Answer:
EXAMPLE 1
Factor the polynomial:
x³-27.
SOLUTION
difference of two cubes factoring formula, and 27 corresponds to b³ so
x-27 is the difference of two cubes. x³ corresponds to a³ in the
b=27=3 and b² = 3²=9.
Write the formula.
Substitute.
x³-27=(x-3)(x²+3x+9)
Because (x²+3x+9) cannot be factored, you are done.
a³-b³=(a-b)(a²+ab+b²)
I want to know how can I find out which can be factored or not? Like I need more explanation and example for this
In general, a polynomial can be factored if it can be written as the product of two or more polynomials of lower degree. For example, x²+3x+9 cannot be factored further because it is already the product of two first-degree polynomials (x+3) and (x+3).
A polynomial of the form a²+2ab+b², where a and b are constants, can also not be factored. This is known as a perfect square trinomial.
A polynomial of the form a³+3a²b+3ab²+b³, where a and b are constants, can also not be factored. This is known as a perfect cube trinomial.
On the other hand, a polynomial of the form x²-y², where x and y are variables, can be factored using difference of squares formula as (x+y)(x-y). Similarly, a polynomial of the form x³-y³ can be factored using difference of cubes formula as (x-y)(x²+xy+y²).
You can also use the rational root theorem to find possible roots of polynomials and factor them. For example if the polynomial is of the form ax²+bx+c, then the roots can only be of the form p/q where p is a factor of c and q is a factor of a.
In general, factoring polynomials can be quite difficult, and many polynomials cannot be factored. However, by using different techniques, like difference of squares and difference of cubes, it is possible to factor some polynomials.
Uday Tahlan