Question

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

Answer 1

Answer:

p ( x ) = x^3 - ax^2 + bx + 3 leaves a remainder -19 when divided by ( x + 2 )

the reminder is

p(-2) =(-2) ^3 - a(-2)^2 + b(-2)+ 3=-19

or, -8-4a-2b+3=-19

or,-4a-2b=(-19+5) =-14

or, 2a+b=14/2=7

or, 2a+b=7___________________(i)

p ( x ) = x^3 - ax^2 + bx + 3 remainder 17 when divided by ( x - 2)

the reminder is

p(2)=(2) ^3 - a(2)^2 + b(2)+ 3=17

or, 8-4a+2b+3=17

or ,-4a+2b=17-11

or, -2a+b=3______________________(ii)

(I) -(ii)

(2a+b+2a-b)=7-3

or, 4a=4

or, a=1

putting the value of a in the equation (ii) we get

-2*1+b=3

or, b=3+2=5

so, a+b=1+5=6

proved

Answer 2

Answer:

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Step-by-step explanation:

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