Question

Let R1 and R 2 be the remainders 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 + R2 + 28=0, find the value of a.

Answer 1

(x-1)=0
=> x= 1
f(X) = 4x^3 +3x^2 -12ax -5
=> f(1) = 4(1)^3 + 3(1)^2 - 12 a(1) -5
 = 4+3-12a -5
= 4+3-5 -12a
= 7-5-12a
= 2-12a
             _______(i)
(x+2)= 0
x= -2
f(x) = 2[tex] x^{3} + a x^{2} -6x +2
f(-2) = 2 (-2)^{3} + a (-2)^{2} - 6(-2) +2[/tex]
 = 2 (-8) +4 a+12 +2
= -16 +12 +2 +4a
= -16 +14 +4a
= -2 +4a
                        __________(ii)
 given,
      3$r_{1} + r_{2} +28 = 0$
=> 3(2-12a) + (-2+4a) +28 = 0
=> 6-36a +(-2+4a) + 28 = 0
=> 6-2+28-36 a+4a = 0
=> 4+28-36 +4a= 0
=> 32 - 32a = 0
=> -32a = -32
=> a= 1

Answer 2

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.