MATH 102 College Algebra

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FACTORING FACTORING FACTORING FACTORING

Factoring polynomia Factoring polynomia Factoring polynomia

Factoring polynomialslslsls 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 To factor polynomials 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 TYPES OF FACTORING TYPES OF FACTORING TYPES OF FACTORING

Type 1. Common Monomial Factor Type 1. Common Monomial Factor Type 1. Common Monomial Factor Type 1. Common Monomial Factor

Examples:

1. 12x² – 9x³ = 3x² (4 – 3x) 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)

Type 2. Difference of Two Squares Type 2. Difference of Two Squares Type 2. Difference of Two Squares 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.

Examples:

1.

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 TrinomialType 3. Perfect Square TrinomialType 3. Perfect Square TrinomialType 3. Perfect Square Trinomial

The square of any binomial is a perfect square trinomialperfect square trinomialperfect square trinomial where the first and the last terms are the square of the perfect square trinomial 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.

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.

ax + ay = a(x + y)

x² – y² = (x + y) (x – y)

3) (x 3) - (x x

x

² ²  = +



 



 +



 



 −

=

−9 9 9

x2

x² + 2xy + y² = (x + y)² x² – 2xy + y² = (x – y)²

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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:

EXERCISES: Factor the following polynomials:Factor the following polynomials:Factor the following polynomials: Factor the following polynomials:

⁴

9.

^x ²⁵

4 1 ₂

−

10. b

²ⁿ

c

ⁿ

– b

ⁿ

c

²ⁿ

C. C.

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Type 4. Sum and Difference of Two Cubes Type 4. Sum and Difference of Two Cubes Type 4. Sum and Difference of Two Cubes Type 4. Sum and Difference of Two Cubes

By long division, we could verify that and

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)Type 5. Other Trinomials (Trial & Error Method)Type 5. Other Trinomials (Trial & Error Method)Type 5. Other Trinomials (Trial & Error Method)

Certain trinomials of the form x² + (a +b)x + ab, can only be factored by trialtrialtrial and error method trialand error method and error method because the sum and error method of the products of the means and the extremes of the factors should be equal to the middle (linear) term of the given trinomial.

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)

2 2

3 3

y xy y x

x y

x = − +

+

+ 2 2

3 3

y xy y x

x y

x = + +

−

x³+ y³ = (x + y)(x² – xy + y²) x³ – y³ = (x – y)(x² + xy + y²)

x² + (a +b)x + ab = (x + a) (x + b)

acx² + (ad + bc)x + bd = (ax + b)(cx + d)

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Type 7. Factoring By Grouping Type 7. Factoring By Grouping Type 7. Factoring By Grouping 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)

2. bx + by + 2hx + 2hy = (bx + by) + (2hx + 2hy) grouping

= b(x + y) + 2h(x + y) removal of common factor from each group

= (x + y)(b+ 2h) Another solution:

= (bx + 2hx) + (by + 2hy)

= x(b + 2h) + y(b + 2h)

= (b + 2h)(x + y)

3. 3x(2a – b) + 4y(b – 2a) = 3x(2a – b) – 4y(2a – b) alteration of sign

= (2a – b)(3x – 4y) 4. xz – kx + kw – wz = (xz – kx) – (wz – kw)

= x(z – k) – w(z – k)

= (z – k)(x – w)

5. ab³ – 3b² – 4a + 12 = (ab³ – 3b²) – (4a – 12)

= b²(a – 3) – 4(a – 3)

= (a – 3) (b² – 4)

= (a – 3) (b – 2) (b+2)

Type 8. Factoring By Addition or Subtraction of Suitable TermsType 8. Factoring By Addition or Subtraction of Suitable TermsType 8. Factoring By Addition or Subtraction of Suitable TermsType 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) 2. z⁴ + 5z² + 9 (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)

EXERCISES:

EXERCISES: Factor the following polynomials completely:Factor the following polynomials completely:Factor the following polynomials completely: Factor the following polynomials completely:

common factor

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9. x³ – x² – x – 1 10. 4x³ – 4x² – x + 1

B. Factoring by Addition or Subtraction of Suitable Terms 1. x⁴ + x² + 1

2. a⁴ + 4a² + 16 3. x⁴ – 10x² + 9 4. b⁴ + 5b² + 9 5. m⁴ – 7m² + 9

6. y⁴ – 14y² + 25 7. 36r⁴ + 15r² + 4 8. k⁴ – 21k² + 36 9. 4b⁴ + 11b² + 9 10. m⁴ + 3m + 4