Question

If the polynomial p (x)=x^4-2x^3+3x^2-ax+8 is divided by (x-2) it leaves remainder=0 find the value of a

Answer 1

Answer:

Hello friend.

Step-by-step explanation:

P(x) = x^4-2x^3+3x^2-ax+8.

g(x) = (x-2).

so x = +2.                                ( since x-2=0, x=+2.)

now put 2 in behalf of x.

P(2) = 2^4-2(2)^3+3(2)^2-a((2)+8 = 0 (since it is a factor.)

P(2) = 16-2(8)+3(4)-2a+8 = 0.

P(2) = 16-16+12-2a+8 = 0.

= 0+20-2a = 0.

= 20-2a = 0.

= -2a = -20.

= a = -20/-2.

  a = +10.

Hope it helps.

Please mark it as the Brainliest.

Answer 2

Answer: p(x)=x^4-2x^3+3x^2-ax+8

It is given that x-2 is a factor of p(x)

x-2=0

=> x=2

Putting the value of x in p(x):-

2^4-2(2)^3+3(2)^2-a(2)+8=0

=> 16-16+12-2a+8=0

=> 18-2a=0

=> a=-18/-2

=> a=9

Hence,a=9