A few examples..
** First, taking out a common factor:
1:
x3 + 3x
x(x2 + 3) -This is factored because you have taken an "x" away from both terms... to put it back into its original form just multiply the "x" by x2 and 3.
2:
6x2 + 16x + 8
2(3x2 + 8x + 4) -All of the numbers were divisible by 2 so you can take 2 out of all of the terms.
** Factoring a Difference of Squares:
x2 - 16 = (x)2 - 42 = (x - 4)(x + 4).
** An 'easy' trinomial (the coefficient of x is 1):
x2 + 5x + 6 = (x + 3)(x + 2). -Notice that 2+3=5 and (2)(3) =6.
** A 'hard' trinomial (the coefficient of x2 is not 1):
8x2 - 2x - 21 = (2x + 3)(4x - 7). -There are several ways of organizing a trial and error process to factor a case like this.
** Factoring by grouping (Check for this when there are 4 terms.)
6x3 - 2x2 + 15x - 5 = 2x2(3x - 1) + 5(3x - 1) = (2x2 + 5)(3x - 1).
A few examples..
** First, taking out a common factor:
1:
x3 + 3x
x(x2 + 3) -This is factored because you have taken an "x" away from both terms... to put it back into its original form just multiply the "x" by x2 and 3.
2:
6x2 + 16x + 8
2(3x2 + 8x + 4) -All of the numbers were divisible by 2 so you can take 2 out of all of the terms.
** Factoring a Difference of Squares:
x2 - 16 = (x)2 - 42 = (x - 4)(x + 4).
** An 'easy' trinomial (the coefficient of x is 1):
x2 + 5x + 6 = (x + 3)(x + 2). -Notice that 2+3=5 and (2)(3) =6.
** A 'hard' trinomial (the coefficient of x2 is not 1):
8x2 - 2x - 21 = (2x + 3)(4x - 7). -There are several ways of organizing a trial and error process to factor a case like this.
** Factoring by grouping (Check for this when there are 4 terms.)
6x3 - 2x2 + 15x - 5 = 2x2(3x - 1) + 5(3x - 1) = (2x2 + 5)(3x - 1).
These are the basic types of factoring in Algebra I. In more advanced courses, there are methods for factoring more complicated polynomials.
We won't be able to answer this accurately without knowing the polynomials.
Other polynomials of the same, or lower, order.
Reducible polynomials.
Whenever there are polynomials of the form aX2+bX+c=0 then this type of equation is know as a quadratic equation. to solve these we usually break b into two parts such that there product is equal to a*c and I hope you know how to factor polynomials.
Yes.
9
Do you care weather a random stranger on their computer cares how to factor polynomials. P.S. i do in fact care how to factor polynomials, but i'm most likely in the minority on this one.
There is no specific term for such polynomials. They may be referred to as are polynomials with only purely complex roots.
10
We can't answer that without some polynomials to choose from.
(2a-3)(a-1)
We won't be able to answer this accurately without knowing the polynomials.
Assuming additive terms, polynomials.6m3 + 50m42m3(3 + 25m)2m3=========common factor
true
x(x+3)(x+5)
12(b + 1)(b + 1)
Other polynomials of the same, or lower, order.