Math Examples of Factoring Algebra

Prime Factorization can be defined as decomposition of Composite Number into non divisible Numbers. Composite numbers are those numbers which can be further classified. Non divisible numbers are called Prime Numbers that cannot be further decomposed such as 2, 3, 5, etc. Product of prime numbers is equals to composite number. Two methods can be used to obtain the prime factorization of any given number. These are division method and factor Tree method. Let's see how prime factorization of a number with exponents can be obtained. Take 45 and find its prime factorization.

Number 2 cannot be prime factor for 45 since 45 cannot be divided by 2. Let’s try 3 which results in 45 as 3 x 15 = 45.
Thus first factor is ‘3'. Now we have to get factor of 15 which can be 3 and 5 since 3 * 5 = 15.
Since both 3 and 5 are prime factors and cannot be further divided and hence we can write
prime factors of 45 are 3, 3, 5 that is 3 x 3 x 5 = 45.
If we have same factor more than once then they can be written in form of exponents that is 3 is used twice hence it can be written as (3)². So answer can be given as 45 = 3² x ⁵.
Let’s see what are the prime factorization of 210 bkt using exponents bkt.

Now to factor 210, first break it into two factors such that:
210 ÷ 3 = 70
70 ÷ 2 = 35
35 ÷ 5 = 7
7 ÷ 7 = 1
Thus prime factors for 210 will be 2 * 3 * 5 * 7.

When two Linear Equations are multiplied then they result in a Quadratic Equation. Suppose we have (x+5)(x-3) Find the product using identities?. To find product of given values we need to know about product identities. Steps to be followed:
Step 1: Here given two terms are (x+5) (x-3) which can also be written as: (x + 5) (x - 3).

Step 2: First of all multiply first term to second polynomial, then multiply second term of first polynomial to all terms of second polynomial. Using these steps it can be written as:
(x + 5) (x - 3) => x² – 3x + 5x – 15,

Step 3: Add like or same terms if present in the equation otherwise this is required solution. In above equation we have two like terms, so on combining both terms it can be written as: x² + 2x – 15. This is required solution. In this way we can easily multiply any terms. Some product identities are given we can also find product using product identities.
p² - q² = (p + q) (p – q),
p³ - q³ = (p - q) (p² + pq + q²),
p³ + q³ = (p + q) (p² - pq + q²),
p⁵ - q⁵ = (p - q) (p⁴ + p³q + p²q² + pq³ + q⁴)
p⁵ + q⁵ = (p + q) (p⁴ - p³q + p²q² - pq³ + q⁴).
 

Here we can see that we have number 12 for which we have to find prime factors. In this we first try to divide 12 by 2,
12 / 2 = 6,
Now try to factor 6 and we find that 2 is the smallest prime number by which 6 divide so,
6 / 2 = 3,
And 3 is itself is a prime number. So the prime Factorization of 12 is,
12 = 2*2* 3.

Here we can see that we have number 147 for which we have to find prime factors. In this we first try to divide 147 by 2 but it is not divided by 2. So next we try to divide that with 3 so
147 / 3 = 49
Now again we try to factor 49 and we find that 7 is the only smallest prime number which divide 49
So,                             
49 / 7 = 7,
And 7 is not further divided because it itself is a prime number. So the prime Factorization of 147 is
147 = 3*7* 7.

Given expression is (m² - 49), which we can factorize using the method given below: 
As we all know from the Algebra that a² – b² = (a + b) (a –b), where a, b are perfect squares. So we use that formula in that expression.
Now according to expression,   
(m² - n²) = (m – n) (m + n),
So the factors of expression (m² - n²) are (m - n) and (m +n).

Given expression is (k⁶ - 16), which we can factorize using the method given below: 
 As we all know that a² – b² = (a + b) (a –b), where a, b are perfect squares. So we use that formula in that expression.
Now according to expression   
(k⁶ - 16) = (k³)² - (4)²= (k³ – 4)(k³ + 4),
 So the factors of expression (k⁶ - 16) are (k³ -4) and (k³ + 4).

