True, it takes a lot of time to eliminate the possibilities. There are indeed two factors, both quadratic.
You have correctly concluded that there are no linear factors.
As in the linear factors, the constant terms of the factors should multiply to 130, which does not give many combinations, namely
1*130
2*65
5*26
10*13
and vice-versa.
Another fact you could use is that the sum of the coefficients in x = -12, since they are required in forming the x³ term , whose coefficient is -12.
You have also correctly concluded that there are no real roots, since the function does not cut the x-axis.
If you need more hints, post.
Write f(x)=x^4-12x^3+59x^2-138x+130 as a product of linear factors.
I listed the factors of 130 and used synthetic division, but none of the remainders came out to be 0. I graphed the function, but the function doesn't pass through the x-axis. I figured I could find the zeros and work out the problem that way, but obviously it didn't work. Does the function have imagenary numbers when factored? I'm stumped on this problem, so if any of you could help me by showing the answer and by providing the steps the took to obtain that answer, I'd really appreciate it! Thanks!
1 answer
1 answer