Someone please help!!!! I get how to factor just not when there are so many numbers.
Can someone show this problem step by step?
x(x^4+2x^3-18x^2-4x+49-30)
since all of the terms at least have 1x in them, you subtract 1x from every term and put it on the outside of the brackets. for example, x^5 now becomes x^4 because you took away 1x from it.
hope i helped. we just finished factoring in class.
The rational root theorem is your friend....
Because the coefficient of the first term is 1, the special case of the rational root theorem
tells you that any rational roots of the equation will be +/- factor of 30.
So, numbers like 1, -1, 2, -2, 3, -3, 5,-5, ....... 30, -30 are candidates.
You can just plug some in and see which ones get you 0.
Pretty quick, I see that 1, -2, 3 are all roots. At this point use should use polynomial
division.
If 1, -2, and 3 are roots then (x-1), (x+2), and (x-3) are factors. You can multiply these
together to get a 3rd degree polynomial and divide this into the original equation.
The result will be a 2nd degree polynomal (and no remainder if you did it right)
You can then use the quadratic equation to find the remaining roots, which will
give you the remaining 2 factors.
Use the rational root theorem and synthetic division
Set it equal to zero and use synthetic substitution
Comments
x(x^4+2x^3-18x^2-4x+49-30)
since all of the terms at least have 1x in them, you subtract 1x from every term and put it on the outside of the brackets. for example, x^5 now becomes x^4 because you took away 1x from it.
hope i helped. we just finished factoring in class.
The rational root theorem is your friend....
Because the coefficient of the first term is 1, the special case of the rational root theorem
tells you that any rational roots of the equation will be +/- factor of 30.
So, numbers like 1, -1, 2, -2, 3, -3, 5,-5, ....... 30, -30 are candidates.
You can just plug some in and see which ones get you 0.
Pretty quick, I see that 1, -2, 3 are all roots. At this point use should use polynomial
division.
If 1, -2, and 3 are roots then (x-1), (x+2), and (x-3) are factors. You can multiply these
together to get a 3rd degree polynomial and divide this into the original equation.
The result will be a 2nd degree polynomal (and no remainder if you did it right)
You can then use the quadratic equation to find the remaining roots, which will
give you the remaining 2 factors.
Use the rational root theorem and synthetic division
Set it equal to zero and use synthetic substitution