Question

When x^2+4x-b is divided by x-a the remainder is 2, find the smallest possible value for b?

Answer 1

The smallest possible value of b is 3.

Step-by-step explanation:

Given:

• When x^2+4x-b is divided by x-a the remainder is 2.

As we know that,

=> Dividend = Divisor * quotient+ remainder

So,

$= > {x}^{2} + 4x - b = (x - a)(x + 4 + a) + 2$

$= > {x}^{2} + 4x - b = {x}^{2} - ax + 4x - 4a + ax - {a}^{2} + 2$

$= > {x}^{2} + 4x - b = {x}^{2} + 4x - ( {a}^{2} + 4a - 2)$

Now comparing the coefficient of RHS and LHS.

${a}^{2} + 4a - 2 = b$

The smallest possible value of b is getting by putting a=1

$= > b = 3$