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(x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5 For integers a,b,c and d, (x^2+ax+b)(x^2+cx+d)=x^4+x^3-2x^2+17x-5. What is the value of a+b+c+d ?

 Aug 28, 2019
 #1
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(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

____________________________________________

   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

 

 

cool cool cool

 Aug 28, 2019

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