zeion said:I just realized I can sub in 2 to get a factor..
But is there a faster way than to guess randomly?
dacruick said:2 is not a factor
willem2 said:I suppose he meant: if you substitute y=2 in the polynomial you get 0, so (y-2) has to be a factor.
You can use the rational root theorem to find all rational numbers that might be a factor, in this case you only have to try -2,-1,1 and 2
Factoring something with a cube power involves finding the common factors of the expression and grouping them together. This is similar to factoring with a square power, but with an extra step of factoring out the cube root.
The main difference is that when factoring with a cube power, you need to factor out the cube root in addition to finding the common factors. This means that the final factors will include both the square root and the cube root of the original expression.
Sure, let's take the expression 27x^3 + 9x^2. The first step is to find the common factor, which in this case is 9. This leaves us with 9(3x^3 + x^2). Next, we factor out the cube root of the remaining terms, which is x^2. This gives us 9x^2(x + 3). Therefore, the final factored expression is 9x^2(x + 3).
If there is a coefficient in front of the variable with the cube power, you will need to factor it out first before finding the common factors. For example, if we have 8x^3 + 4x^2, we can factor out 4x^2 to get 4x^2(2x + 1). Then, we can proceed with factoring out the cube root and find the common factor as mentioned before.
No, not all expressions with cube powers can be factored. Some expressions may already be fully factored, while others may not have any common factors that can be factored out. In these cases, the expression cannot be factored any further.