# FACTORING TRINOMIALS - METHOD OF FACTORING QUADRATIC EXPRESSIONS, EXAMPLES, EXERCISE

Algebra Factoring before Factoring Trinomials,
if you have not already done so.

There, we explained Factors or Divisors,
Factoring Polynomials, Prime Polynomial etc.

That knowledge is a prerequisite here.

There, we have seen some special products as Algebra
Formulas and we have also seen their proofs.

Now we make use of two of those Formulas
to find the factors, given the product.

Observing those two Formulas, leads us to the
method of Factorization of Quadratic Expressions.

We discuss the steps involved in the method and
apply it to solve a number of problems.

We present a number of solved examples
and give problems for practice in exercise.

Look at the Formulas 10 and 11 in
Factoring Special Products
which are reproduced below.

Formula 10 in Factoring Special Products :

Simple Quadratic Polynomial as Product of Two Factors:

x2 + x(a + b) + ab = (x + a)(x + b)

Formula 11 in Factoring Special Products :

General Quadratic Polynomial as Product of Two Factors:

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

Observing these two Formulas, we can say that if
the numerical product of the cofficient of x2 and constant term
be expressed as two numerical factors such that
their sum is equal to the coefficient of x, then
we can write the factors of the Quadratic Polynomial.

This leads us to a method.

## Method of Factoring Trinomials (Quadratics)

Step 1 :

Multiply the coefficient of x2 by the constant term.

Step 2 :

Resolve this product into two factors such that
their sum is the coefficient of x

Step 3 :

Rewrite the x term as the sum of
two terms with these coeffiecients.

Step 4 :

Then take the common factor in the
first two terms and the last two terms.

Step 5 :

Then take the common factor from the two terms thus formed.
What you get in step 5 is the product of the required two factors.

The method will be clear by the following Soved Examples.

### Example 1 of Factoring Trinomials (Quadratics)

Factorize 9x2 + 26x + 16

Solution to Example 1 of Factoring Trinomials (Quadratics) :

Let P = 9x2 + 26x + 16

In Factoring of Trinomials (Quadratics) , follow the five steps listed above.

Step 1: Coefficient of x2 x constant term
= 9 x 16 = 144

Step 2: We have to express 144 as two factors whose sum
= coefficient of x = 26;
144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 8 x 18; (8 + 18 = 26)

Step 3: P = 9x2 + 26x + 16 = 9x2 + 8x + 18x+ 16

Step 4: P = x(9x + 8) + 2(9x + 8)

Step 5: P = (9x + 8)(x + 2)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 9x2 + 26x + 16, we get the Factors as (9x + 8)(x + 2) Ans.

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### Example 2 of Factoring Trinomials (Quadratics)

Factorize 2 - 5x - 18x2

Solution to Example 2 of Factoring Trinomials (Quadratics) :

Let P = 2 - 5x - 18x2 = -18x2 - 5x + 2

In Factoring of Trinomials (Quadratics) , follow the five steps listed above.

Step 1: Coefficient of x2 x constant term
= -18 x 2 = -36

Step 2: We have to express -36 as two factors whose sum
= coefficient of x = -5 ;
-36 = -2 x 18 = -2 x 2 x 9 = 4 x -9; [4 + (-9) = -5]

Step 3: P = -18x2 - 5x + 2 = -18x2 + 4x - 9x+ 2

Step 4: P = 2x(-9x + 2) + 1(-9x + 2)

Step 5: P = (-9x + 2)(2x + 1)

Thus, Applying the Method of Algebra Factoring of Quadratic Polynomial to 2 - 5x - 18x2, we get the Factors as (-9x + 2)(2x + 1) Ans.

### More Solved Examples on Factoring Trinomials (Quadratics)

The following Links take you to
more solved Examples and Exercise
problems for practice.

Set 2 of Solved Examples

Set 3 of Solved Examples

Set 4 of Solved Examples

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## Exercise : Factoring Trinomials (Quadratics)

1. Factorize 8x2 + 25x + 18
2. Factorize 3 - 2x - 21x2

For Answers See at the bottom of the Page..

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