Question

If 3x³+ax²-6x+5 when divided by 3x+2 gives remainder 5 then find a​

Answer 1

Answer:

(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) = 2x^{3} + a x^{2} -6x +2 < br / > f(-2) = 2 (-2)^{3} + a (-2)^{2} - 6(-2) +2x3+ax2−6x+2<br/>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,

      3r_{1} + r_{2} +28 = 0r1+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

Step-by-step explanation:

hope it helps you