# FACTOR POLYNOMIALS by SPLITTING

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1 FACTOR POLYNOMIALS by SPLITTING THE IDEA FACTOR POLYNOMIALS by SPLITTING The idea is to split the middle term into two pieces. Say the polynomial looks like c + bx + ax 2. Further more suppose the DL METHOD fails, as well as the GROUPING METHOD... thus far these are the only two methods we have learned to factor polynomials, yet it is not enough. Consider for example x 2 + 5x + 6 on this polynomial DL METHOD to factor fails since the gcd of the terms is 1, moreover, GROUPING METHOD fails, since all grouping combinations lead nowhere. In deed, it is time for a new idea. Suppose we go on a do something amazingly creative and bold. Suppose we SPLIT the MIDDLE. That is, the middle term is 5x but 5x can be written.. oohh.. so many ways... 5x = x+4x OR 5x = 20x 15x OR 7x 2x OR 2x+3x OR 100x+105x, etc... etc.. the possibilities are endless... but... Suppose we go on and split 5x as 5x = 2x + 3x then.. observe..., if we choose just the right way to split up 5x, we might just get lucky and proceed factor by grouping!!! x 2 + 5x + 6 x 2 + ( 5x ) + 6 = x 2 + ( 2x + 3x ) + 6 (BI, the creative, bold splitting the middle idea) = ( x 2 + 2x ) + ( 3x + 6 ) (ALA) = (x) [ x + 2 ] + (3) [ x + 2 ] (DL, BI, note: looks like someone distributed [x + 2]) = (x + 3) [ x + 2 ] Incidentally, rather than splitting 5x as 2x + 3x, we could have also split it up as 3x + 2x. Observe... x 2 + 5x + 6 x 2 + ( 5x ) + 6 = x 2 + ( 3x + 2x ) + 6 (BI, the creative, bold splitting the middle idea) = ( x 2 + 3x ) + ( 2x + 6 ) (ALA) = (x) [ x + 3 ] + (2) [ x + 3 ] (DL, BI, note: looks like someone distributed [x + 3]) = (x + 2) [ x + 3 ] Now, while this method has its limitation, it is a powerful one that can help factor more than degree two polynomials. Observe, 3x x c

2 3x 4 + ( 41x 2) + 60 = 3x 4 + ( 5x x 2) + 60 (BI, the creative, bold splitting the middle idea) = ( 3x 4 + 5x 2) + ( 36x ) (ALA) = (x 2 ) [ 3x ] + (12) [ 3x ] (DL, BI, note: looks like someone distributed [3x 2 + 5]) = (x ) [ 3x ] It even works with more than one variable, 6x yx y 2 6x 6 + ( 31yx 3) + 18y 2 = 6x 6 + ( 27yx 3 + 4yx 3) + 18y 2 (BI, the creative, bold splitting the middle idea) = ( 6x yx 3) + ( 4yx y 2) (ALA) = (3x 3 ) [ 2x 3 + 9y ] + (2y) [ 2x 3 + 9y ] (DL, BI, note: looks like someone distributed [2x 3 + 9y]) = (3x 3 + 2y) [ 2x 3 + 9y ] Of course, at this point, the author is enjoying a tiny charlatan moment carrying on the mysteriously effective splitting-the-middle strategy and ignoring the 1000 lbs. gorilla in the discussion. Namely, amongst the millions of ways to split up 5x for example, how would one know that 2x + 3x and 3x + 2x are good ways to split the middle and not 15x 10x nor the other millions of ways to split 5x?? Some Hints 1. This works well for trinomials (three terms) 2. Write the polynomial in descending order (of the exponet on the undeterminant) 3. If there is a common factor pull it out first (using D.L.) THE IDEA behind THE IDEA: HOW to SPLIT the MIDDLE Consider the polynomial x 2 + 7x + 10 the secret to the splitting the middle method is to try to see a few moves ahead. Imagine we have already split up the middle, and grouped and factored so that x 2 + 7x + 10 = (x + b)(x + d) For some appropriate constants a and b. Now we play detective to see how much we can guess about these unknown numbers a and b. If we work our way back, we can write (x + a)(x + b) = x 2 + ax + bx + ab (FOIL) = x 2 + (a + b)x + ab (ALA,DL) in theory this should be equal to our polynomial = x 2 + 7x + 10 c

