Factoring is the breaking apart of a polynomial into a product of other smaller polynomials. Consider the equation to be of the form $ax^{2}$ + bx + c.

While factoring find two numbers that will multiply to equal the constant term 'c', and when the two terms are added it should be equal to 'b' (coefficient of x term).

A cubic equation has the form: $ax^{3}$ + $bx^{2}$ + cx + d = 0, (a$\neq$ 0)
A cubic equation will have at least one real root it can also have three real roots but not zero roots.To factorize a cubic equation first find a factor, where in you substitute the value to see which value turns the equation to zero.

Then apply synthetic division to get a polynomial which we multiply by.

Example: Solve : $x^{3}$ - $5x^{2}$ -2x + 24 = 0

Solution: To find a factor for the given equation substitute x = - 2 we see that the equation turns out to be zero. So x = - 2 is a solution to this equation and x + 2 is a factor for the given expression.

Now $x^{3}$ - $5x^{2}$ -2x + 24 = 0 can be written in the form (x + 2) ($x^{2}$ + ax + b). Where a and b are numbers.

To find a and b we use a process called synthetic division.

1. Write down the coefficients of a cubic equation in the first row of a table and draw a vertical line and write down the known root x = -2. Bring down the number 1 from the first row.
Factoring Equations

2. Multiply 1 by -2. Put the result in the second position of the blank column. Add the numbers in the second column and put in the bottom row.
Factoring Cubic Equations

3. Now -7 is multiplied by -2 and the result 14 is placed in the third position of the blank row.add the numbers of the third row and multiply the result by -2.
Factoring Polynomial Equations

4. The process continues,
Factoring Equations with Exponents

We see that the final element in the bottom row is zero. The coefficients of a quadratic are 1, -7 and 12
So the quadratic is $x^{2}$ - 7x + 12

The cubic has been reduced to (x + 2) $x^{2}$ - 7x + 12

The quadratic can be factorized to give (x + 2) (x - 3) (x - 4) = 0

Therefore, the solutions are x = -2, 3 or 4.
A polynomial is an expression of finite length consisting of variables and constants. The polynomial coefficients may be only integer numbers.
Standard form of polynomial equation is $c_{n}x^{n}$ + $c_{n-1}x^{n-1}$+ ....+ $c_{1}$x + $c_{0}$ = 0. Given below are the steps to be considered while solving a polynomial equation by factoring.
    1. The given equation must be in standard form.
    2. Factor the left hand side as zero is on the right.
    3. Each factor should be set equal to zero.
    4. Solve to determine the roots.

Solved Example

Question: Solve $x^{2}$ - 3x - 4 = 0
Solution:
 
We see that the given equation is in the standard form.
$x^{2}$ - 3x - 4 = 0

Factor the polynomial equation
$x^{2}$ - 4x + x - 4 = 0
x(x - 4) + 1 (x - 4 ) = 0
(x - 4) (x + 1) = 0

Set each factor to zero
(x - 4) = 0 or (x + 1) = 0
Determine the roots
(x - 4) = 0   $\Rightarrow$ x = 4 or
(x + 1) = 0  $\Rightarrow$ x = -1

Thus x = 4 or x = -1
 

In exponents we multiply the expressions with the same base and add exponents. To solve the exponents factor the like terms and group them. A fundamental exponential rule is $x^{y} \times x^{z}$ = $x^{y+z}$.

Example: $x^{5}$ = $2^3$, $2^{2}$ = b, are exponents equations.

Solved Examples

Question 1: Factorize the polynomial $x^{2}$ - 13x + 36 = 0.
Solution:
 
We see that the given equation is in the standard form.
$x^{2}$ - 13x +36 =0

Factor the polynomial equation
$x^{2}$ - 9x -4x +36 = 0
x(x - 9) -4 (x - 9) = 0
(x - 9) (x - 4) = 0

Set each factor to zero
(x - 9) = 0 or (x -4) = 0
Determine the roots
(x - 9) =0   $\Rightarrow$ x = 9 or
(x - 4) = 0  $\Rightarrow$ x = 4
Thus x = 4 or x = 9
 

Question 2: Solve $2^{13}$ + $2^{13}$ 
Solution:
 
$2^{13}$ (1) +  $2^{13}$ (1) 
              = $2^{13}$ (1 + 1)
              = $2^{13}$ * (2}
              = $2^{13}$ * $2^{1}$
              = $2^{13+1}$
               = $2^{14}$
               = 16384