How would I factor x^4+3x^2+6x+10?
Which work? Where did you get that trinomial?
Same question. Where did you get that trinomial?
If you already had the trinomial from DenisB, I suppose you might have gotten it by doing polynomial division.
For instance, it does not factor by grouping.
So, if someone were to use the Rational Root Theorem, they would see that it has no rational roots. So, either it factors as
two binomials with integer coefficients or it doesn't factor (over integer coefficients).
But how did the initial quadratic factor come about from DenisB?
Well, you can't just stare at a polynomial and expect to suddenly see its factors! You have to try things and do some calculations. First, when we talk about "factoring" a polynomial, with integer coefficients we generally mean into lower degree polynomials with integer coefficients. Here the polynomial is $x^4+ 3x^2+ 6x+ 10$. 10 factors as (1)(10) or (2)(5).
So we might try $(x^2+ ax+ 1)(x^2+ bx+ 10)= x^2(x^2+ bx+ 10)+ax(x^2+ bx+ 10)+ 1(x^2+ bx+ 10)= x^4+ (a+ b)x^3+ (10+ ab+ 1)x^2+ (10a+ b)x+ 10$. We want that to equal to $x^4+ 3x^2+ 6x+ 10$ so we must have a+ b= 0, 11+ ab= 3, and 10a+ b= 6. From a+ b= 0, b= -a, of course. Then 10a+ b= 10a- a= 9a= 6 so that a= 2/3 not an integer! Further, if a= 2/3, b must equal -2/3 so 11+ ab becomes 11- 4/9= 95/9, not 3.
If we try $(x^2+ ax+ 2)(x^2+ bx+ 5)= x^2(x^2+ bx+ 5)+ ax(x^2+ bx+ 5)+ 2(x^2+ bx+ 5)= x^4+ bx^3+ 5x^2+ ax^3+ abx^2+ 5ax+ 2x^2+ 2bx+ 10= x^4+ (a+b)x^3+ (5+ab+ 2)x^2+ (5a+ 2b)x+ 10$. We want that to equal to $x^4+ 3x^2+ 6x+ 10$ so we must have a+ b= 0, 5+ ab= 3, and 5a+2b= 6. Again, a+ b= 0 gives b= -a so 5a+ 2b= 5a- 2a= 3a= 6 so a= 2 and b= -2. Then 7+ ab= 7- 4= 3!
Yes, $(x^2+ 2x+ 2)(x^2- 2x+ 5)= x^4+ 3x^w+ 6x+ 10$.
Of course, there are other possibilities: $(x^2+ ax- 1)(x^2+ bx- 10)$ or $(x^2+ ax- 2)(x^2+ bx- 5)$ or a linear term times a cubic but fortunately we don't need to look at those.