0
1 - y^(3)=
1^(3) - (3) =
This factors to
(1 - y)(1^(2) + y + y^(2))
More simply written as
( 1 - y)(1 + y + y^(2))
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What is completely factored form?
A completely factored form is one which is composed of product of factors and can't be factorized further. Let us consider two examples: x2 - 4x + 4 is not a factored form because it can be factored as (x - 2)(x - 2). (x +1)(x2 - 4x + 4) is also not a factored form because x2 - 4x + 4 can be factored further as (x - 2)(x - 2). So, the completely factored form is (x + 1)(x - 2)(x - 2).
What is the factored form of 2x-10x?
2x(1-5)
What is x2 - 2x plus 1 factored?
It is (x-1)(x-1) when factored
What is 2a plus 2b in factored form?
2(a+b) is 2a plus 2b in factored form.
What is the factored form of 3x2 plus 21x minus 24?
3(x - 1)( x + 8)
What is the factored form of 1-y3?
-2
What is 27-y3 factored completely?
27-y3 factored completely = 24
What is the factored form of 8-y3?
5
What is completely factored form?
A completely factored form is one which is composed of product of factors and can't be factorized further. Let us consider two examples: x2 - 4x + 4 is not a factored form because it can be factored as (x - 2)(x - 2). (x +1)(x2 - 4x + 4) is also not a factored form because x2 - 4x + 4 can be factored further as (x - 2)(x - 2). So, the completely factored form is (x + 1)(x - 2)(x - 2).
What is the factored form of 6x to the second power plus 5x plus 1?
6x2+5x+1 = (3x+1)(2x+1) when factored
What is the factored form of 1-y squared?
(1-y)(1+y)
What is the factored form of 3x2-2x-1?
It is: (3x+1)(x-1)
How do you factor x3-x?
That can't be factored. In general, a binomial (i.e., a polynomial with two parts) can only be factored if both parts have a common factor, or if you can express it as the sum or difference of squares, cubes, etc.
What is 6a2b3-3a2b in factored form?
3a2b(2b2-1)
What is the factored form of 2x-10x?
2x(1-5)
How do you know if a polynomial is in factored form?
You can't know if a general polynomial is in factored form.
How do you find the equation when all zeros are given?
Let's say y1, y2, and y3 are zeros. Set up an expression like this (x-y1)(x-y2)(x-y3) [This is factored form] and then multiply carefully. That works with any number of roots. If 0 is a root, add an x by itself to the beginning of the factored form expression. Also, imaginary roots come in twos. Therefore, if given i as a root, you need (x-i)(x+i). If you have something in x + yi form for a root, you'll need the complex conjugate. That would make (x-(a+bi))(x-(a-bi))
