Name: Date: 1 Math 2 Variable Manipulation Part 4 Polynomials B MULTIPLYING TRINOMIALS Multiplying trinomials is the same process as multiplying binomials except for there are more terms to multiply than just four. Multiply each term in the first polynomial by each term in the second polynomial. This is the distributive method. Box or Vertical Methods could also be used. Example: (2x + 3)(6x 2 7x 5) Solution: Example: (7k 3)(k 2 2k + 7) Solution: Multiply 7k by k 2, -2k and +7, then multiply -3 by k 2, -2k and +7 To get 7k 3-14k + 49k and -3k 2 + 6k 21 Combine like terms to get: 7k 3 17k 2 + 55k 21 1. (4a + 2)(6a 2 a + 2) 2. (6n 2 6n 5)(7n 2 + 6n 5) 3. (7k 3)(k 2 2k + 7) 4. (n 2 + 6n 4)(2n 4) 5. (x 2 2x 8)( x 2 + 3x 5)
2 FACTORING QUADRATIC FORM Some quadratic equations have higher exponent values, but can be factored simply by factoring out a common divisor and then factoring the equation. Or it may be a factored as normal just with the higher exponent value. Example: 3x 3 + 24x 2 + 21x Solution: 3x is common to each of the terms so first factor out 3x to get 3x(x 2 + 8x + 7x) 3x(x + 1)(x + 7) Example: x 4 + 9x 2 + 20 Solution: The first term is x 4 and the second term is x 2, so we can just factor as normal except for factoring x 2 instead of x and x 4 instead of x 2. x 4 + 9x 2 + 20 (x 2 + 4)(x 2 + 5) 6. Factor: x 4 8x 2 + 15 7. x 4 7x 2 + 10 8. x 4 + 4x 2 + 3 9. a 4 + 6a 2 + 5 10. x 4 + x 2 12
3 FACTORING TRINOMIALS WHEN a > 1 There are several methods to use when quadratic equations have a > 1. One way is trial and error. First, guess that the first term in one binomial is just the unknown and the other is a times the unknown. Then you can guess the second terms in both binomials by listing the factors that multiply to the last number and then and use trial and error to find the answer. Note: The sign of terms in the polynomial can give clues. If the last number is positive, then you know the signs of both binomials are the same (+ and + or and -) since when multiplied, you get a positive number. If the middle number is negative, you know both are negative and if the middle number is positive, you know they are both positive. If the last number is negative, then you know that one binomial is positive and the other is negative. Example: 2n 2 + 13n + 6 Solution: First guess that the first terms are 2n and n since 2n*n = 2n 2 (2n )(n ) Since the last number is positive, we know that both binomials will be the same and since the middle number is positive, we know that both binomials are positive. (2n + )(n + ) Then start to guess numbers that multiply to be 6. Since they are both positive numbers, the only choices are 2 and 3 or 6 and 1. If we try 2 and 3, we get (2n + 2)(n + 3) FOILing this we get 2n 2 + 8n + 6. That is not correct and we know that the middle number needs to be a higher number, so try 6 and 1. (2n + 6)(n + 1) FOILing this guess we get 2n 2 + 8n + 6. That is not correct and we still know that number needs to be higher. If we put the 6 on the outside spot, it would multiply by the 2, so that would give a higher number so try 1 and 6. (2n + 1)(n + 6) FOILing this guess we get 2n 2 + 13n + 6 so (2n + 1)(n + 6) is the correct answer. This may seem like a lot of work, but it gets the answer and it works semi-easily when a is a prime number and there is only one choice of the first terms in the binomials. However, if there are multiply choices to factor a or many choices when factoring c, factoring by grouping is another way that is more efficient. Steps to Factor by Grouping when a > 1 1. Multiply a*c = d 2. Find two factors which multiply to d and add to b 3. Split the second term into two parts using the two factors from step 2. Rewrite the entire polynomial with four terms. Make sure to place the middle terms nearest the first and last terms which correlate the best. 4. Group the polynomial into two separate binomials with parentheses. 5. Factor out a common divisor from both binomials. You know you are correct when a common binomial is left in each section. 6. Factor out the common binomial to get the final factored binomials.
