Question

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

Answer 1

Answer:

Step-by-step explanation:

4x^3+3x^2-12ax-5

divide by x-1

4x^2+7x +(7-12a)

2-12a = R1

2x^3 +ax^2 -6x +2

divide by x+2

2x^2 +(a-4)x + (2-a)

2a-2 = R2

3R1+3R2+28=0

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

-30a =-28

a=28/30

a=14/15

Answer 2

Answer:

7/6

Step-by-step explanation:

Let the given polynomials be p₁(x) = 4x³ + 3x² - 12ax - 5 and p₂(x) = 2x³ + ax² - 6x + 2.

Let g₁(x) = x - 1 and g₂(x) = x + 2 are divisors of p₁(x) and p₂(x) respectively.

(i)

Given R₁ is the remainder when p₁(x) is divided by g₁(x) = x - 1.

By remainder theorem, R₁ = p₁(1)

⇒ R₁ = p₁(1)

        = 4(1)³ + 3(1)² - 12a - 5

        = 4 + 3 - 12a - 5

        = 2 - 12a

(ii)

Let R₂ is the remainder when p₂(x) is divided by g₂(x) = x + 2.

By remainder theorem, R₂ = p₂(-2)

⇒ R₂ = p₂(-2)

        = 2(-2)³ + a(-2)² - 6(-2) + 2

        = -16 + 4a + 12 + 2

        = 4a - 2

Now,

∴ 3R₁ + 3R₂ + 28 = 0

⇒ 3(2 - 12a) + 3(4a - 2) + 28 = 0

⇒ 6 - 36a + 12a - 6 + 28 = 0

⇒ -24a + 28 = 0

⇒ -24a = -28

⇒ a = 28/24

⇒ a = 7/6.

Therefore, the value of a = 7/6.

Hope this helps!