Question

When x3-ax2+11x-6 is divided by x-4 the remainder is 6 . Find the value of a

Answer 1

GIVEN :

When $x^3-ax^2+11x-6$ is divided by x-4 the remainder is 6 . Find the value of a

TO FIND :

The value of a

SOLUTION :

Given that when $x^3-ax^2+11x-6$ is divided by x-4 the remainder is 6 .

By using the Remainder theorem we can have that

f(4) = 6 and remainder=6

Put x=4 in $x^3-ax^2+11x-6$

$4^3-a(4)^2+11(4)-6=6$

64-16a+44-6 = 6

-16a= 6-64-44+6

-16a=-96

16a=96

$a=\frac{96}{16}$

a = 6

∴ the value of a in the given expression is 6

Answer 2

Answer:

= a = 6

solution is in the attached photo.