FACTORING CUBIC POLYNOMIALS

Every cubic polynomial will have 3 factors. To find those factors, we follow the following steps.

Step 1 :

We can find one linear factor of the given cubic polynomial using synthetic division.

Step 2 :

At the end of the first step, we will have quadratic factors. By factoring the quadratic equation, we can get other two factors.

Step 3 :

List all three factors.

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 x- x - 6.

x- 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 4x+ 4x -3.

  =  4x+ 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 :

x- 23x+ 142x - 120

Solution :

Step 1 :

Let p(x) = x- 23x+ 142x - 120

Step 2 :

(x - 1) is a factor.

x- 22x + 120 = (x - 10)(x - 12)

Step 3 :

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

Example 4 :

4x- 5x+ 7x - 6

Solution :

Step 1 :

Let p(x) = 4x- 5x+ 7x - 6

Step 2 :

(x - 1) is one of the factors.

4x- 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 :

x+ x - 6 = (x + 3)(x - 2)

Step 3 :

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

Example 6 :

 x3 + 13x+ 32x + 20

Solution :

Step 1 :

Let p(x) = x3 + 13x+ 32x + 20

Step 2 :

(x + 1) is one of the factors.

x+ 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 + 7x+ 11x + 5 = 0

Solution :

Step 1 :

Let p(x) =  x3 + 7x+ 11x + 5

factoring-cubic-polynomial-q1

-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 + 2x- 2x = 0

Solution :

Step 1 :

Let p(x) =  4x3 + 2x- 2x

 4x3 + 2x- 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

2x = 0

x = 0/2

x = 0

2x - 1 = 0

2x = 1

x = 1/2

x + 1 = 0

x = -1

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

Example 9 :

-x3 - 3x+ x + 3 = 0

Solution :

Step 1 :

Let p(x) =  -x3 - 3x+ x + 3

 -x3 - 3x+ 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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