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P.5 Factoring Polynomials

P.5 Factoring Polynomials. Factoring.

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P.5 Factoring Polynomials

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  1. P.5 Factoring Polynomials

  2. Factoring Factoringis the process of writing a polynomial as the product of two or more polynomials. The factors of 6x2 – x – 2 are 2x + 1 and 3x – 2. In this section, we will be factoring over the integers. Polynomials that cannot be factored using integer coefficients are called irreducible over the integers or prime. The goal in factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors is prime or irreducible. In this situation, the polynomial is said to be factored completely.

  3. Common Factors In any factoring problem, the first step is to look for the greatest common factor. The greatest common factor is an expression of the highest degree that divides each term of the polynomial. The distributive property in the reverse direction ab + ac = a(b + c) can be used to factor out the greatest common factor. The GCF will be the “largest” integer (only negatively signed if the leading coefficient is negatively signed) that can divide without remainder into each term, and the lowest power for any one variable that appears in each term. (Examples to follow.)

  4. Example • Solution • We begin by determining the greatest common factor. 9 is the greatest integer that divides 18 and 27. Furthermore, x2is the greatest expression (SMALLEST exponent for the variable x that appears in each term) that divides x3 and x2. Thus, the greatest common factor of the two terms in the polynomial is 9x2. • 18x3 + 27x2 • = 9x2(2x) + 9x2(3) Express each term with the greatest common factor as a factor. • = 9x2(2 x + 3) Factor out the greatest common factor. Factor: a. 18x3 + 27x2 b. x2(x + 3) + 5(x + 3) b. In this situation, the greatest common factor is the common polynomial factor (x + 3). We factor out this common factor as follows. x2(x + 3) + 5(x + 3) = (x + 3)(x2 + 5) Factor out the greatest common factor polynomial (GCFP).

  5. Factoring using the sum/product method. Solution a. The factors of the first term are x and x: (x )( x ) To find the second term of each factor, we must find two numbers whose product is 8 and whose sum is 6. (Always start with finding the integers for the product, then select the ones that add to the sum.) EX: Factor a. x2 + 6x + 8 b. x 4 + 3x 2y – 18 y 2 SUM PRODUCT 6 8 9 8*1 6 4*2 From the table above, we see that 4 and 2 are the required integers. Thus, x2 + 6x + 8 = (x + 4)( x + 2) or (x + 2)( x + 4).

  6. Solution b. We begin with x 4 + 3x 2y – 18 y 2 = ( )( ). To find the second term of each factor, we must find two numbers whose sum is __________, and whose product is __________. Ex., con’t.: Factor: b. x 4 + 3x 2y – 18 y 2 SUM PRODUCT From the table above, we see that _______ and ________ are the required integers. Thus, (ans) x 4 + 3x 2y – 18 y 2= or

  7. (Assume, for the moment, that there is no greatest common factor.) 1. Find two First terms whose product is ax2: ( x + )( x + ) = ax2 + bx + c 2. Find two Last terms whose product is c: (x + )(x + ) = ax2 + bx + c 3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx: ( x + )( x + ) = ax2 + bx + c I O (sum of O + I) A Strategy for Factoring ax2 + bx + c If no such combinations exist, the polynomial is prime.

  8. Solution (using “sum and product WITH GROUPING”). Example Factor: 8x2 – 10x – 3.

  9. Possible Factors of 8x2 – 10x – 3 Sum of Outside and Inside Products (Should Equal –10x) (8x + 1)(x – 3) -24x + x = -23x (8x – 3)(x + 1) 8x – 3x = 5x (8x – 1)(x +3) 24x – x = 23x (8x + 3)(x – 1) -8x + 3x = -5x This is the required middle term. (4x + 1)(2x – 3) -12x + 2x = -10x (4x – 3)(2x + 1) 4x – 6x = -2x (4x – 1)(2x + 3) 12x – 2x = 10x (4x + 3)(2x – 1) -4x + 6x = 2x Thus, 8x2 – 10x – 3 = (4x + 1)(2x – 3) or (2x – 3)(4x + 1). (using text method) Text Example cont.

  10. Factoring by the ac method • For a polynomial in the form: ax2 + bx + c • 1. Multiply a*c (product) • 2. Find factors of the product whose sum is b • 3. Arrange the factors of a and c so that they will yield those factors when the binomials are multiplied. • 4. Check by multiplying the factors. Ex: Factor 4x2 + 16 x + 15 (on board) S P Will do if asked, otherwise do your homework.

  11. Let A and B be real numbers, variables, or algebraic expressions, 1. A2 + 2AB + B2 = (A + B)2 2. A2 – 2AB + B2 = (A – B)2 If we recognize that we have a perfect square trinomial (look for “perfect squares” at the beginning and end), we can use this first before doing too much work. Factoring Perfect Square Trinomials

  12. Text Example • Factor: x2 + 6x + 9. Solution: x2 + 6x + 9 = x2 + 2 · x · 3 + 32 = (x + 3)2 • Factor: 81x2 - 90x + 25 • (ans:

  13. The Difference of Two Squares • If A and B are real numbers, variables, or algebraic expressions, then • A2 – B2 = (A + B)(A – B). • In words: The difference of the squares of two terms factors as the product of a sum and the difference of those terms. (That is, “the difference of square can be factored as the product of CONJUGATES”.)

  14. Text Example Solution: • Factor: 81x2 - 49 81x2 – 49 = (9x)2 – 72 = (9x + 7)(9x – 7). (You will notice when we check this by multiplying it out, the middle terms are additive inverses (add to zero) because they have opposite signs.) Q: Would it be possible to factor 81x2 + 49? A: NO. There is no real number factorization for the SUM of squares.

  15. Type Example A3 + B3 = (A + B)(A2 – AB + B2) x3 + 8 = x3 + 23 = (x + 2)( x2 – x·2 + 22) = (x + 2)( x2 – 2x + 4) A3 – B3 = (A – B)(A2 + AB + B2) 64x3 – 125 = (4x)3 – 53 = (4x – 5)((4x)2 + (4x)(5) + 52) = (4x – 5)(16x2 + 20x + 25) Factoring the Sum and Difference of 2 Cubes Note that there is NO FACTORIZATION for the SUM of SQUARES!

  16. Factoring by Grouping (by way of Example) Factor: x3 – 5x2 – 4x + 20 (DO NOT LOOK AT YOUR NOTES, no need to write.) Solution x3 – 5x2 – 4x + 20 (Note that there is no gcf among all 4 terms). = (x3 – 5x2) + (-4x + 20) Group the terms with common factors. = x2(x – 5) – 4(x – 5)Factor from each group. (Note that the negative “leading coefficient” must be factored out.) = (x – 5)(x2 – 4) Factor out the common binomial factor, (x – 5). = (x – 5)(x + 2)(x – 2) Factor completely by factoring x2 – 4 as the difference of two squares.

  17. A Strategy for Factoring a Polynomial • Factor out the GCF, if there is one. • Determine the number of terms in the polynomial and try factoring as follows: • 2 terms: can the binomial be factored by one of the special forms including difference of two squares, sum of two cubes, or difference of two cubes? • 3 terms: is the trinomial a perfect square trinomial? If the trinomial is not a perfects square trinomial, try factoring by trial and error (or the ac method). • 4 or more terms: try factoring by grouping. • Check to see if any factors with more than one term in the factored polynomial can be factored further. (Look at exponents.) If so, factor completely.

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