4 Example: 2n 2 + 13n + 6 Solution: This is the same question from the trial and error method above. This time we use the steps. Step 1: Multiply 2*6 = 12. Step 2: Find two numbers that Multiply to 12 and add to 13. They are 1 and 12. Step 3. Split the second term into + n and + 12n and rewrite the equation replacing the middle term with the two new terms. Notice that the equation is still equivalent just the middle term is split. 2n 2 + n + 12n + 6 Step 4. Group the polynomials into two separate binomials with parentheses. (2n 2 + n) + (12n + 6) Step 5. Factor out a common divisor from both binomials. n(2n + 1) + 6(2n + 1) We know we are right because we have (2n + 1) in both parts. Step 6. Factor out the common binomial to get the final factored binomials. We will factor out (2n + 1) from both parts leaving n + 6. (2n + 1)(n + 6) For this simple example, both methods are about the same amount of work/time. However, this method works for all polynomials with a > 1. 11. 2n 2 n 6 12. 20v 2 + 11v 3 13. 9x 2 3x 30 14. 2r 2 12r + 14
5 MIX OF ALL FACTORING TECHNIQUES 15. 15n 2 27n 6 16. 6x 9 n 30x 5 n 300xn 17. 6a 2 25a 25 18. 5nu 8 15nu 4 + 40n 19. 16b 2 + 60b 100 20. 2x 6 + 13x 4 + 6x 2
6 SOLVING A QUADRATIC EQUATION To solve a quadratic equation, put it in the form of ax 2 +bx + c = 0 -- in other words, set it equal to 0. Then factor the left side (if you can), and set each factor equal to 0 separately to get the two solutions. Sometimes the left side might not be obviously factorable. You can always use the quadratic formula. Just plug in the coefficients a, b, and c from ax 2 + bx+ c = 0 into the formula: b ± b 2 4ac 2a To solve x 2 + 4x + 2 = 0, plug a = 1, b = 4, and c = 2 into the formula: 4 ± 4 2 4 1 2 2 1 4 ± 8 2 = -2 ± 2 Whether you use reverse-foil or the quadratic equation, you will almost always get two solutions, or roots, to the equation. 21. 2x 2 + 3x 20 = 0 22. 4b 2 + 8b + 7 = 4 23. 5x 2 + 9x = 4 24. 9n 2 = 4 + 7n 25. 2x 2 36 = x 26. k 2 31 2k = 6 3k 2 2k
7 OTHER PROBLEMS WITH QUADRATICS 27. For a certain quadratic equation, ax 2 + bx + c = 0, the solutions are x = 0.75 and x = - 0.4. Which of the following could be factors of ax 2 + bx + c? a. (4x - 3) AND (5x + 2) b. (4x - 2) AND (5x + 3) c. (4x + 2) AND (5x - 3) d. (4x + 3) AND (5x - 2) e. (4x + 3) AND (5x + 2) 28. For what nonzero whole number k does the quadratic equation y 2 + 2ky + 4k = 0 have exactly one real solution for y? a. 8 b. 4 c. 2 d. -4 e. -8 29. In the equation x 2 + mx + n = 0, m and n are integers. The only possible value for x is 3. What is the value of m? 30. Let S be the set of all integers that can be written as 2n 2 6n, where n is a nonzero integer. Which of the following integers is in S? a. 6 b. 30 c. 46 d. 64 e. 80
1. 24a 3 + 8a 2 + 6a + 4 2. 42n 4 6n 3 101n 2 + 25 3. 7k 3 17k 2 + 55k 21 4. 2n 3 + 8n 2 32n + 16 5. x 4 + 5x 3 3x 2 14x + 40 6. (x 2 3)(x 2 5) 7. (x 2 2)(x 2 5) 8. (x 2 + 3)(x 2 + 1) 9. (a 2 + 1)(a 2 + 5) 10. (x 2 3)(x 2 + 4) 11. (2n + 3)(n 2) 12. (5v 1)(4v + 3) 13. 3(3x + 5)(x 2) 14. (2r 2)( r 7) 15. (15n + 3)(n 2) 16. 6xn(x 4 10)(x 4 + 5) 17. (2a + 5)(3a + 5) 18. 5n(u 8 3u 4 + 8) 19. 4(b + 5)(4b 5) 20. x 2 (2x 2 + 1)(x 2 + 6) 21. { 5 2, 4} 22. {- 1 2, - 3 2 } 23. {- 4 5, 1} 24. { 7+ 193, 7 193 } 18 18 25. {9/2, -4} 26. {-5/2, 5/2} 27. (4x - 3) AND (5x + 2) 28. 4 29. 6 30. 80 Answer Key 8