Question

If (x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by(x-2) leaves the remainder 4. Prove that a=0 and b=2 Please answer ASAP..i am getting graded on this Irrelevant answer will be reported

Answer 1

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Answer 2

Given :

(x+2) is a factor of p(x) = ax^3 + bx^2 + x - 6

Reaminder is 4 when p(x) is divided by (x-2) .

To find :

To prove a = 0 and b = 2 .

Solution :

(x+2) is a factor of p(x) = ax^3 + bx^2 + x - 6

then x = -2 will make the equation equal to 0 .

Putting x = -2 into p(x) ,

p(-2) = -8a + 4b - 8 = 0

=> 2a = -2 + b        ....(i)

Reaminder is 4 when p(x) is divided by (x-2) .

Then , at x = 2 , the equation will be equal to 4 .

Putting x = 2 into p(x) ,

p(2) = 8a + 4b - 4 = 4

=> 2a = 2 - b       .......(ii)

from (i) and (ii) ,

b = 2

putting value of b into (i) ,

2a = -2 + 2

=> a = 0

Hence proved that a = 0 and b = 2 .