Factor x^4 + x^3 + 2x^2 - x + 3 completely over the integers.

Step-by-step solution

1. 1

   Rule out linear factors first. By the Rational Root Theorem any rational root divides $3$, so the only candidates are $\pm1,\pm3$. Testing: $f(1)=1+1+2-1+3=6$, $f(-1)=1-1+2+1+3=6$, $f(3)=81+27+18-3+3=126$, $f(-3)=81-27+18+3+3=78$. None vanish, so there is no linear factor and the only possible integer factorisation is into two quadratics.

2. 2

   Set up the undetermined-coefficient template. Because the leading coefficient is $1$, both quadratics can be taken monic:

   $x^4+x^3+2x^2-x+3=(x^2+ax+b)(x^2+cx+d)$

   Expanding gives $x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$.

3. 3

   Match coefficients to get four equations.

   $a+c=1,\qquad ac+b+d=2,\qquad ad+bc=-1,\qquad bd=3$

   The constant equation $bd=3$ is the one with fewest options, so it drives the search: $(b,d)\in\{(1,3),(3,1),(-1,-3),(-3,-1)\}$.

4. 4

   Test (b, d) = (1, 3). Then $ac+4=2$, so $ac=-2$, and with $a+c=1$ the pair $\{a,c\}$ are roots of $t^2-t-2=(t-2)(t+1)$, giving $\{2,-1\}$. Check the linear coefficient: $ad+bc=2(3)+1(-1)=5\neq-1$. Swapping to $a=-1,c=2$ gives $-3+2=-1$... but that is the same as relabelling, i.e. case $(b,d)=(3,1)$ below. Taken in the stated order this case fails.

5. 5

   Test (b, d) = (3, 1). Again $ac+4=2$ so $ac=-2$ and $\{a,c\}=\{2,-1\}$. Now $ad+bc=2(1)+3(-1)=2-3=-1$ ✓ — all four equations are satisfied by $a=2,\ b=3,\ c=-1,\ d=1$.

6. 6

   Write the factorisation and confirm it is complete.

   $x^4+x^3+2x^2-x+3=(x^2+2x+3)(x^2-x+1)$

   Neither factor splits further over the reals: their discriminants are $2^2-4(1)(3)=-8$ and $(-1)^2-4(1)(1)=-3$, both negative, so the quartic has no real roots at all.

7. 7

   Verify by expanding back. $(x^2+2x+3)(x^2-x+1)=x^4-x^3+x^2+2x^3-2x^2+2x+3x^2-3x+3=x^4+x^3+2x^2-x+3$ ✓, matching every coefficient.