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# help

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The polynomial \(x^6 + ax + b\) is divisible by \(x^2 - 2x - 1.\) Find \(a + b.\)

Apr 15, 2019

#1
+26007
+1

Here is my scrawled answer..... basically I did the long division and found what values of A and B produce a zero remainder....

a+b = -99

Apr 15, 2019
#2
+111433
+1

x^4  + 2x^3  + 5x^2   + 12x  + 29

x^2 - 2x - 1   [   x^6  +  0x^5  + 0x^4 + 0x^3 + 0x^2 + ax + b ]

x^6   -  2x^5   - x^4

________________________

2x^5    + x^4  + 0x^3

2x^5   -  4x^4 -  2x^3

_______________________

5x^4 + 2x^3  + 0x^2

5x^4 -10x^3  - 5x^2

___________________

12x^3 + 5x^2 + ax

12x^3 -24x^2 -12x

__________________________

29x^2 + ( a + 12)x + b

29x^2  -    58x      -29

____________________________

(a + 12 + 58)x + (b + 29)

(a + 70)x  + (b + 29)  =  0

ax + 70x + b + 29  =  0

ax + b   =   -70x - 29

So

a + b     =  -70 + (-29)     =  -99

Apr 15, 2019
#3
+26007
+1

A LOT more legible than my answer !  LOL.....

ElectricPavlov  Apr 15, 2019
#4
+111433
0

We still arrived at the same station.....that's encouraging  !!!!

CPhill  Apr 15, 2019
edited by CPhill  Apr 15, 2019