Question

let R and R are the remainder when 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 . find the value of a​

Answer 1

Answer:

Value of a is 1.

Step-by-step explanation:

Given polynomials,

f(x) = 4x³ + 3x² - 12ax - 5       and  g(x) = 2x³ + ax² - 6x + 2

R1 and R2 is remainder when polynomials divided by  x - 1  and x + 2

3 × R1 + R2 + 28 = 0

To find: Value of a.

Using Remainder theorem which states that if a polynomial p(x) is divisible by polynomial of form x - a then remainder is given by p(a).

According tot remainder theorem,

R1 = f( 1 ) = 4(1)³ + 3(1)² - 12a(1) - 5 = 4 + 3 - 12a - 5 = 2 - 12a

R2 = g( -2 ) = 2(-2)³ + a(-2)² - 6(-2) + 2 = -16 + 4a + 12 + 2 = 4a - 2

Now,

3 × R1 + R2 + 28 = 0

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

6 - 36a + 4a + 26 = 0

-32a = -32

a = 1

Therefore, Value of a is 1.