# FACTORING BINOMIALS USING ALGEBRA FORMULAS, SOLVED EXAMPLES, EXERCISE PROBLEMS WITH ANSWERS

Algebra Factoring before Factoring Binomials,
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 some of those Formulas
to find the factors, given the product.

We list out the required Formulas from there
with L.H.S. and R.H.S. reversed.

We have to remember these Formulas from
L.H.S. to R.H.S. and from R.H.S. to L.H.S.
and be able to apply them in both directions.

## Algebra Formulas used for Factoring Binomials

Formula 1 in Algebra Factoring:

Difference of Two Squares as Product of Sum and Difference:

a2 - b2 = (a + b)(a - b)

Formula 2 in Algebra Factoring:

Sum of Two Cubes as Product of Two Factors:

a3 + b3 = (a + b)(a2 - ab + b2)

Formula 3 in Algebra Factoring:

Difference of Two Cubes as Product of Two Factors:

a3 - b3 = (a - b)(a2 + ab + b2)

Here a, b, c, d, x are all real numbers.

Each of the letters in fact represent a TERM.

e.g. The above Formula 1 can be stated as

(First term)2 - (Second term)2
= (First term + Second term)(First term - Second term)

Similarly in other Formulae also, we can
replace each of the letters by a TERM.

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## Example 1 of Factoring Binomials :

Solved Example 1 of Factoring Binomials :

Factorize 7a4 + 28

solution to Example 1 of Factoring Binomials :

Let P = 7a4 + 28 = 7(a4 + 4) = 7{(a2)2 + 22}
Here, inside the bracket, (first term)2 + (second term)2 is present.
Let us add and subtract 2(first term)(second term),
so that the value does n't change.
i.e. Adding and subtracting 2(a2)(2), we get
P = 7{(a2)2 + 22 + 2(a2)(2) - 2(a2)(2)}
= 7{(a2 + 2)2 - 2(a2)(2)} = 7 {(a2 + 2)2 - (2a)2}
This is the difference of the squares of two terms which is equal to theproduct of the sum and difference of the terms.[See Formula 3]
∴ P = 7{(a2 + 2) + (2a)}{(a2 + 2) - (2a)}
P = 7(a2 + 2a + 2)(a2 - 2a + 2).

Thus Algebra Factoring of 7a4 + 28 by using Algebra Formulas gave the factors as 7(a2 + 2a + 2)(a2 - 2a + 2). Ans.

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## Example 2 of Factoring Binomials :

Solved Example 2 of Factoring Binomials :

Factorize x12 - 4096

solution to Example 2 of Factoring Binomials :

Let P = x12 - 4096
We know 4096 = 8 x 512 = 8 x 8 x 64 = 84 = (23)4 = 2(3 x 4) = 212
∴ P = x12 - 4096 = x12 - 212 = (x6)2 - (26)2
This is difference of squares of two terms which is equal to the product of the sum and difference of the two terms [See Formula 3].
∴ P = (x6 + 26)(x6 - 26)= {(x2)3 + (22)3}{(x3)2 - (23)2}
The first bracket is sum of two cubes to which we can apply Formula 4 and the second bracket is difference of two squares to which we can apply Formula 3.
We have a3 + b3 = (a + b)(a2 - ab + b2) [See Formula 4]
a2 - b2 = (a + b)(a - b) [See Formula 3]
Applying these here, we get
P = {(x2) + 22)}{(x2)2 - (x2)(22) + (22)2}{(x3) + (23)}{(x3) - (23)}
The last two brackets are sum of two cubes (Formula 4)
and difference of two cubes (Formula 5)
a3 - b3 = (a - b)(a2 + ab + b2) [See Formula 5]
Applying Formula 4 and 5 to last two brakets, we get
{(x3) + (23)} = (x + 2)(x2 - x(2) + 22) = (x + 2)(x2 - 2x + 4)
{(x3) - (23)} = (x - 2)(x2 + x(2) + 22) = (x - 2)(x2 + 2x + 4)
∴ P = (x2 + 4)(x4 - 4x2 + 16)(x + 2)(x2 - 2x + 4)(x - 2)(x2 + 2x + 4).

Thus Algebra Factoring of x12 - 4096by using Algebra Formulas gave the factors as (x2 + 4)(x4 - 4x2 + 16)(x + 2)(x2 - 2x + 4)(x - 2)(x2 + 2x + 4). Ans.

## Exercise : Factoring Binomials

Solve the following problems on Factoring Binomials.

1. Factorize x4 + 4
2. Factorize 729a6 - 64b6
For Answers See at the bottom of the Page.

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## Answers to Exercise : Factoring Binomials

Answers to Problems on Factoring Binomials :

1. (x2 + 2x + 2)(x2 - 2x + 2)
2. (3a + 2b)(9a2 - 6ab + 4b2) (3a - 2b)(9a2 + 6ab + 4b2)

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