PRE-CALCULUS FACTORING REVIEW |

**A/ FACTORING**

**Factoring is a basic skill** and must be mastered with efficiency. The easiest way to do that is to remember and **use the words that describe the different factoring methods** we use to turn messy polynomials into pretty products. Once we know that P(*x*) = (*x* – a)(*x* – b)(*x* – c), we know the zeros of the polynomial and can graph it quite easily. We can also easily locate the maximums and minimums of the function. This lessons covers 5 different factoring methods.

1) Common Factor |

2) Difference of Squares |

3) Sum or Difference of Cubes |

4) Trinomials |

5/ The Factor Theorem |

Exercise 1 |

Solutions for Exercise 1 |

.

**ex 1a:** a*x* – a*y* + a*z* = a(*x* – *y* + *z*)

**ex 1b:** 2b*x* + 4b*y* – 18b*z* = 2b(*x* + 2*y* – 9*z*)

**Group Common Factor**

**ex 2a:** a*x* + a*y *– b*x* – b*y* = a(*x* +* y*) – b(*x* + *y*) = (a – b)(*x* + *y*)

**ex 2b:** 8*x*^{2} + 16*y*^{2} – *x*^{3} – 2*xy*^{2} = 8(*x*^{2} + 2*y*^{2}) – *x*(*x*^{2} + 2*y*^{2}) = (8 – *x*)(*x*^{2} + 2*y*^{2})

.

**ex 3a:** *x*^{2} – *y*^{2} = (*x* + *y*)(*x* – *y*)

difference of squares = (sum of the roots)(difference of the roots)

**ex 3b:** 4*a*^{2} – 16*b*^{2} = 4(*a*^{2} – 4*b*^{2}) = 4(*a* – 2*b*)(*a* + 2*b*) -- (a combo of common and diff of sqs.)

.

**3) Sum or Difference of Cubes:**

**ex 4:** *x*^{3} + *y*^{3} = (*x* + *y*)(x^{2} – *xy* + *y*^{2})

**ex 5:** *x*^{3} – *y*^{3} = (*x* – *y*)(*x*^{2} + *xy* + *y*^{2})

**n.b.:** to factor a sum or difference of cubes we make 2 brackets.

in the first: put the **cube roots** and the **same sign**

in the 2nd: **square** the 1st term, **multiply** the terms and **change the sign**,

**square** the last term in the first bracket.

**Examples**:

a) sum of cubes 27 (3 |
b) difference of cubes 64 – 125 (4 – 5 |

c) difference of squares, then sum and difference of cubes.
( ( |

.

There are 3 types of trinomials: **perfect square, simple** and **complex trinomials**.

A **perfect square trinomial** is the square of a binomial.

*x*² – 8*x* + 16 = (*x* – 4)² and 4*x*² + 12*x* + 9 = (2*x* + 3)²

In a **simple trinomial** the coefficient of *x*² is always equal to 1.

In a **complex trinomial** the coefficient of *x*² is not equal to 1.

Perfect squares can be either simple or complex trinomials.

.

**4a) Perfect Square Trinomials:**

The **1st and last** terms are **perfect squares** and the **middle** term is **double** the product of **the roots** of those perfect squares. Since all perfect squares are positive in the set of Real numbers, the sign before the 3rd term will always be +. The sign of the middle term tells us if the binomial is a sum or difference.

**ex 6:** *x*² + 6*xy* + 9*y* ² = (*x* + 3*y*)(*x* + 3*y*) = (*x* + 3*y*)² : the binomial is a sum.

but: 4*x* ² – 20*xy* + 25*y* ² = (2*x* – 5*y*)² : the binomial is a difference.

**ex 7a:** Here, the first 3 terms are a perfect square binomial.

*x* ² + 6*xy* + 9*y* ² – 16 = (*x* + 3*y*)² – 4^{2} = (*x* + 3*y* + 4)(*x* + 3*y* – 4)

Then it's a difference of squares.

**ex 7b:** Here, the last 3 terms are a perfect square binomial when we factor out – 1.

4*a* ² – 9*b* ² – 24*b* – 16 = 4*a* ² – (9*b* ² + 24*b* + 16) = 4*a* ² – (3*b* + 4)²

- we get [2

**n.b.** when the perfect square trinomial is after the minus sign, use inner brackets in your expression of the factors.

.

**4b/ Simple Trinomials**

The coefficient of the squared term is always = 1

example 8:
we need factors of 6 that |
example 9:
we need factors of 6 that |
example 10:
we need factors of 6 that |

**Notes:**

if the **3rd term's sign is positive**, both brackets have the **same sign** as the middle term. The inners and outers **ADD** to give the middle term.