3 Here lies the secret on how to split up the 7 in 7x successfully. It should be split up into pieces a + b with the BIG hint that when you multiply ab you get the coefficient 10. Now there aren t that many ways to multiply two integers to get 10. We can list them all, First we multiply the leading coefficient, 1 times the last coefficient, 10 to obtain, 10. We now consider all the possible ways to factor 10 into the product of two integers: The ways to factor 10 = ab are: (1)(10) (-1)(-10) (2)(5) (-2)(-5) But, we also know that a + b = 7. This leads the elimination of most of these candidates for ab, leaving only the factors that add up to 7, thus the ones that will effectively help factor the polynomial. (1)(10) (-1)(-10) (2)(5) (-2)(-5) x 2 + ( 7x ) + 10 = x 2 + ( 2x + 5x ) + 10 (BI, THE split-the-middle idea!) = ( x 2 + 2x ) + ( 5x + 10 ) (ALA) = (x) [ x + 2 ] + (5) [ x + 2 ] (DL, BI, note: looks like someone distributed [x + 2]) = (x + 5) [ x + 2 ] The method just needs a slight modification when the leading coefficient is not 1. In this case, we multiply the leading coefficient and the last coefficient, and the factors of this product become the candidates to split the middle term. For example: Factor 6x 2 + ( 19x ) + 10 First we multiply the leading coefficient, 6 times the last coefficient, 10 to obtain, 60. We now consider all the possible ways to factor 60 into the product of two integers: The ways to factor 60 = ab are: (1)(60) (-1)(-60) (2)(30) (-2)(-30) (3)(20) (-3)(-20) (4)(15) (-4)(-15) (5)(12) (-5)(-12) (6)(10) (-6)(-10) But, we also know that a + b = 19. This leads the elimination of most of these candidates for ab, leaving only the factors that add up to 19, thus the ones that will effectively help factor the polynomial. (1)(60) (-1)(-60) (2)(30) (-2)(-30) (3)(20) (-3)(-20) (4)(15) (-4)(-15) (5)(12) (-5)(-12) (6)(10) (-6)(-10) 6x 2 + ( 19x ) + 10 = 6x 2 + ( 4x + 15x ) + 10 (BI, THE split-the-middle idea!) = ( 6x 2 + 4x ) + ( 15x + 10 ) (ALA) = (2x) [ 3x + 2 ] + (5) [ 3x + 2 ] (DL, BI, note: looks like someone distributed [3x + 2]) = (2x + 5) [ 3x + 2 ] c

4 Yet another example Factor 10x 2 + ( 31x ) + 15 First we multiply the leading coefficient, 10 times the last coefficient, 15 to obtain, 150. We now consider all the possible ways to factor 150 into the product of two integers: The ways to factor 150 = ab are: (1)(150) (-1)(-150) (2)(75) (-2)(-75) (3)(50) (-3)(-50) (5)(30) (-5)(-30) (6)(25) (-6)(-25) (10)(15) (-10)(-15) But, we also know that a + b = 31. This leads the elimination of most of these candidates for ab, leaving only the factors that add up to 31, thus the ones that will effectively help factor the polynomial. (1)(150) (-1)(-150) (2)(75) (-2)(-75) (3)(50) (-3)(-50) (5)(30) (-5)(-30) (6)(25) (-6)(-25) (10)(15) (-10)(-15) 10x 2 + ( 31x ) + 15 = 10x 2 + ( 6x + 25x ) + 15 (BI, THE split-the-middle idea!) = ( 10x 2 + 6x ) + ( 25x + 15 ) (ALA) = (2x) [ 5x + 3 ] + (5) [ 5x + 3 ] (DL, BI, note: looks like someone distributed [5x + 3]) = (2x + 5) [ 5x + 3 ] Another example; Factor 2x 2 + ( 13x ) + 15 First we multiply the leading coefficient, 2 times the last coefficient, 15 to obtain, 30. We now consider all the possible ways to factor 30 into the product of two integers: The ways to factor 30 = ab are: (1)(30) (-1)(-30) (2)(15) (-2)(-15) (3)(10) (-3)(-10) (5)(6) (-5)(-6) (6)(5) (-6)(-5) But, we also know that a + b = 13. This leads the elimination of most of these candidates for ab, leaving only the factors that add up to 13, thus the ones that will effectively help factor the polynomial. (1)(30) (-1)(-30) (2)(15) (-2)(-15) (3)(10) (-3)(-10) (5)(6) (-5)(-6) (6)(5) (-6)(-5) c

5 2x 2 + ( 13x ) + 15 = 2x 2 + ( 3x + 10x ) + 15 (BI, THE split-the-middle idea!) = ( 2x 2 + 3x ) + ( 10x + 15 ) (ALA) = (x) [ 2x + 3 ] + (5) [ 2x + 3 ] (DL, BI, note: looks like someone distributed [2x + 3]) = (x + 5) [ 2x + 3 ] One more example; Factor 7x 2 + ( 31x ) + 12 First we multiply the leading coefficient, 7 times the last coefficient, 12 to obtain, 84. We now consider all the possible ways to factor 84 into the product of two integers: The ways to factor 84 = ab are: (1)(84) (-1)(-84) (2)(42) (-2)(-42) (3)(28) (-3)(-28) (4)(21) (-4)(-21) (6)(14) (-6)(-14) (7)(12) (-7)(-12) But, we also know that a + b = 31. This leads the elimination of most of these candidates for ab, leaving only the factors that add up to 31, thus the ones that will effectively help factor the polynomial. (1)(84) (-1)(-84) (2)(42) (-2)(-42) (3)(28) (-3)(-28) (4)(21) (-4)(-21) (6)(14) (-6)(-14) (7)(12) (-7)(-12) 7x 2 + ( 31x ) + 12 = 7x 2 + ( 3x + 28x ) + 12 (BI, THE split-the-middle idea!) = ( 7x 2 + 3x ) + ( 28x + 12 ) (ALA) = (x) [ 7x + 3 ] + (4) [ 7x + 3 ] (DL, BI, note: looks like someone distributed [7x + 3]) = (x + 4) [ 7x + 3 ] c

