how to factorize a cubic polynomial ?

Factoring By Grouping

1. 1
   Group the polynomial into two sections. Grouping the polynomial into two sections will let us attack each section individually.
    
   • Say we're working with the polynomial x³ + 3x² - 6x - 18 = 0. Let's group it into (x³ + 3x²) and (- 6x - 18)
      
    
2. 2
   Find what's common in each section.
   • Looking at (x³ + 3x²), we can see that x² is common.
      
   • Looking at (- 6x - 18), we can see that -6 is common.
      
    
3. 3
   Factor the commonalities out of the two terms.
   • Factoring out x² from the first section, we get x²(x + 3).
      
   • Factoring out -6 from the second section, we get -6(x + 3).
      
    
4. 4
   If each of the two terms contains the same factor, you can combine the factors together.
   • This gives us (x + 3)(x² - 6).
      
    
5. 5
   Find the solution by looking at the roots. If you have an x² in your roots, remember that bothnegative and positive numbers fulfill that equation.
   • The solutions are 3, and √6.
      
    

Part 2: Factoring Using the Free Term

1. 1
   Rearrange the expression so it's in the form of aX³+bX²+cX+d.
   • Let's say we're working with the equation: x³ - 4x² - 7x + 10 = 0.
      
    
2. 2
   Find the all of the factors of "d". The constant "d" is going to be the number that doesn't have any variables, such as "x," next to it.
   • Factors are the numbers you can multiply together to get another number. In our case, the factors of 10, or "d," are: 1, 2, 5, and 10.
      
    
3. 3
   Find one factor that causes the polynomial to equal zero. We want to determine which factor makes the polynomial equal zero when we substitute the factor for each "x" in the equation.
   • Let's start by using our first factor, 1. Let's substitute "1" for each "x" in the equation:
     (1)³ - 4(1)² - 7(1) + 10 = 0
      
   • This gives us: 1 - 4 - 7 + 10 = 0.
      
   • Because 0 = 0 is a true statement, we know that x = 1 is a solution.
      
    
4. 4
   Do a little rearranging. If x = 1, we can rearrange the statement to look a bit different without changing what it means.
   • "x = 1" is the same thing as "x - 1 = 0" or "(x - 1)". We've just subtracted a "1" from each side of the equation.
      
    
5. 5
   Factor your root out of the rest of the equation. "(x - 1)" is our root. Let's see if we can factor it out of the rest of the equation. Let's take it one polynomial at a time.
   • Can we factor (x - 1) out of the x³? No we can't. Be we can borrow a -x² from the second variable; then we can factor it: x²(x - 1) = x³ - x².
      
   • Can we factor (x - 1) out of what remains from our second variable? No, again we can't. We need to borrow another little bit from the third variable. We need to borrow a 3x from -7x. This gives us -3x(x - 1) = -3x² + 3x.
      
   • Since we took a 3x from -7x, our third variable is now -10x and our constant is 10. Can we factor this? We can! -10(x - 1) = -10x + 10.
      
   • What we did was rearrange the variables so that we could factor out a (x - 1) out of the entire equation. Our rearranged equation looks like this: x³ - x² - 3x² + 3x - 10x + 10 = 0, but it's still the same thing as x³ - 4x² - 7x + 10 = 0.
      
    
6. 6
   Continue to substitute by the factors of the free term. Look at the numbers that we factored out using the (x - 1) in Step 5:
   • x²(x - 1) - 3x(x - 1) - 10(x - 1) = 0. We can rearrange this to be a lot easier to factor one more time: (x - 1)(x² - 3x - 10) = 0.
      
   • We're only trying to factor (x² - 3x - 10) here. This factors down into (x + 2)(x - 5).
      
    
7. 7
   Your solutions will be the factored roots. You can check whether your solutions actually work by plugging each one, individually, back into the original equation.
   • (x - 1)(x + 2)(x - 5) = 0 This gives us solutions of 1, -2, and 5.
      
   • Plug -2 back into the equation: (-2)³ - 4(-2)² - 7(-2) + 10 = -8 - 16 + 14 + 10 = 0.
      
   • Plug 5 back into the equation: (5)³ - 4(5)² - 7(5) + 10 = 125 - 100 - 35 + 10 = 0.