Question

If p(x) = x3 ax2 + bx + 3 leaves a remainder 19 when divided by (x + 2) and a remainder 17 when divided by (x 2), prove that a + b = 6.

Answer 1

Answer:

gddykxclhxhkdyldulclj

Answer 2

Answer:

Question :-- if p(x)= x^3-ax^2+bx+3 leaves a remainder -19 when divided by (x+2) and a remainder 17 when divided by (x-2), prove that a+b=6 ?

Concept used :--- Let p(x) be any polynomial of degree greater than or equal to one and let ‘a’ be any real number. If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).

Solution :--

Case (1) :-- p(x)= x^3-ax^2+bx+3 leaves a remainder -19 when divided by (x+2) ...

So, when we put x = (-2) , it will gives Remainder as (-19) .

→ p(-2) = x^3-ax^2+bx+3 = (-19)

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

→ (-8) -4a -2b +3 = (-19)

→ 4a + 2b = -5 + 19

→ 4a + 2b = 14

→ 2a + b = 7 ------------------- Equation (1) .

_____________________________

Case (2) :---- p(x)= x^3-ax^2+bx+3 leaves a remainder 17 when divided by (x-2)..

So, when we put x = 2 , we get remainder as 17.

→ p(2) = x^3-ax^2+bx+3 = 17

→ (2)³ - a(2)² + 2b + 3 = 17

→ 8 - 4a + 2b + 3 = 17

→ 4a - 2b = 11-17

→ 4a -2b = -6

→ 2a - b = (-3) ----------------- Equation (2) .

_________________________

Adding Equation (1) and (2) now we get,

→ 4a = 7 + (-3)

→ 4a = 4

→ a = 1 .

Putting value of a in Equation (1) , we get,

→ 2*1 + b = 7

→ b = 7-2 = 5 ..

___________________________

So,

→ a + b = 1 + 5 = 6 . .

✪✪ Hence Proved ✪✪