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Factoring

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MATH 102 – College Algebra

FACTORING polynomials Factoring polynomials is simply the reverse process of the special product formulas. Thus, the reverse process of special product formulas will be used to factor polynomials. To factor polynomials will mean to express it as a product of positive integral powers of distinct prime factors. TYPES OF FACTORING Type 1. Common Monomial Factor Examples: 3x2 (4 – 3x) 1. 12x2 – 9x3 = 2. -10a6b5 – 15a4b6 – 20a3b4 Solution: The common factor of -10, -15 and -20 is -5; For a6, a4, a3, the common factor is a3 For b5, b6 and b4, the common factor is b4 Therefore, the common monomial factor of the given polynomial is -5a3b4 After getting the common monomial factor, divide the given polynomial by this to get the other factor. -10a6b5 – 15a4b6 – 20a3b4 = -5a3b4(2a3b + 3ab6 + 4)

ax + ay = a(x + y)

Type 2. Difference of Two Squares The difference of two squares is equal to the product of the sum and difference of the square roots of the terms.

x – y = (x + y) (x – y)

2

2

Examples: 1.

x 2 − 9 =  x 2 − 9   x 2 + 9  = (x - 3) (x + 3)      

2. 16x4 – 81y4 = (4x2)2 – (9y2)2 = (4x2 – 9y2) (4x2 + 9y2) = (2x – 3y) (2x – 3y) (4x2 + 9y2) 3. –ax2 + 9a = -a (x2 – 9) = -a (x – 3)(x + 3)

Type 3. Perfect Square Trinomial The square of any binomial is a perfect square trinomial where the first and the last terms are the square of the first and square of last term of the binomial and the other term is a plus (or minus) the product of the first and the product of the first and last term of the binomial. x + 2xy + y = (x + y) 2 2 2 x – 2xy + y = (x – y)
2 2 2

To check if the trinomial is a perfect square trinomial, • two terms should be perfect squares • the other term should be a plus or minus twice the product of the square roots of the other terms.

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Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I

| Bataan Peninsula State University

MATH 102 – College Algebra

Examples: 1. Find the factors of 4x2 – 20xy + 25y2. Solution: Check if the given is perfect square trinomial: Two terms are perfect squares, 4x2 and 25y2. The other term, -20xy is the product of the square root of 4x2, which is 2x and the square root of 25y2 which is 5y. Therefore, 4x2 – 20xy + 25y2 = (2x – 5y)2 2. 9a2 + 30ab + 25b2 = (3a + 5b)2 3. 100x3 – 220x2 + 121x (get the common factor first) = x (100x2 – 220x + 121) = x (10x – 11)2 EXERCISES: Factor the following polynomials:

A. 1. 2. 3. 4. 5. 6. 7. 8. 9. 3x2 – 9x 6x2y – 12xy2 4a5b2 – 16a3b4 15x3y2 + 20xy3 – 5xy2 3a3 – 9ab2 2x2 – 8x4y2 + 2xy3 3x2y – 9xy2 + 36xy -2x2y3 + 8xy2 – 4x3y2 12xy3 + 24xy2 – 30x2y4

B. 1. 2. 3. 4. 5. 6. 7. 8. 9. y2 – 169 a6 – 16c6 a2b2 – 121 2k2 – 8 m4 – b4 12x3 – 75x x2k – 1 (a + b)2 – 25
1 2 x − 25 4

C. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. m2 – 6m + 9 2k3 -4k2 +2k 4y2 – 20yz + 25z2 25d2 – 80d + 64 2y2 – 20x + 50 9v4 – 24v2w + 16w2 9x4 – 6x2y2 + y4 98x2 – 84xy + 18y2 100p3 – 220p2 + 121p
1 2 x + 3x + 9 4

10. b2ncn – bnc2n

6
Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I

| Bataan Peninsula State University

MATH 102 – College Algebra

Type 4. Sum and Difference of Two Cubes By long division, we could verify that

x3 + y 3 = x2 − xy + y 2 and x+y
2 2

x3 − y 3 = x2 + xy + y 2 x−y

x + y = (x + y)(x – xy + y ) 3 3 2 2 x – y = (x – y)(x + xy + y ) Examples: 1. 27x3 – 8y3 = (3x – 2y) (9x2 + 6xy + 4y2) 2. a3 + 64 = (a + 4)(a2 – 4a + 16) 3. 125b4c – bc4 = bc(125b3 – c3) = bc(5b – c)(25b2 + 5bc + c2) Type 5. Other Trinomials (Trial & Error Method) Certain trinomials of the form x2 + (a +b)x + ab, can only be factored by trial and error method because the sum of the products of the means and the extremes of the factors should be equal to the middle (linear) term of the given trinomial. x + (a +b)x + ab = (x + a) (x + b) 2 acx + (ad + bc)x + bd = (ax + b)(cx + d)
2

