All Polynomials are a specific type of Multinomial and Polynomials are where Factoring is applied to solve math problems. Polynomials, like integers, can be decomposed into Prime Factors:

5x2 − 25 = 5 (x2 − 5) = 5 (x − √5) (x + √5)

9x2 − 16y2 = (3x + 4y) (3x − 4y)

Keep in mind that we (often) work using quantities in Algebra. The (3x + 4y) is a quantity and so is (3x − 4y). They cannot be decomposed farther, so they are Prime, even though the quantities contain integers that are not prime numbers.

x2 − y2 = (x +y) (x − y)

This Factoring Method is known as the Difference of Squares.

To demonstrate this method, factor: 3x3 − 48x

3x3 − 48x = 3x (x2 − 16) = 3x (x + 4) (x − 4)

You might be asked to factor Polynomials of the form:

a2x2 − b2y2

a2x2 = aaxx = (ax) (ax) = (ax)2

Similar:

b2y2 = (by)2

Thus,

a2x2 − b2y2 = (ax)2 − (by)2 = (ax + by) (ax − by)

As long as the power of a number is even the Difference of Squares method can be applied.

The Difference of Squares method can also apply twice in the same math problem:

x4 − y4 = (x2 + y2) (x2 − y2) = (x2 + y2) (x + y) (x − y)

Trinomial Factoring Form: x2 + Mx + N

(Where M and N are integers greater than zero),

find integers “a” and “b” so that: a + b = M and ab = N

x + a

x + b

x2 + ax

+ bx + ab

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

Example: x2 + 5x + 4

Look at the constant 4 and find two positive integers, “a” and “b”, whose product is 4:

{2, 2} and {4, 1}

Next, look at the coefficient of the middle term, it is 5. Take the positive integer set that when its integers are added provide a sum equal to the middle term coefficient:

4 + 1 = 5

{4, 1}

Result:

(x + 4) (x + 1) = x2 + 5x + 4

“Whenever we have more than one term in any quantity, each term of each quantity must be multiplied by each term of the other quantity once to obtain the product.”

To Factor a Monomial and a Binomial:

(2x) (x + 2) = 2xx + (2x) (2) = 2x2 + 4x

To Factor Two Binomials:

(x + 4) (x + 1) = xx + (x) (1) + (x) (4) + 4 = x2 + 1x + 4x + 4 = x2 + 5x + 4

By factoring x2 + Mx + N so that: x2 + (a + b) x + ab

This table is useful to determine the signs of “a” and “b”:

M = a + b

positive

negative

positive

negative

N = ab

positive

positive

negative

negative

a and b

both positive

both negative

one positive, one negative

one positive, one negative

Trinomial Factoring Form: Lx2 + Mx + N

(Where L, M and N are integers and L not equal 1)

Example: 2x2 + 5x + 2

(L = 2, M = 5, N = 2 of factoring form Lx2 + Mx + N)

First, find the factors of the x2 coefficient (L):

{2, 1} {1, 2}

Next, try each set substituting factor set values into “a” and “b”.

{2, 1}:

(2x + 2) (x + 1) = 2x2 + (2 + 2) x + 2 ≠ 2x2 + 5x + 2

The correct solution set is:

{1, 2}:

(2x + 1) (x + 2) = 2x2 + (4 + 1) x + 2 = 2x2 + 5x + 2

Example: 6x2 − 17 + 5

(L = 6, M = −17, N = 5 of factoring form Lx2 + Mx + N)

Note that M < 0 and N > 0 so “a” and “b” must both be negative

First, find the factors of the x2 coefficient (L):

{6, 1} {3, 2}

Next, find the factors of the constant N where ab = N:

(Note that M < 0 and N > 0 so “a” and “b” must both be negative [From table above])

{−5, −1} {−1, −5}

Next, try each set substituting factor set values into “a” and “b”.

