Today our topic of discussion is Factorization of Mathematics.
Factorization of Mathematics
Factorization
If an expression is equal to the product of two or more expressions, each of the latter expressions is called a factor of the former expression. After finding the possible factors of any algebraic expression and then expressing the expression as the product of these factors are called factorization or resolution into factors.
The algebraic expressions may consist of one or more terms. So, the factors may also contain one or more terms. Some process of resolving expressions into factors will be discussed here.
Common Factor: If any polynomial expression has common factor in every term, at first they are to be found out. For example,
Example 18. 3a² * b + 6a * b² + 12a² * b² = 3ab(a + 2b + 4ab)
Example 19. 2ab(x – y) + 2bc(x – y) + 3ca(x – y) = (x – y)(2ab + 2bc + 3ca)
Perfect Square: square. An expression can be expressed in the form of a perfect
Example 20. Resolve into factors: 4x ² + 12x + 9
Solution: 4x²+ 12x + 9 = (2x)² + 2 * 2x * 3 + (3)²
= (2x + 3) ^ 2 = (2x + 3)(2x + 3)
Example 21. Resolve into factors: 9x² – 30xy + 25y²
Solution: 9x² – 30xy + 25y ²
= (3x)² – 2 * 3x * 5y + (5y)²
= (3x – 5y)² = (3x – 5y)(3x – 5y)
Difference of two squares: Expressing an expression as the difference of two squares and then applying the formula a² – b ² = (a + b)(a – b) .
Example 22. Resolve into factors: a ² – 1 + 2b – b ²
Solution: a ² – 1 + 2b – b² = a ²- (b²- 2b + 1) = a ² – (b – 1) ² = \{a + (b – 1)\}\{a – (b – 1)\} = (a + b – 1)(a – b + 1)
Example 23.
Resolve into factors: a4 + 64b4
Solution: a4 + 64b4 = (a ²) ²+ (8b²) ²
= (a ²) ² + 2a ² * 8b ²+ (8b ²) ² – 16a² * b ²
= (a ² + 8b ²)² – (4ab)²
= (a²+ 8b²+ 4ab)(a ² + 8b² – 4ab)
= (a²+ 4ab + 8b ^ 2)(a²- 4ab + 8b ²)
Work: Resolve into factors:
1) abx² + acx²+adz 2) xa2-144xb2 3) x2-2xy-4y-4
Middle term factorization: Factors can be determined by applying the formula x ² + (a + b) * x + ab = (x + a)(x + b) In this method, a polynomial of the form x² + px + q can be factorized, if two integers a and b can be found so that, a + b = p and ab = q
For this, two factors of q with their signs are to be taken whose algebraic sum is p. If q > 0 , a and b will be of same sign and if q < 0 , a and b will be of opposite sign. To be noted p and q may not be integers.
Example 24. Resolve into factors: x ² + 12x + 35
Solution: x ²+ 12x + 35 = x ²+ (5 + 7) * x + 5 * 7 = (x + 5)(x + 7)
Example 25. Resolve into factors: x ² + x – 20
Solution: x ²+ x – 20 = x ^ 2 + (5 – 4) * x + (5)(- 4) = (x + 5)(x – 4)
Middle Term Break-Up: By middle term break-up method of polynomial of the form of a * x ^ 2 + bx + c would be a * x²+ bx + c = (rx + p)(sx + q) if a * x² + bx + c = Tsx² + (rq + sp)x + pq. That is, a = rs b = rq + sp and c = pq
Hence, ac = rspq (rq) (sp) and b = rq + sp Therefore, to determine factors of the polynomial a * x ^ 2 + bx + c ac, that is, the product of the coefficient of x ^ 2 and the term free from a are to be expressed into two such factors whose algebraic sum is equal to b, the coefficient of x.
Example 26. Resolve into factors: 3x² – x – 14
Solution: 3x² – x – 14 = 3x² – 7x + 6x – 14
= x(3x – 7) + 2(3x – 7) = (3x – 7)(x + 2)
Work: Resolve into factors:
1) (x² +x-56 2) (16x³ – 46x² +15x 3)(12x² +17x+6)
Perfect Cube Form: Factors can be determined by expressing an expression in the form of perfect cube.
Example 27. Resolve into factors: 8x³+ 36x² * y + 54x * y ² + 27y ³
Solution: 8x ³+ 36x ² * y + 54x * y² + 27y³
= (2x) ³+ 3 * (2x)² * 3y + 3 * 2x * (3y)² + (3y) ³
= (2x + 3y)³ = (2x + 3y)(2x + 3y)(2x + 3y)
Formulae of addition or subtraction of two cubes: Factors can be determined by applying the formulae a³ + b³ = (a + b)(a ² – ab + b ²) and a³ – b³= (a – b)(a²+ ab + b²)
Example 28. Resolve into factors: 1) 8a ³+ 27b³ * 2 )a6 -64
Solution:
1) 8a³ + 27b ³ = (2a)³ + (3b)³
= (2a + 3b)\{(2a) ²- 2a * 3b + (3b) ²}
= (2a + 3b)(4a ² – 6ab + 9b ²)
2) a6 – 64 = (a²)³ – (4) ³ = (a ² – 4)\{(a ²) ² + a ² * 4 + (4)²\}
= (a²- 4)(a 4 + 4a² + 16)
but a² – 4 = a² – 2 ²= (a + 2)(a – 2) and a ^ 4 + 4a² + 16 = (a²)² + (4)² + 4a²
= (a² + 4)² – 2(a²)(4) + 4a ²
= (a ² + 4) ² – 4a ²
= (a ² + 4) ²- (2a) ²
= (a ² + 4 + 2a)(a ²+ 4 – 2a)
= (a ²+ 2a + 4)(a² – 2a + 4)
∴a-64 = (a+2)(a-2) (a²+2a + 4) (a² – 2a+4)
Alternative method: a ^ 6 – 64 = (a ³ ) ^ 2 – 8²
= (a³ + 8)(a ³ – 8)
= (a ³ + 2³)(a³ – 2³)
= (a + 2)(a²- 2a + 4)(a – 2)(a ² + 2a + 4)
= (a + 2)(a – 2)(a² + 2a + 4)(a² – 2a + 4)
Work: Resolve into factors:
1) 2x ^ 4 + 16x
2) 8 – a³ + 3a² * b – 3a * b² + b³
3) (a + b)³ + (a – b)³
Factors of the expression with fractional coefficients: Factors of the expression with fractional coefficients can be expressed in several ways. For
example, a³ + 1/27 = a ³ + 1/(3 ³) = (a + 1/3)(a² – a/3 + 1/9) 1 Again, a+7=(274³+1) = {(3a)³ + (1)³) = (3a+1)(9a² – 3a+1) 27 27
In the second solution, the factors involving the variables are with integral coefficients but the two solutions are same. 1/27 * (3a + 1)(9a² – 3a + 1) = 1/3 * (3a + 1) * 1/9 * (9a² – 3a + 1)
= (a + 1/3)(a² – a/3 + 1/9)
Example 29. Resolve into factors: x³ + 6x² * y + 11x * y ² + 6y³
Solution: x³ + 6x²* y + 11x * y² + 6y ³
= \{x³ + 3x ²* 2y + 3x * (2y)²+ (2y)³\} – x * y²- 2y³
= (x + 2y)³ – y² * (x + 2y) = (x + 2y)\{(x + 2y)² – y²\}
= (x + 2y)(x + 2y + y)(x + 2y – y) = (x + 2y)(x + 3y)(x + y) = (x + y)(x + 2y)(x + 3y)
Read more: