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FACTORING SPECIAL PRODUCTS - FACTORING EXPRESSIONS USING ALGEBRA FORMULAS

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Please study
Algebra Factoring before Factoring Special Products,
if you have not already done so.

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

That knowledge is a prerequisite here.

Also, please study Algebra Formulas.

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

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

We list out the same Formulas
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 Special Products

Formula 1 in Algebra Factoring:

Polynomial expressible as Square of Sum of Two Terms:

a2 + 2ab + b2 = (a + b)2

Formula 2 in Algebra Factoring:

Polynomial expressible as Square of Difference of Two Terms:

a2 - 2ab + b2 = (a - b)2

Formula 3 in Algebra Factoring:

Difference of Two Squares as Product of Sum and Difference:

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

Formula 4 in Algebra Factoring:

Sum of Two Cubes as Product of Two Factors:

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

Formula 5 in Algebra Factoring:

Difference of Two Cubes as Product of Two Factors:

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

Formula 6 in Algebra Factoring:

Polynomial expressible as Cube of Sum of Two Terms:

a3 + 3a2b + 3ab2 + b3 = a3 + 3ab(a + b) + b3 = (a + b)3

Formula 7 in Algebra Factoring:

Polynomial expressible as Cube of Difference of Two Terms:

a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3 = (a - b)3

Formula 8 in Algebra Factoring:

Polynomial expressible as Square of Sum of Three Terms:

a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2

Formula 9 in Algebra Factoring:

Sum of Cubes of Three Terms Minus three times the product of the three terms as Product of Two Factors:

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Formula 10 in Algebra Factoring:

Simple Quadratic Polynomial as Product of Two Factors:

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

Formula 11 in Algebra Factoring:

General Quadratic Polynomial as Product of Two Factors:

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

Formula 12 in Algebra Factoring:

Simple Cubic Polynomial as Product of Three Factors:

x3 + x2(a + b + c) + x(ab + bc + ca) + abc = (x + a)(x + b)(x + c)

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

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








Example 1 of Factoring Special Products

Factorize 16x2y2 + 24xyz + 9z2

solution to Example 1 of Factoring Special Products :

Let P = 16x2y2 + 24xyz + 9z2
= (4xy)2 + 2(4xy)(3z) + (3z)2

This is like a2 + 2ab + b2
with (4xy) in place of a and 3z in place of b

We have a2 + 2ab + b2 = (a + b)2 [ See Formula 1]

∴ P = (4xy + 3z)2

Thus Algebra Factoring of 16x2y2 + 24xyz + 9z2
by using Algebra Formulas gave the factors as
(4xy + 3z)2. Ans.




Example 2 of Factoring Special Products

Factorize (2a + 3b)2 - 2(2a + 3b)(a - b) + (a - b)2

solution to Example 2 of Factoring Special Products :

Let P = (2a + 3b)2 - 2(2a + 3b)(a - b) + (a - b)2
= (first term)2 - 2(first term)(second term) + (second term)2
= (first term - second term)2 [See Formula 2]

= {(2a + 3b) - (a - b)}2 = {2a + 3b - a + b}2
= {a + 4b}2

Thus Algebra Factoring of (2a + 3b)2 - 2(2a + 3b)(a - b) + (a - b)2
by using Algebra Formulas gave the factors as
{a + 4b}2 Ans.





Example 3 of Factoring Special Products

Factorize 7a4 + 28

solution to Example 3 of Factoring Special Products :

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 the product 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.



Example 4 of Factoring Special Products

Factorize 16x4 + 4x2 + 1

solution to Example 4 of Factoring Special Products :

Let P = 16x4 + 4x2 + 1 = (4x2)2 + 12 + (4x2)(1)
This looks like a2 + b2 + ab with (4x2) in place of a and 1 in place of b
To make it (a + b)2, we have to add one more ab { = (4x2)(1)} [See Formula 1]
Not to change the value, let us add and subtract.
Adding and subtracting (4x2)(1) to the R.H.S. of P, we get
P = (4x2)2 + 12 + (4x2)(1) + (4x2)(1) - (4x2)(1)
= (4x2)2 + 12 + 2(4x2)(1) - (4x2)(1)
= (4x2 + 1)2 - (2x)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 = {(4x2 + 1) + (2x)}{(4x2 + 1) - (2x)}
= (4x2 + 2x + 1)(4x2 - 2x + 1).