Given equation is (m² - 64) which we can factorize using the method given below: 
As we all know that a² – b² = (a + b) (a – b), where ‘a’, ‘b’ are perfect squares. So we use that formula in that expression.
 Now according to expression,   
(m² - 64) = (m)² - (8)² = (m – 8) (m + 8),
So the factors of expression (m² - 64) are (m - 8) and (m + 8).

Given equation is (m² - 49) which we can factorize using the method given below: 
As we all know from the Algebra that a² – b² = (a + b) (a –b), where ‘a’, ‘b’ are perfect squares. So we use that formula in that expression.
 Now according to expression   
(m² - 49) = (m)² – (7)²,
(m - 7) (m + 7),
So the factors of expression (m² - 49) are (m + 7) and (m – 7).

Here we can easily see that we have an expression (p² - q²) which we have to factorize.
As we all know from the Algebra that a² – b² = (a + b) (a –b), where a, b are perfect squares. So we use that formula in that expression.
Now according to expression   
(p² – q²) = (p – q) (p + q),
So the factors of expression (p² - q²) are (p + q) and (p – q).

We can solve the given expression as:
Here we can easily see that we have a trinomial expression x² + 18x+ 81 which we have to factorize.
To factorize that we separate 18x term like that the addition become 18x and multiplication becomes 81x².
x² + 18x + 81,
So,
x² + 9x +9x + 81,
x(x + 9) + 9(x+9),
(x + 9) (x + 9),
so the factors of expression x² + 18x+ 81 are (x + 9) (x + 9).

The solution of above example is as follows:
Here we can easily see that we have a trinomial expression k² + 10k+ 25 which we have to factorize.
To factorize that we separate 10k term like that the addition become 10k and multiplication becomes 25k².
k²+ 10k + 25,
So,
k²+ 5k +5k + 25,
k(k + 5) + k( k + 5),
(k + 5) (k + 5),
So, the factors of expression k² + 10k + 25 are (k + 5) (k + 5).

The solution of above example is as follows:
Here we can easily see that we have a trinomial expression a² + 12a+ 36 which we have to factorize.
To factorize that we separate 12a term like that the addition become 12a and multiplication becomes 36a².
a² + 12a + 36,
So,
a² + 6a + 6a + 36,
a(a + 6) + 6(a+ 6),
(a + 6) (a + 6),
So, the factors of given expression a² +12a+ 36 are (a + 6) (a + 6).

The solution of above question is as follows:
Here we can easily see that we have a trinomial expression k² + 4k + 4 which we have to factorize.
To factorize that we separate 4k term like that the addition become 4k and multiplication becomes 4k².
k²+ 4k + 4,
So,
k²+ 2k + 2k + 4,
k(k + 2) + 2( k+ 2),
(k + 2) (k + 2),
So, the factors of expression k²+ 4k+ 4 are (k + 2) (k + 2).

The solution of above problem is as follows:
 Here we can easily see that we have a trinomial expression a² + 8a+ 16 which we have to factorize.
 To factorize that we separate 8a term like that the addition become 8a and multiplication becomes 16a².
a² + 8a + 16,
So,
a² + 4a +4a + 16,
a (a + 4) + 4( a+ 4),
(a + 4) (a + 4),
So the factors of expression a² + 8a+ 16 are (a + 4) (a + 4).

Here we can see that two equations are present (x + 4)(x + 7) and (x² +6x +8) for which we have two find G.F.C. To find G.F.C we have two first factorize both equations
So first we factor                              (x + 4)(x + 7) = (x + 4) (x + 7)
Now we factor x² + 6x + 8
                                                x²+ 6x + 8 = x² +4x + 2x + 8,
So we take suitable term common from both x² +4x and 2x + 8 terms.
                                                x(x+ 4) + 2(x+4),
                                                (x +4)(x+2),
 We can clearly see that in both equations we have common term i.e. (x + 4) this is the G.C.F of both equations.