Question

Let R1 and R2 are the remainders e=when the polynomials
f(x) = 4x3+3x2-12ax -5 and g(x) = 2x3+ax2-6x +2 are divided by (x-1) and (x+2) respectively. If 3R1 + R2 + 28 = 0, find the value of a.

Answer 1

Question:

Let R₁ and R₂ be the remainders when the polynomials   f(x) = 4x³ + 3x² - 12ax - 5 = 0  and  g(x) = 2x³ + ax² - 6x + 2 = 0  are divided by  (x - 1)  and  (x + 2)  respectively. If  3R₁ + R₂ + 28 = 0,  find the value of a.

Solution:

Consider the polynomial  f(x) = 4x³ + 3x² - 12ax - 5 = 0  and its divisor,  (x - 1).

If f(x) leaves remainder R₁ on division by (x - 1), then f(1) will be R₁.  So,

    f(1) = R₁

⇒  4(1)³ + 3(1)² - 12a(1) - 5 = R₁

⇒  4 + 3 - 12a - 5 = R₁

⇒  2 - 12a = R₁

Now, consider the polynomial  g(x) = 2x³ + ax² - 6x + 2 = 0  and its divisor, (x + 2).

If g(x) leaves remainder R₂ on division by (x + 2), then g(-2) will be R₂.  So,

    g(-2) = R₂

⇒  2(-2)³ + a(-2)² - 6(-2) + 2 = R₂

⇒  2(-8) + 4a + 12 + 2 = R₂

⇒  - 16 + 4a + 12 + 2 = R₂

⇒  4a - 2 = R₂

So we wrote R₁ and R₂ in terms of 'a', right?!

Now consider  3R₁ + R₂ + 28 = 0.

    3R₁ + R₂ + 28 = 0

⇒  3(2 - 12a) + (4a - 2) + 28 = 0

⇒  6 - 36a + 4a - 2 + 28 = 0

⇒  32 - 32a = 0

⇒  32(1 - a) = 0

⇒  1 - a = 0

⇒  a = 1

Hence, 1 is the answer.

Answer 2

Polynomials :

Solution :

Given, f(x) = $\mathsf{4x^3\:+\:3x^2\:-12ax\:-5}$

Factor of f(x) is ( x - 1 ).

(x - 1 ) = 0

x = 1

So, f( 1 ) = 4 + 3 - 12a - 5

f( 1 ) = 2 - 12a

R1 = 2 - 12a

Now, For g(x ) = $\mathsf{2x^3\:+\:ax^2\:-6x\:+2}$

Factor of g(x) is ( x +2 ).

( x +2 ) = 0

x = - 2

So, g( - 2 ) = $\mathsf{2(-2)^3\:+\:a(-2)^2\:-6(-2)\:+2}$

g ( - 2 ) = - 16 + 4a +12 +2 = 4a - 2

R2 = 4a - 2

Given, 3R1 + R2 + 28 = 0

3( 2 - 12a) + ( 4a - 2) + 28 = 0

6 - 36a + 4a - 2 + 28 = 0

- 32a + 32 = 0

a = $\mathsf{\dfrac{-32}{-32}}$

a = 1

So, the value of 'a' is 1.