Question

if (ax^3+bx^2+x-6) has (x+2) as a factor and leaves remainder 4,when divided by (x-2), find the value of a and b

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Answer 1

Answer:

a=0 and b = 2

Step-by-step explanation:

Given ,

p(x) =ax³+bx²+x-6

p(-2) = 0

p(2) = 4

Main solution :

p(-2)=0

-8a+4b-2-6=0

-8a+4b=8

2a-b = -2 ----------(i)

Now ,

p(2) = 4

8a+4b+2-6=4

8a+4b = 8

2a+b = 2-------------(ii)

(i) +(ii)

4a = -2+2

4a= 0

=>>>> a = 0

Putting this value of a in (i) , we get :

0-b=-2

-b=-2

=>>>>>>b=2

Answer 2

answer is a is equal to 0 and b is equal to 2

Step-by-step explanation:

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