Question

R1 and R2 are the remainders when the polynomials f(x)=x^3+2ax^2-5x-7 and g(x) = x^3+x^2-12x+6a are divided by
(x+1) & (x-2) respectively. If 2R1 +R2= 12.
Find 'a' .

Answer 1

Answer:

Make x+1, zero of the polynomial, therefore x = -1

R1 =  -1+2a+5-7 

R1 = -3+2a

Now, make x-2 a zero of the polynomial, therefore x=2

R2 = -12+6a

therefore, R1+R2= 6 , (took 2 common)

substitte the values

-3+2a+6a-12=6

8a=21

a=21/8

Answer 2

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.