If the **3rd term's sign is negative**, the brackets have **opposite sign**s. The inner and outer products **SUBTRACT** to give the middle term.

**4c/ Complex Trinomials**

The coefficient of the squared term is not 1.

Watch for prime numbers, they're easy to factor.

example 11: 2 outers + inners = – 5 |
example 12: 3 outers + inners = 11 |
example 13: 5 difference of outers & inners |

.

When we need to factor a polynomial of degree 3 or higher, we use the factor theorem. It says if we can **find a value for x that makes the polynomial equal 0**, then

andThe Factor TheoremIf P( x) = a_{0} + a_{1 }x + a_{2 }x ² + ..... + a_{n }x^{ n} P(thena) = 0, (andx – a) is a factor of P(x) P(x) = (x – a) [g(x)] |

**example 14:** factor 3*x *³ + 7*x*² – 2*x* – 8

If we set *a = *–* 2* the expression = 0. This means that it's divisible by (*x + 2*).

Now the quotient, 3*x*² + *x* – 4 = (3*x* + 4)(*x* – 1)

so, 3*x*³ + 7*x*² – 2*x* – 8 = (*x + *2)(3*x* + 4)(*x* – 1)

**Note:** if P(*x*) is missing powers, make sure to enter 0*x *^{n} where *n* is a missing power when dividing,. For example, 3* x *³ – 2*x* – 8 will become 3*x*³ + **0***x***²** – 2*x* – 8 so that we can line up the like terms that result from multiplying divisor by quotient.

.

Factor completely:

1) cd + cdx |
2) x³ + 3x² + x |
3) tx² – 4txy + 4txy² |

4) 2m ² – 6m + mn – 3n |
5) 9x ² + 6xy + y ² – 16 |
6) b^{4} – 16 |

7) 81 – m^{4} |
8) a²x ² – x ² – a ² + 1 |
9) 9a^{2} – 12ab + 4b² |

10) 6 – 19bc + 3b^{2}c² |
11) 6y^{2} + 11y + 3 |
12) 3t^{2} – 22t + 7 |

13) x^{3} – 27y³ |
14) 24c^{4} – 81cd³ |
15) 64 – x^{6} |

16) 9 – x^{2} – 2xy – y² |
17) 16r^{2} + 28st – 4s^{2} – 49t² |
18) 9a^{2} + 4bc – 4c^{2} – b² |

19) 30xy – 16z + 9^{2}x^{2} + 25y² |

20) Use the factor theorem to find the zeros of these polynomials.

a) P(x) = x³ + 4x² + x – 6 |
b) P(x) = x³ + 5x² – 2x – 24 |
c) P(x) = x³ – 6x² + 3x + 10 |

.

1) cd(1 + x) |
2) x(x² + 3x + 1) |
3) tx(x – 4y + 4y²) |

4) (2m + n) (m – 3) |
5) (3x + y + 4)(3x + y – 4) |
6) (b – 2)(b + 2)(b² + 4) |

7) (3 – m)(3 + m)(9 + m²) |
8) (a – 1)(a + 1)(x – 1)(x + 1) |
9) (3a – 2b)² |

10) (6 – bc)(1 – 3bc) |
11) (2y + 3)(3y + 1) |
12) (3t – 1)(t – 7) |

13) (x – 3y)(x² + 3xy + 9y²) |
14) 3c(2c – 3d)(4c² + 6cd + 9d²) | |

15) (2 – x)(4 + 2x + x²)(2 + x)(4 – 2x + x²) |
16) (3 + x + y)(3 – x – y) | |

17) (4r + 2s – 7t)(4r – 2s + 7t) |
18) (3a – b + 2c)(3a + b – 2c) |
19) (3x + 5 + 4y)(3x + 5 – 4y) |

20) Use the factor theorem to find the zeros of these polynomials.

a) P(1) = 0 so (*x* – 1) is a factor.

the factors are (*x* – 1)(*x* + 3)(*x* + 2) and the zeros are 1, – 3 and – 2.

b) P(2) = 0 so (*x* – 2) is a factor.

the factors are (*x* – 2)(*x* + 3)(*x* + 4) and the zeros are 2, – 3 and – 4.

.

c) P(2) = 0 so (*x* – 2) is a factor.

the factors are (*x* – 2)(*x* + 1)(*x* – 5) and the zeros are 2, – 1 and 5.

.

*(all content of the MathRoom Lessons **© Tammy the Tutor; 2004 - ).*