Question

Let R1 and R2 are the remainders e=when the polynomials<br />
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.
Pls answer!! ❤️❤️❤️​

Answer 1

$\mathfrak{The\:Answer\:is}$

Given,

(x-1)=0

=> x= 1

Putting value of 'x' in f(x), we get

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)

Given,

(x+2)= 0

x= -2

Putting value of 'x' in f(x), we get

f(x) = 2x^{3} + a x^{2} -6x +2

=> f(-2) =2 (-2)^{3} + a (-2)^{2} - 6(-2) +2

= 2 (-8) +4 a+12 +2

= -16 +12 +2 +4a

= -16 +14 +4a

= -2 +4a __________(ii)

given,

3R1 + R2 +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

Therefore, value of a is 1.

$\boxed{Hope\:This\:Helps}$

Answer 2

Answer:

TheAnsweris

Given,

(x-1)=0

=> x= 1

Putting value of 'x' in f(x), we get

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)

Given,

(x+2)= 0

x= -2

Putting value of 'x' in f(x), we get

f(x) = 2x^{3} + a x^{2} -6x +2

=> f(-2) =2 (-2)^{3} + a (-2)^{2} - 6(-2) +2

= 2 (-8) +4 a+12 +2

= -16 +12 +2 +4a

= -16 +14 +4a

= -2 +4a __________(ii)

given,

3R1 + R2 +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

Therefore, value of a is 1.

\boxed{Hope\:This\:Helps}

HopeThisHelps