Question

when a polynomial f(x) is divisible. by x-3 and x+6, the respective remainders are 7and 22 What is the remainder when f(x) is divided by (x-3)( (x+6)​

Answer 1

$\large\underline{\sf{Solution-}}$

Let g(x) be the quotient and r(x) be remainder when f(x) is divided by (x - 3)(x + 6).

Since,

We know that,

When ever a polynomial of degree n is divided by another polynomial of degree 2, the remainder will always be a polynomial degree 1 less than degree of denominator.

So, r(x) = ax + b

Thus,

f(x) is defined as

$\rm :\longmapsto\:f(x) = (x - 3)(x + 6)g(x) + ax + b - - (1)$

Now,

Given that,

• On dividing f(x) by (x - 3), the remainder is 7.

We know,

Remainder Theorem

• It states that if a function f(x) is divided by (x-a), then f(a) is the remainder.

So,

$\red{\rm :\longmapsto\: \: f(3) = 7}$

$\rm :\longmapsto\:(3 - 3)(3 + 6)g(3) + 3a + b = 7$

$\rm :\longmapsto\:3a + b = 7 - - - (2)$

Also,

Given that,

• On dividing f(x) by (x + 6), the remainder is 22.

We know,

Remainder Theorem

• It states that if a function f(x) is divided by (x-a), then f(a) is the remainder.

So,

$\red{\rm :\longmapsto\: \: f( - 6) = 22}$

$\rm :\longmapsto\:( - 6 - 3)( - 6 + 6)g( - 6) - 6a + b = 22$

$\rm :\longmapsto\: - 6a + b = 22 - - - (2)$

☆ On Subtracting equation (1) from equation (2) we get

$\rm :\longmapsto\: - 6a - 3a = 22 - 7$

$\rm :\longmapsto\: - 9a = 15$

$\bf :\implies\:a = - \: \dfrac{5}{3}$

☆ On substituting the value of a in equation (1), we get

$\rm :\longmapsto\:3 \times \dfrac{( - 5)}{3} + b= 7$

$\rm :\longmapsto\: -5 + b= 7$

$\bf :\longmapsto\:b = 12$

Hence,

Remainder is

$\bf :\longmapsto\:r(x) = - \: \dfrac{5}{3}x + 12$