FACTORING CUBIC POLYNOMIALS

Factor the following cubic polynomials :

Example 1 :

x³ - 2x² - 5 x + 6

Solution :

Step 1 :

Let p(x)  =  x³ - 2x² - 5 x + 6

Step 2 :

By dividing the cubic polynomial by 1, we get 0 as remainder. So (x - 1) is a factor. 

We can get the other two factors, by factoring the quadratic polynomial x² - x - 6.

x² - x - 6 = (x - 3)(x + 2)

Step 3 :

The required factors are (x - 1) (x - 3) and (x + 2).

Example 2 :

4x³ - 7x + 3

Solution :

Step 1 :

Let p(x)  =  4x³ - 7x + 3

Step 2 :

(x-1) is one of the factors.

We get the other two factors, by factoring the quadratic polynomial 4x² + 4x -3.

  =  4x² + 6x - 2x - 3 (decomposing the middle term)

  = 2x(2x + 3) - 1(2x + 3)

= (2x - 1)(2x + 3)

Step 3 :

So, the factors are (x - 1)(2x - 1)(2x + 3).

Example 3 :

x³ - 23x² + 142x - 120

Solution :

Step 1 :

Let p(x) = x³ - 23x² + 142x - 120

Step 2 :

(x - 1) is a factor.

x² - 22x + 120 = (x - 10)(x - 12)

Step 3 :

So, the factors are (x - 1)(x - 10)(x - 12).

Example 4 :

4x³ - 5x² + 7x - 6

Solution :

Step 1 :

Let p(x) = 4x³ - 5x² + 7x - 6

Step 2 :

(x - 1) is one of the factors.

4x² - x + 6 is not factorable.

Step 3 :

So, the factors are (x-1)(4x²-x+6).

Example 5 :

 x³ - 7x + 6

Solution :

Step 1 :

Let p(x) = x³ - 7x + 6

Step 2 :

x² + x - 6 = (x + 3)(x - 2)

Step 3 :

So, the factors are (x - 1)(x + 3)(x - 2).

Example 6 :

 x³ + 13x² + 32x + 20

Solution :

Step 1 :

Let p(x) = x³ + 13x² + 32x + 20

Step 2 :

(x + 1) is one of the factors.

x² + 12x + 20 = (x + 10)(x + 2)

Step 3 :

So, the factors are (x + 1)(x + 10)(x + 2).

Solving the following cubic equation.

Example 7 :

 x³ + 7x² + 11x + 5 = 0

Solution :

Step 1 :

Let p(x) =  x³ + 7x² + 11x + 5

-1 is one of the solution.

Step 2 :

x² + 6x + 5 = 0

x² + 1x + 5x + 5 = 0

x (x + 1) + 5 (x + 1) = 0

(x + 5)(x + 1) = 0

Step 3 :

x + 5 = 0 and x + 1 = 0

x = -5 and x = -1

So, the three solutions are -1, -1 and -5.

Example 8 :

 4x³ + 2x² - 2x = 0

Solution :

Step 1 :

Let p(x) =  4x³ + 2x² - 2x

 4x³ + 2x² - 2x = 0

Factoring 2x, we get

2x(2x² + x - 1) = 0

2x(2x² + 2x - 1x - 1) = 0

2x[2x (x + 1) - 1 (x + 1)] = 0

2x (2x - 1)(x + 1) = 0

Equating each factor to 0, we get

So, the solutions are -1, 0 and 1/2

Example 9 :

-x³ - 3x² + x + 3 = 0

Solution :

Step 1 :

Let p(x) =  -x³ - 3x² + x + 3

 -x³ - 3x² + x + 3 = 0

Factoring -x², we get

 -x² (x + 3) + 1(x + 3) = 0

(x + 3)(-x² + 1) = 0

Equating each factor to 0, we get

x + 3 = 0

x = -3

-x² + 1 = 0

-x² = -1

x² = 1

x = -1 and 1

So, the solutions are -3, -1 and 1.

Example 10 :

The volume (in cubic feet) of a room in the shape of a rectangular prism is represented by 12z³ − 27z. Find expressions that could represent the dimensions of the room.

Solution :

Volume of rectangular prism = 12z³ − 27z

= 3z (4z² - 9)

= 3z (2²z² - 3²)

= 3z [(2z)² - 3²]

= 3z (2z + 3)(2z - 3)

Length = 3z, width = 2z + 3 and height = 2z - 3

Example 11 :

Factor the polynomial completely.

x³ + 2x²y − x − 2y

Solution :

= x³ + 2x²y − x − 2y

= x² (x + 2y) - 1(x + 2y)

= (x² - 1) (x + 2y)

= (x + 1)(x - 1) (x + 2y)

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

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