Thus Algebra Factoring of 16x4 + 4x2 + 1 by using Algebra Formulas gave the factors as (4x2 + 2x + 1)(4x2 - 2x + 1). Ans.





Example 5 of Factoring Special Products

Factorize 27x3 + 27x2 + 9x + 1

solution to Example 5 of Factoring Special Products :

Let P = 27x3 + 27x2 + 9x + 1 = (3x)3 + 3(3x)2(1) + 3(3x)(1)2 + (1)3
This looks like a3 + 3a2b + 3ab2 + b3
with (3x) in place of a and 1 in place of b.
We have a3 + 3a2b + 3ab2 + b3 = a3 + 3ab(a + b) + b3 = (a + b)3 [See Formula 6]
Applying this Formula here, we get
P = (3x + 1)3.

Thus Algebra Factoring of 27x3 + 27x2 + 9x + 1 by using Algebra Formulas gave the factors as (3x + 1)3. Ans.




Example 6 of Factoring Special Products

Factorize 64a3 - 336a2 + 588a - 343

solution to Example 6 of Factoring Special Products :

Let P = 64a3 - 336a2 + 588a - 343
We know
64 = 4 x 4 x 4 = 43;
336 = 7 x 48 = 7 x 3 x 16 = 3 x 42 x 7;
588 = 7 x 84 = 7 x 7 x 12 = 72 x 3 x 4 = 3 x 4 x 72;
343 = 7 x 49 = 7 x 7 x 7 = 73
∴ P = (4a)3 - 3(4a)2(7) + 3(4a)(7)2 - 73
This looks like
(first term)3 - 3(first term)2(second term)
+ 3(first term)(second term)2 - (second term)3
which is equal to
(first term - second term)3 [See Formula 7],
with (4a) in place of first term and 7 in place of second term.
Applying this Formula here, we get
P = (4a - 7)3.

Thus Algebra Factoring of 64a3 - 336a2 + 588a - 343 by using Algebra Formulas gave the factors as (4a - 7)3. Ans.




Example 7 of Factoring Special Products

Factorize a6 + 54a3 + 729

solution to Example 7 of Factoring Special Products :

Let P = a6 + 54a3 + 729
We know 729 = 9 x 81 = 9 x 3 x 27 = 27 x 27 = 272; 54 = 2 x 27;
Let t = a3; then a6 = (a3)2 = t2
∴ P = t2 + 2(27)t + (27)2
This looks like a2 + 2ab + b2 which is equal to (a + b)2 [ See Formula 1], with (t) in place of a and 27 in place of b
∴ P = (t + 27)2
(t + 27) = a3 + 33 = (a + 3){a2 - a(3) + 32} [See Formula 4]
= (a + 3)(a2 - 3a + 9)
∴ P = (t + 27)2 = {(a + 3)(a2 - 3a + 9)}2 = (a + 3)2(a2 - 3a + 9)2.

Thus Algebra Factoring of a6 + 54a3 + 729 by using Algebra Formulas gave the factors as (a + 3)2(a2 - 3a + 9)2. Ans.




