1 thought on “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 divi”
[tex]\large\underline{\sf{Solution-}}[/tex]
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.
[tex]\large\underline{\sf{Solution-}}[/tex]
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
[tex]\rm :\longmapsto\:f(x) = (x – 3)(x + 6)g(x) + ax + b – – (1)[/tex]
Now,
Given that,
We know,
Remainder Theorem
So,
[tex] \red{\rm :\longmapsto\: \: f(3) = 7}[/tex]
[tex]\rm :\longmapsto\:(3 – 3)(3 + 6)g(3) + 3a + b = 7[/tex]
[tex]\rm :\longmapsto\:3a + b = 7 – – – (2)[/tex]
Also,
Given that,
We know,
Remainder Theorem
So,
[tex] \red{\rm :\longmapsto\: \: f( – 6) = 22}[/tex]
[tex]\rm :\longmapsto\:( – 6 – 3)( – 6 + 6)g( – 6) – 6a + b = 22[/tex]
[tex]\rm :\longmapsto\: – 6a + b = 22 – – – (2)[/tex]
☆ On Subtracting equation (1) from equation (2) we get
[tex]\rm :\longmapsto\: – 6a – 3a = 22 – 7[/tex]
[tex]\rm :\longmapsto\: – 9a = 15[/tex]
[tex]\bf :\implies\:a = – \: \dfrac{5}{3} [/tex]
☆ On substituting the value of a in equation (1), we get
[tex]\rm :\longmapsto\:3 \times \dfrac{( – 5)}{3} + b= 7[/tex]
[tex]\rm :\longmapsto\: -5 + b= 7[/tex]
[tex]\bf :\longmapsto\:b = 12 [/tex]
Hence,
Remainder is
[tex]\bf :\longmapsto\:r(x) = – \: \dfrac{5}{3}x + 12 [/tex]