Question

Let R1 and R2 are the remainders when the polynomials x2 + 2x2 - 5ax -7 and x3 + ax2 -12x + 6 are divided by x +1 and x – 2 respectively. If 2R1+ R2 =6, find the value of a.

Answer 1

3x^2-5ax-7  and x+1
now

remainder is 3+5a-7=r1   2r1=6+10a-14

and 
x^3+ax^2-12x+6 and x-2
now 

8+4a-24+6=r2


now adding both terms 

8+4a-24+6+6+10a-14=14a-4=6

now a=10/14

Answer 2

Answer:

Value of a is 2.

Step-by-step explanation:

Given polynomials,

x³ + 2x² - 5ax - 7      and  x³ + ax² - 12x + 6

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

2 × R1 + R2 = 6

To find: Value of a.

Let, p(x) = x³ + 2x² - 5ax - 7

q(x) = x³ + ax² - 12x + 6

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 = p( -1 ) = (-1)³ + 2(-1)² - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6

R2 = q( 2 ) = 2³ + a(2)² - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10

Now,

2 × R1 + R2 = 6

2( 5a - 6 ) + 4a - 10 = 6

10a + 4a - 12 -10 = 6

14a = 28

a = 2

Therefore, Value of a is 2.