x 5 f102 3102 2 8102 5 f102 5 The y-intercept is 10, 52.