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Question

If p(x)=x^3-ax^2+bx+3 leaves a remainder -19 whendivided by (x+2) and a remainder 17 when divided by (x-2), then prove that a+b=6


Solution

The given polynomial p(x)= x³ - ax² + bx +3
When p(x) divide by (x+2) remainder= -19
By remainder theorem,

as it is x+2.  we take -2
ie , P(-2)= ( -2³) - a(-2²)+b(-2) +3 =  -19
-8 - 4a-2b+3 = -19

4a + 2b -3 +8 = 19
4a + 2b +5 = 19
4a +2b = 14
2a + b = 7......eqn(1)

When p(x) divide by (x-2) remainder= 17
As it is x-2  we take +2
P(2) = 2³ - a(2²) + b(2) +3 = 17
 8 - 4a + 2b + 3 = 17
-4a + 2b +11 =17
-4a + 2b = 6
-2a + b = 3....eqn(1)
Adding eqn (1) & (2)

2a + b  +  -2a + b = 7+3
2b  = 10 
b = 5

2a + b = 7
2a + 5 = 7
2a = 7-5
2a = 2
a = 1

As a = 1, b= 5
a+b= 1+5=6

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