6 1. By splitting the middle term, factor the polynomial: x 2 + ( 2x ) By splitting the middle term, factor the polynomial: x 2 + ( 3x ) By splitting the middle term, factor the polynomial: x 2 + ( x ) By splitting the middle term, factor the polynomial: x 2 + (0) By splitting the middle term, factor the polynomial: x 2 + ( 4x ) By splitting the middle term, factor the polynomial: x 2 + ( x ) By splitting the middle term, factor the polynomial: 3x 2 + ( 7x ) By splitting the middle term, factor the polynomial: 6x 2 + ( 5x ) By splitting the middle term, factor the polynomial: x 2 + ( 2x ) By splitting the middle term, factor the polynomial: x 2 + ( 2x ) By splitting the middle term, factor the polynomial: 2x 2 + ( 3x ) By splitting the middle term, factor the polynomial: 2x 2 + ( 13x ) By splitting the middle term, factor the polynomial: 2x 2 + ( 13x ) + 15 c

7 14. By splitting the middle term, factor the polynomial: 6x 2 + ( x ) By splitting the middle term, factor the polynomial: 9x 2 + (0) By splitting the middle term, factor the polynomial: 6y 2 + ( 7y ) By splitting the middle term, factor the polynomial: 10y 2 + ( y ) By splitting the middle term, factor the polynomial: 15y 2 + ( 4y ) By splitting the middle term, factor the polynomial: 15x 4 + ( 13x 2) By splitting the middle term, factor the polynomial: 15x 4 + ( 8x 2) By splitting the middle term, factor the polynomial: 2x 4 + ( 7x 2) By splitting the middle term, factor the polynomial: 15y 2 + ( 16y ) By splitting the middle term, factor the polynomial: 3x 4 + ( x 2) By splitting the middle term, factor the polynomial: 3x 4 + ( 2x 2) By splitting the middle term, factor the polynomial: 2x 4 + ( 5x 2) By splitting the middle term, factor the polynomial: 2x 4 + ( 5yx 2) + 3y 2 c

8 27. By splitting the middle term, factor the polynomial: 2x 6 + ( 5x 3) By splitting the middle term, factor the polynomial: 2x 6 + ( y 4 x 3) + y By splitting the middle term, factor the polynomial: 9 + (0) + y By splitting the middle term, factor the polynomial: 25 + (0) + y By splitting the middle term, factor the polynomial: 9 + (0) + x By splitting the middle term, factor the polynomial: 5 + ( 4x 3) + x 6 c

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### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4) ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

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### x 4-1 = (x²)² - (1)² = (x² + 1) (x² - 1) = (x² + 1) (x - 1) (x + 1) Factoring Polynomials EXAMPLES STEP 1 : Greatest Common Factor GCF Factor out the greatest common factor. 6x³ + 12x²y = 6x² (x + 2y) 5x - 5 = 5 (x - 1) 7x² + 2y² = 1 (7x² + 2y²) 2x (x - 3) - (x - 3) =

### Math 25 Activity 6: Factoring Advanced Instructor! Math 25 Activity 6: Factoring Advanced Last week we looked at greatest common factors and the basics of factoring out the GCF. In this second activity, we will discuss factoring more difficult

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### HIBBING COMMUNITY COLLEGE COURSE OUTLINE HIBBING COMMUNITY COLLEGE COURSE OUTLINE COURSE NUMBER & TITLE: - Beginning Algebra CREDITS: 4 (Lec 4 / Lab 0) PREREQUISITES: MATH 0920: Fundamental Mathematics with a grade of C or better, Placement Exam,

### Factoring Polynomials UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials A polynomial is an expression that has variables that represent numbers. A number can Factorising quadratics An essential skill in many applications is the ability to factorise quadratic expressions. In this unit you will see that this can be thought of as reversing the process used to

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### SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014)) SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014)) There are so far 8 most common methods to solve quadratic equations in standard form ax² + bx + c = 0.

### 3 1. Note that all cubes solve it; therefore, there are no more Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

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### Unit 3: Day 2: Factoring Polynomial Expressions Unit 3: Day : Factoring Polynomial Expressions Minds On: 0 Action: 45 Consolidate:10 Total =75 min Learning Goals: Extend knowledge of factoring to factor cubic and quartic expressions that can be factored

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### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form Goal Graph quadratic functions. VOCABULARY Quadratic function A function that can be written in the standard form y = ax 2 + bx+ c where a 0 Parabola Quarter I: Special Products and Factors and Quadratic Equations Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Time Frame: 20 days Time Frame: 3 days Content Standard: 7-2 Factoring by GCF Warm Up Lesson Presentation Lesson Quiz Algebra 1 Warm Up Simplify. 1. 2(w + 1) 2. 3x(x 2 4) 2w + 2 3x 3 12x Find the GCF of each pair of monomials. 3. 4h 2 and 6h 2h 4. 13p and 26p