FACTORING CUBIC POLYNOMIALS USING IDENTITIES

Factor the following cubic polynomials. 

Example 1 :

8x³ + 125y³

Solution :

 8x³ + 125y³ = (2x)³ + (5y)³

a³ + b³ = (a + b)(a² - ab + b²)

= (2x + 5y)[(2x)² - (2x)(5y) + (5y)²]

= (2x + 5y)(4x² - 10xy + 25y²) 

Example 2 :

27x³ - 8y³

Solution :

27x³ - 8y³ = (3x)³ - (2y)³

 a³ - b³ = (a - b)(a² + ab + b²)

= (3x - 2y)[(3x)² + (3x)(2y) + (2y)²] 

  = (3x - 2y)(9x² + 6xy + 4y²) 

Example 3 :

a⁶ - 64

Solution :

a⁶ - 64 = (a²)³ - 4³

 a³ - b³ = (a - b)(a² + ab + b²)

= (a² - 4)[(a²)² + a²(4) + 4²]

= (a² - 4)[(a²)² + 4a² + 4²]

= (a² - 4)[(a²)² + 4² + 4a²]

a² + b² = (a + b)² - 2ab

= (a² - 4)[(a² + 4)² - 2(a²)(4) + 4a²]

= (a² - 4)[(a² + 4)² - 8a² + 4a²]

= (a² - 4)[(a² + 4)² - 4a²]

= (a² - 2²)[(a² + 4)² - (2a)²]

a² - b² = (a + b)(a - b)

= (a + 2)(a - 2)(a² + 4 + 2a)(a² + 4 - 2a)

= (a + 2)(a - 2)(a² + 2a + 4)(a² - 2a + 4)

Example 4 :

x³ + 8y³ + 6 xy - 1

Solution :

= x³ + 8y³ + 6xy - 1

= x³ + (2y)³ + (-1)³ - 3(x)(2y)(-1)

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

= (x + 2y - 1)(x² + 4y² + 1 - 2xy + 2y - x)

Example 5 :

l³ - 8m³ - 27n³ - 18 lmn

Solution :

= l³ - 8m³ - 27n³ - 18 lmn

= l³ + (-2m)³ + (-3n)³ - 3(1)(-2m)(-3n)

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

= (l - 2m - 3n)(l² + 4m² + 9n² + 2lm - 6mn + 3ln)

So, the factors are (l - 2m - 3n)(l² + 4m² + 9n² + 2lm - 6mn + 3ln)

Example 6 :

16r³ - 6r² - 56r + 21

Solution :

= 16r³ - 6r² - 56r + 21

Factoring 2r² from the first two terms and factoring -7 from the last two terms.

= 2r² (8r - 3) - 7(8r - 3)

= (2r² - 7)(8r - 3)

So, the factors are (2r² - 7)(8r - 3)

Example 7 :

42x³ + 24x² + 49x + 28

Solution :

= 42x³ + 24x² + 49x + 28

Factoring 6x² from the first two terms and factoring 7 from the last two terms.

= 6x² (7x + 4) + 7(7x + 4)

= (6x² + 7)(7x + 4)

So, the factors are (6x² + 7)(7x + 4).

Example 8 :

x³ + 5x² + 6x

Solution :

= x³ + 5x² + 6x

Factoring x, we get

= x (x² + 5x + 6)

= x (x² + 3x + 2x + 6)

= x [x (x + 3) + 2(x + 3)]

= x (x + 3) (x + 2)

So, the factors are x (x + 3) (x + 2).

Example 9 :

2x³ + 6x² + 5x + 15

Solution :

= 2x³ + 6x² + 5x + 15

Factoring 2x², we get

= 2x² (x + 3) + 5(x + 3)

= (x + 3) (2x² + 5)

So, the factors are (x + 3) (2x² + 5).

Example 10 :

3a³ - 7a² - 9a + 21

Solution :

= 3a³ - 7a² - 9a + 21

Factoring a², we get

= a² (3a - 7) - 3(3a - 7)

= (3a - 7) (a² - 3)

So, the factors are (3a - 7) (a² - 3).

Example 11 :

10n³ - 2n² - 25n + 5

Solution :

= 10n³ - 2n² - 25n + 5

Factoring n², we get

= 2n² (5n - 1) - 5(5n - 1)

= (5n - 1) (2n² - 5)

So, the factors are (5n - 1) (2n² - 5).

Example 12 :

16x⁴ - 2x

Solution :

= 16x⁴ - 2x

= 2x (8x³ - 1)

= 2x (2³x³ - 1)

= 2x [(2x)³ - 1³]

= 2x (2x - 1) ((2x)² - (2x)(1) + 1²)

= 2x (2x - 1) (4x² - 2x + 1)

So, the factors are 2x (2x - 1) (4x² - 2x + 1).

Example 13 :

10x⁴ - 10

Solution :

= 10x⁴ - 10

= 10(x⁴ - 1)

= 10((x²)² - (1²)²)

= 10[(x² + 1)(x² - 1²)]

= 10(x² + 1)(x + 1)(x - 1)

So, the factors are 10(x² + 1)(x + 1)(x - 1).

Example 14 :

x²(x - 1) - 9(x - 1)

Solution :

= x²(x - 1) - 9(x - 1)

= (x² - 9) (x - 1)

= (x² - 3²) (x - 1)

= (x + 3)(x - 3)(x - 1)

So, the factors are (x + 3)(x - 3)(x - 1).

Example 14 :

x³ + 4x² - 36x - 144

Solution :

= x³ + 4x² - 36x - 144

= x² (x + 4) - 36(x + 4)

= (x² - 36)(x + 4)

= (x² - 6²)(x + 4)

= (x + 6)(x - 6)(x + 4)

So, the factors are (x + 6)(x - 6)(x + 4).

Example 15 :

2x³ - 14x² + 24x

Solution :

= 2x³ - 14x² + 24x

= 2x³ - 14x² + 24x

= 2x(x² - 7x² + 12)

= 2x(x² - 3x - 4x + 12)

= 2x[x(x - 3) - 4(x - 3)]

= 2x(x - 4) (x - 3)

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