You're on the right track. Notice your 3rd row is a linear combination of the 1st 2? What if rows 2 and 3 are multiples of row 1?

5x^4 - 14x ^3 + 18x^2 + 40x + sixteen = (x^2 -4x +eight)(ax^2 +bx+c) expand RHS and simplify ax^4 - 4ax^three + 8ax^2 + bx^three - 4bx^2 + 8bx + cx^2 - 4cx + 8c =>x^four(a) -x^3(4a - b) + x^2(8a-4b + c) + x(8b -4c) + 8c evaluating with coreesponding terms in LHS a = 5 -------------------eqn(1) 4a - b = 14------------eqn(2) 8a-4b+c = 18------------eqn(3) 8b-4c=40--------------eqn(4) 8c = 16-----------------eqn(5) from(1) a = 5: from (5) c = 2 plugging in a value in (2) --four*5 - b = 14 <=>20 - b = 14 <-> b = 6 so a = 5: b = 6 : c = 2 so 5x^4 - 14x ^3 + 18x^2 + 40x + 16 = (x^2 -4x +8)(5x^2 +6x+2) to seek out all solutions equate RHS = zero (x^2 -4x +8)(5x^2 +6x+2) = zero when x^2 - 4x + 8 = 0, making use of quadratic system x = [4 ± sqrt(16 -32)]/2 = [4± sqrt(-16)]/2 = [4 ± 4i]/2 x = 2 ± 2i x = 2+ 2i or 2 -2i when 5x^2 + 6x + 2 = 0 x = [-6±sqrt(36 -40)]/10 x = [-6 ± sqrt(-4)]/10 = [-6 ± 2i]/10 x = (1/5)(-three + i) or (1/5)(-three - i)

Yes, that is correct.

(By the way, Philo, it only takes one counterexample for an "always" statement to be false. So the fact that there are examples in which there are infinitely many solutions does not really matter.)

Lord bless you today!