{6, 1} and {−5, −1} gives: (6x − 5) (1x − 1) = 6x2 − 6x − 5x + 5 = 6x2 − 11x + 5

{6, 1} and {−1, −5} gives: (6x − 1) (1x − 5) = 6x2 − 30x − 1x + 5 = 6x2 − 31x + 5

{3, 2} and {−5, −1} gives: (3x − 5) (2x − 1) = 6x2 − 3x − 10x + 5 = 6x2 − 13x + 5

{3, 2} and {−1, −5} gives: (3x − 1) (2x − 5) = 6x2 − 15x − 2x + 5 = 6x2 − 17x + 5

From this factoring exercise we see that 6x2 − 17x + 5 = (3x − 1) (2x − 5)

Sometimes by grouping terms provides insight so that we can factor polynomials. To accomplish this rearrange terms and group together terms having a common factor.

Factor: ax + ay + bx + by

First, regroup terms together with common factors.

ax + ay + bx + by = (ax + ay) + (bx + by) = a (x + y) + b (x + y)

Now we see the common factor of “a” and “b” is (x + y).

It enables us to again group factors as:

(a + b) (x + y)

To better understand this group factoring method,

Factor: a2x2 − 9a2 − 4x2 − 36

a2 (x2 − 9) − 4 (x2 − 9) =

(a2 − 4) (x2 − 9) =

(a + 2) (a − 2) (x − 3) (x + 3)

The result is prime factors.

“Do not confuse the Sum and Difference of Cubes with a Perfect Binomial Cube!”

Both the sum and difference of cubes can be factored. Additionally, perfect binomial cubes can be factored. This makes the Cube Formula in actual fact four formulas. With squares, the difference of squares can be factored: a2 − b2 = (a + b) (a − b). The sum of squares cannot be factored: a2 + b2

Sum of Cubes:

a3 + b3 = (a + b) (a2 − ab + b2)

Difference of Cubes:

a3 − b3 = (a − b) (a2 + ab + b2)

To factor these cubes we first write the quantity, either sum (a + b) or difference (a − b), then when writing the second quantity reverse the first math sign in that quantity. Thus, for the sum of cubes the second quantity is always (a2 − ab + b2) and for the difference of cubes the second quantity is always (a2 + ab + b2).

Sum of Cubes example

Factor: 64x3 + 8y3

The cube sum formula is: a3 + b3 = (a + b) (a2 − ab + b2):

Let a3 = 64x3 and b3 = 8y3, then a = 4x and b = 2y

64x3 + 8y3 = (4x + 2y) (16x2 − 4x 2y + 4y2)

64x3 + 8y3 = (4x + 2y) (16x2 − 8xy + 4y2)

… Depending on the circumstance we sometimes need to factor farther:

(4x + 2y) (16x2 − 8xy + 4y2) =

(4x + 2y) (16x2 − 8xy + 4y2) =

2 (2x + y) 4 (4x2 − 2xy + y2) =

8 (2x + y) (4x2 − 2xy + y2)

Difference of Cubes example

Factor: x3 − 27

The cube difference formula is:

a3 − b3 = (a − b) (a2 + ab + b2)

Let a3 = x3 and b3 = 27, then a = x and b = 3

x3 − 27 = (x − 3) (x2 + 3x + 9)

If we raise the power of a number to three we are computing the cube of that number. For example: 33 = 9 and 53 = 125. Thus, the cube of a binomial (a + b) is (a + b)3. These binomials can be factored by the following formulas that give us special products called Perfect Binomial Cubes:

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

(a − b)3 = a3 − 3a2b + 3ab2 − b3

Factor the Binomial Cube: (2x + 3)3

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

Let a = 2x and b = 3

Then,

(2x + 3)3 =

(2x)3 + 3 (2x)2 3 + 3 2x 32 + 33 =

8x3 + 36x2 + 54x + 27

Find the Binomial Cube of: x3 − 6x2 + 12x − 8

If (a − b)3 = a3 − 3a2b + 3ab2 − b3

Let a = 3√x3 = x and b = 3√8 = 2

Then,

x3 − 6x2 + 12x − 8 = (x − 2)3

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