Maths Questions

(1)
f(x) = x^3 - 7x + 6 is divisible by x^2 - 3x + k
We can use the following equation,
x^3 - 7x + 6 = ( x^2 - 3x + k ) ( x + a )
x^3 - 7x + 6 = x^3 - 3x^2 + kx + ax^2 - 3ax + ak
By comparing the coefficient
0 = -3 + a ... (1)
-7 = k - 3a ... (2)
6 = ak ... (3)
from (1), a = 3
from (3) and a = 3, k = 2

(2)
2x^3 = 12x - 9
2x^3 - 12x + 9 = 0
Let f(x) = 2x^3 - 12x + 9
For 0th interval,
f(-3) = 2(-3)^3 - 12(-3) + 9 = -9
f(-2) = 2(-2)^3 - 12(-2) + 9 = 17
therefore, there is a root in this interval.

For 1st interval,
f(-3) = 2(-3)^3 - 12(-3) + 9 = -9
f(-2.5) = 2(-2.5)^3 - 12(-2.5) + 9 = 7.75
therefore, there is a root in the this interval.

For 2rd interval,
f(-2.75) = 2(-2.75)^3 - 12(-2.75) + 9 = 21.2
f(-2.5) = 2(-2.5)^3 - 12(-2.5) + 9 = 7.75
therefore, there is no root in the this interval.
f(-3) = 2(-3)^3 - 12(-3) + 9 = -9
f(-2.75) = 2(-2.75)^3 - 12(-2.75) + 9 = 21.2
therefore, there is no root in the this interval.

For 3rd interval,
f(-2.875) = 2(-2.875)^3 - 12(-2.875) + 9 = 19.73
f(-2.75) = 2(-2.75)^3 - 12(-2.75) + 9 = 21.2
therefore, there is no root in the this interval.

when f(x) is divided by x^2-3x+k,

the remainder in the x term is (2-k)

the remainder in the digit term is (6-3k)

both of them should be zero. since x^2-3x+k is a factor of f(x)

(2-k)=0 and (6-3k)=0

we can get k=2