Example 8 of Factoring Special Products

Factorize x12 - 4096

solution to Example 8 of Factoring Special Products :

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




Example 9 of Factoring Special Products

If x3 + y3 + z3 = 3xyz, show that either x + y + z = 0 or x = y = z

solution to Example 9 of Factoring Special Products :

By data x3 + y3 + z3 = 3xyz ⇒ x3 + y3 + z3 - 3xyz = 0
We have a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) [See Formula 9]
Applying this here, we get
(x + y + z)(x2 + y2 + z2 - xy - yz - zx) = 0
⇒ either (x + y + z) = 0 or (x2 + y2 + z2 - xy - yz - zx) = 0
Let P = (x2 + y2 + z2 - xy - yz - zx)
Multiplying both sides with 2, we get
2P = (2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx)
Writng 2x2 as x2 + x2, 2y2 as y2 + y2 and 2z2 as z2 + z2 and regrouping the terms, we get
2P = {(x2 + y2 - 2xy) + (y2 + z2 - 2yz) + (z2 + x2 - 2zx)}
= {(x - y)2 + (y - z)2 + (z - x)2}
P = 0 ⇒ 2P = 2 x 0 = 0 ⇒ {(x - y)2 + (y - z)2 + (z - x)2} = 0
SUM OF SQUARES OF THREE TERMS IS ZERO ⇒ EACH TERM IS ZERO.
∴ (x - y) = 0, (y - z) = 0 and (z - x) = 0 ⇒ x = y, y = z and z = x
⇒ x = y = z
Thus, If x3 + y3 + z3 = 3xyz, then either x + y + z = 0 or x = y = z.

Thus, by Algebra Factoring of x3 + y3 + z3 - 3xyz,
by using Algebra Formulas, we have proved that
either x + y + z = 0 or x = y = z if x3 + y3 + z3 = 3xyz.





Example 10 of Factoring Special Products

Factorize (4x - 3y)3 + (y - 3x)3 + (2y - x)3

solution to Example 10 of Factoring Special Products :

Let P = (4x - 3y)3 + (y - 3x)3 + (2y - x)3
Let A = (4x - 3y), B = (y - 3x) and C = (2y - x)
Then A + B + C = (4x - 3y) + (y - 3x) + (2y - x) = 4x - 4x + 3y - 3y = 0
We can prove If A + B + C = 0, A3 + B3 + C3 = 3ABC
For proof, see the Example 12 of Algebra Formulas.
∴ A + B + C = 0 ⇒ A3 + B3 + C3 = 3ABC
∴ P = (4x - 3y)3 + (y - 3x)3 + (2y - x)3 = A3 + B3 + C3 and also A + B + C = 0
∴ P = 3ABC = 3(4x - 3y)(y - 3x)(2y - x).

Thus Algebra Factoring of (4x - 3y)3 + (y - 3x)3 + (2y - x)3 by using Algebra Formulas gave the factors as 3(4x - 3y)(y - 3x)(2y - x). Ans.




Exercise : Factoring Special Products

  1. Factorize (ab + cd)2 + 18(ab + cd) + 81
  2. Factorize 49x2 - 28xyz + 4y2z2
  3. Factorize x4 + 4
  4. Factorize 625a4 + 25a2b2 + b4
  5. Factorize 8a3 + 60a2 + 150a + 125
  6. Factorize 125x3 - 75x2 + 15x - 1
  7. Factorize 64x6 + 16x3 + 1
  8. Factorize 729a6 - 64b6
  9. Factorize 27x3 + 64y3 + 36xy -1
  10. Factorize (x - y)3 + (y - z)3 + (z - x)3

For Answers See at the bottom of the Page.

Answers to Exercise : Factoring Special Products

  1. (ab + cd + 9)2
  2. (7x - 2yz)2
  3. (x2 + 2x + 2)(x2 - 2x + 2)
  4. (25a2 + b2 + 5ab)(25a2 + b2 - 5ab)
  5. (2a + 5)3
  6. (5x - 1)3
  7. (2x + 1)2(4x2 - 2x + 1)2
  8. (3a + 2b)(9a2 - 6ab + 4b2) (3a - 2b)(9a2 + 6ab + 4b2)
  9. (3x + 4y - 1)(9x2 + 16y2 + 1 + 12xy - 4y -3x)
  10. 3(x - y)(y - z)(z - x)












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