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When f(x) = ax^3 - 6x^2 + bx - 5 is divided by x - 1, the remainder is -5. When f(x) is divided by x + 2, the remainder is 53. Find the ordered pair (a,b).

 Apr 23, 2021
 #1
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Its given that 

\(f(x)= ax^3-6x^2+bx-5\)

 

When f(x) is divided by x-1 we get remainder 

\(r=b-a+1\)

\(b-a+1=-5\)

\(a-b=6\)                                           ...(1)                           

 

Now, when f(x) is divided by x+2 we get remainder 

\(r' = -8a-2b-29\)

\(-8a-2b-29=53\)

\(-8a-2b=82\)

\(-4a-b=41\)                                   ...(2)

 

Subtracting eq (1) from (2)

\(-4a-b-a+b=41-6\)

\(-5a=35\)

∴ \(a=-7\)

 

From eq (1)

\(b=-13\)

 

Hence the ordered pair (a,b) is (-7,-13). 

 

~Thank You :)

 Apr 23, 2021

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