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 .
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Answer 1

Using remainder theorem:  If f(x) leaves remainder r when divided by (x - a), then f(a) must be equal to r. In question:

Case1: When p(x) is divided by (x + 2)

=> p(-2) = r

=> (-2)³ - a(-2)² + b(-2) + 3 = -19

=> -8 - 4a - 2b + 3 = -19

=> 7 = 2a + b      ...(1)     [simplified]

Case2: When p(x) is divided by (x - 2)

=> p(2) = 17

=> (2)³ - a(2)² + b(2) + 3 = 17

=> 8 - 4a + 2b + 3 = 17

=> b - 2a = 3      ...(2)     [simplified]  

On adding (1) and (2), we get b = 5

Substituting in (1), 7 = 2a + 5

=> 1 = a

Hence, a + b = 1 + 5 = 6

Answer 2

~Solution :-

$\Large \mathfrak \green{Case\: 1 :}$

• Here, p(x) is divided by (x + 2)

$\sf{p(-2) = r}$

$\to \sf{ (-2 {)}^{3} - a(-2 {)}^{2} + b(-2) + 3 = -19}$

$\to \sf{-8 - 4a - 2b + 3 = -19 }$

$\to \bf \red{7 = 2a + b}$

• Hence Simplified!

$\Large \mathfrak \green{Case \:2 :}$

• Here, p(x) is divided by (x - 2)

$\sf{ p(2) = 17 }$

$\to \sf{ (2 {)}^{3} - a(2 {)}^{2} + b(2) + 3 = 17}$

$\to \sf{ 8 - 4a + 2b + 3 = 17}$

$\to \bf \red{ b - 2a = 3}$

• Hence Simplified!

$\leadsto$ On adding case 1 and case 2, we get the final solution $\tt \blue{6}$.

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