Question 1039605
#1 is quadratic form.  Substitute for t^2 and follow what you know.


#2 will have negative roots, and maybe some positive roots.  Check using synthetic division for pos & neg of 1, 2, 3, 4, 6.


The only REAL root is {{{-4}}} for the linear binomial factor, {{{x+4}}}.   ( I used Google Search Engine's graphing of function feature instead of actually doing synthetic division).   "product of linear binomials"?   No.  Not unless you want two complex factors, too.  You can use general solution for quadratic equation for that.


Synthetic Division checking  root  {{{-4}}} gives you coefficients <pre>1, 2, 3</pre>
Meaning the factorization {{{(x+4)(x^2+2x+3)}}}.  


Roots for the quadratic factor are  {{{-1+i*sqrt(2)}}}  and {{{-1-i*sqrt(2)}}}.   General solution for quadratic formula will give those.


Full factorization of f is  {{{highlight((x+4)(x+1-i*sqrt(2))(x+1+i*sqrt(2)))}}}.