## Trinomials

There are two different kinds of trinomials that require similar techniques but one method is easier and the other is harder.

If a=1, then we refer to it as **monic** or an “**easy trinomial**” because the factoring method is easier.

If a\neq 1, then we refer to it as **non-monic** or an “**hard trinomial**” because the factoring method is more difficult.

#### Easy Trinomials

##### How to Factor an Easy Trinomial

Make sure that the trinomial is in **standard form**: ax^2+bx+c and that a=1.

Then, find two numbers that ** multiply to** c and

**b.**

*add to*The easy trinomial will factor to be:

(x+ first number )(x+ second number )

For example: x^2+7x+12

In this case, we must find two numbers that multiply to c=12 and add to b=7. The two numbers would be 3 and 4. So: x^2+7x+12=(x+3)(x+4)

#### Hard Trinomials

##### How to Factor a Hard Trinomial

Make sure that the trinomial is in **standard form**: **ax^2+bx+c** and that a is positive.

A hard trinomial is more difficult to factor than an easy trinomial. There are many different techniques.

One method is called “**decomposition**” which involves splitting the middle ‘bx‘ into two terms with new coefficients. These two coefficients must *multiply to***a \times c** and *add to***b**.

You’ll then be able to “**group/triple common factor**” the four terms.

For example: 3x^2-8x+4

In this case, we must find two numbers that multiply to a \times c=12 and add to b=-8. The two numbers would be -2 and -6. So, we split the -8x term to be -2x and -6x, then group: \begin{array}{rl} & 3x^2-2x-6x+4 \\ = & x(3x-2)-2(3x-2) \\ = & (3x-2)(x-2) \end{array}

#### Perfect Square Trinomials

A **perfect square trinomial** can be factored as a regular trinomial but there is a faster way to factor this special type. If the first term and last term are **perfect squares**, there’s a good chance you have a perfect square trinomial.

##### How to Factor a Perfect Square Trinomial

First, test to see if you have a perfect square trinomial. Take the square root of the first term and square root of the last term. If: |middle \; term| =2 \times \sqrt{first \; term} \times \sqrt{last \; term}

Then, you have a perfect square trinomial. It factors to: \Big( \sqrt{first \; term} + sign \; of \; b + \sqrt{last \; term} \Big)^2

For example: 16x^2-24x+9

Notice that: \sqrt{16x^2}=4x and /sqrt{9}=3. Also, |-24x|=2(4x)(3). Therefore, it is a perfect square trinomial and: 16x^2-24x+9=(4x-3)^2

### Practice

Factor 2x^2-10x+12

###### SOLUTION

Using the Law of Division: 2^{11} div 2^8=2^{11-8}=2^3

Evaluating: 2^3=8

Factor: 4x^2-20xy+25y^2

###### SOLUTION

Using the Law of Powers: (m^0)^{10}=m^0

We should more properly use the Law of Zero Exponents to write: m^0=1

Factor: 4x^2-5xy-6y^2

###### SOLUTION

Since we are multiplying the two powers, we use the Law of Multiplication: x+4=7

Therefore: x=3

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