3

3

Examples: 1. Factor x2 – 2x – 8. Solution: Find the value of a and b where x2 + (a +b)x + ab = (x + a) (x + b) = x2 – 2x – 8 ab = - 8 , thus a and b has opposite signs (a + b) = - 2, the factors of – 8 that will give a sum of 2 are – 4 and 2. The factors, therefore, are (x – 4) (x + 2). x2 – 2x – 8 = (x – 4) (x + 2) 2. 15x2 + 2x – 8 Solution: Find the value of a,b, c and d where acx2 + (ad + bc)x + bd = (ax + b)(cx + d) = 15x2 + 2x – 8 ac = 15 bd = - 8 (ad + bc) = 2 Possible factors for ac are 3 & 5, -3 & -5, 15 & 1, -15 & -1 Possible factors for bd are -4 & 2, 4 & -2, 8 & -1 and -8 & 1 From these pairs of factors, the pair that will give (ad + bc) = 2 are 3 & 5 and -2 & 4 Therefore, 15x2 + 2x – 8 = (3x – 2)(5x + 4) 3. 6x4 – x2 – 15 Solution:

Find the value of a,b, c and d where acx2 + (ad + bc)x + bd = (ax + b)(cx + d) = 6x4 – x2 - 15 ac = 6 bd = - 15 (ad + bc) = -1 Possible factors for ac are 3 & 2, 6 & 1 Possible factors for bd are -3 & 5, 3 & -5, 15 & -1, -15 & 1 From these pairs of factors, The pair that will give (ad + bc) = -1 are 3 & 2 and -5 & 3

Therefore, 6x4 – x2 – 15 = (3x2 – 5)(2x2 + 3)

7
Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I

| Bataan Peninsula State University

MATH 102 – College Algebra
Type 7. Factoring By Grouping Sometimes proper grouping of terms is necessary to make the given polynomial factorable. After terms are grouped, a complicated expression may be factored easily by applying Types 1 to 5 formulas. This type of factoring is usually applied to algebraic expressions consisting of at least four terms. Examples: 1. 3x(a – b) + 4y(a – b) =(a – b)(3x + 4y)

common factor 2. bx + by + 2hx + 2hy = = = Another solution: = = = 3. 3x(2a – b) + 4y(b – 2a) 4. xz – kx + kw – wz = = = = = = = = = (bx + 2hx) + (by + 2hy) x(b + 2h) + y(b + 2h) (b + 2h)(x + y) 3x(2a – b) – 4y(2a – b) (2a – b)(3x – 4y) (xz – kx) – (wz – kw) x(z – k) – w(z – k) (z – k)(x – w) (ab3 – 3b2) – (4a – 12) b2(a – 3) – 4(a – 3) (a – 3) (b2 – 4) (a – 3) (b – 2) (b+2) alteration of sign (bx + by) + (2hx + 2hy) b(x + y) + 2h(x + y) (x + y)(b+ 2h) grouping removal of common factor from each group

5. ab3 – 3b2 – 4a + 12

Type 8. Factoring By Addition or Subtraction of Suitable Terms This type of factoring is usually applied to polynomials of degree 4 with two terms being perfect squares and both preceded by positive sign. Through addition or subtraction of suitable terms, the given will always lead to the difference of two squares. Examples: 1. v4 + 4 (a polynomial of degree 4, with two terms being perfect square and are both preceded positive sign.) to factor: we add + 4v2 and – 4v2 , thus, the value of the polynomial is not changed because + 4v2 and – 4v2 is equal to 0. = (v4 + 4v2 + 4) – 4v2 v4 + 4v2 + 4 – 4v2 = (v2 + 2)2 – (2v)2 (difference of two squares) = [(v2 + 2) + 2v] [(v2 + 2) – 2v] = (v2 +2v + 2)(v2 – 2v + 2) (z4 and 9 are perfect squares) we add + z2 and – z2 to make z4 + 5z + 9 a perfect trinomial square z4 + 5z2 + z2 + 9 – z2 = (z4 + 6z + 9) – z2 = (z2 + 3)2 – z2 = (z2 + 3 + z)(z2 + 3 – z) = (z2 + z + 3)(z2 –z + 3)

2. z4 + 5z2 + 9

EXERCISES: Factor the following polynomials completely: A. Factoring by Grouping 1. 2. 3. 4. xy + 2x + y + 2 uv +3u + 2v + 6 8rs + 4r + 2s + 1 ac – ad + bc – bd 5. 6. 7. 8. x3 + 2x2 + x + 2 a2 – ab + a – b 2c2 + 4cd – 3c – 6d xy + x – 2y2 – 2y

8

Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I

| Bataan Peninsula State University

MATH 102 – College Algebra
9. x3 – x2 – x – 1 10. 4x3 – 4x2 – x + 1

B. Factoring by Addition or Subtraction of Suitable Terms 1. 2. 3. 4. 5. x4 + x2 + 1 a4 + 4a2 + 16 x4 – 10x2 + 9 b4 + 5b2 + 9 m4 – 7m2 + 9 6. 7. 8. 9. 10. y4 – 14y2 + 25 36r4 + 15r2 + 4 k4 – 21k2 + 36 4b4 + 11b2 + 9 m4 + 3m + 4

9
Prepared by Mrs. Koni Gutierrez Cruz Assistant Professor I

| Bataan Peninsula State University

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