(x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5
(x^2 + ax + b) ( x^2 + cx + d) =
x^4 + cx^3 + dx^2
+ +ax^3 + acx^2 + adx
+ + bx^2 + bcx + bd
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x^4 + (a + c)x^3 + ( ac + b + d)x^2 + (bc + ad)x + bd
Equating coefficients we have this system of equations
a + c = 1
ac + b + d = -2
ad + bc = 17
bd = - 5
Since bd = - 5.....then we have the following possibilities for ( b, d) = (-1, 5), ( 5, -1) , ( 1, -5) or ( -5, 1)
Check (b, d) = ( -1, 5)
ad + bc = 17 ⇒ a(5) + ( -1)c =17 ⇒ 5a - c = 17 (1)
a + c = 1 (2)
Add (1) and (2) and we get that 6a = 18b⇒ a = 3
And using (1), 5(3) - c = 17 ⇒ -c = 2 ⇒ c =-2
So...in this situation a = 3, b = -1 , c = -2 , d = 5 and their sum = 5
Check (b,d) = (5, -1)
a(-1) + (5)c = 17 ⇒ -a + 5c = 17 (3)
a + c = 1 (4)
Add (3) and (4) and we get that
6c = 18 ⇒ c = 3
And using (3), -a + 5(3) = 17 ⇒ -a + 15 = 17 ⇒ -a = 2 ⇒ a =-2
So....in this situation a = -2, b = 5, c = 3 d = -1 and the sum is still = 5
Check for yourself that (b,d) = (1, - 5) and ( -5, 1) don't provide integer solutions
Check
(x^2+3x-1)(x^2-2x+5) = x^4+x^3-2x^2+17x-5
(x^2 -2x + 5) ( x^2 + 3x -1) = x^4+x^3-2x^2+17x-5
