find T so x-3 is a factor of f(x)=x^3-4x^2+Tx+3
a little synthetic division reveals that
f(x) = (x-3)(x^2 - x + T-3)+ (3T-6)
If we want f(3) to be zero, then 3T-6=0, so T=2
f(x) = x^3 - 4x^2 + 2x + 3
= (x-3)(x^2 - x - 1)
or ...
if x-3 is a factor, then f(3) = 0
f(3) = 27 - 36 + 3T + 3 = 0
3T = 6
T = 2
To find the value of T such that x-3 is a factor of f(x), we can use the factor theorem. According to the factor theorem, if x-3 is a factor of f(x), then f(3) should be equal to zero.
So, to find the value of T, we need to substitute x=3 into f(x) and set it equal to zero.
Let's calculate f(3):
f(x) = x^3 - 4x^2 + Tx + 3
Substituting x=3:
f(3) = (3)^3 - 4(3)^2 + T(3) + 3
= 27 - 36 + 3T + 3
= 3T - 6
Now, we set f(3) equal to zero:
3T - 6 = 0
Adding 6 to both sides:
3T = 6
Dividing both sides by 3:
T = 2
So, the value of T that makes x-3 a factor of f(x) is T = 2.