Each of the following is a factor of x^2 +3x-4 except
A. x+4
B. x-1
C. x-4
D. 1
well, x^2+3x-4 = (x+4)(x-1), so ...
So it's 1?
To determine which of the given options is not a factor of the polynomial x^2 + 3x - 4, we can use the factor theorem. According to the factor theorem, if a polynomial P(x) has a factor of (x - a), then P(a) = 0.
To check if an option is a factor, substitute the value represented by the option into the polynomial and verify if the result is zero. If it is not zero, then the option is not a factor.
Let's substitute the values from the given options into the polynomial:
A. x + 4:
When we substitute x = -4 into the polynomial, we get (-4)^2 + 3(-4) - 4 = 16 - 12 - 4 = 0. Therefore, A is a factor.
B. x - 1:
Substituting x = 1 into the polynomial gives us (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0. Therefore, B is a factor.
C. x - 4:
For x = 4, the polynomial becomes (4)^2 + 3(4) - 4 = 16 + 12 - 4 = 24, which is not zero. Therefore, C is not a factor.
D. 1:
When we substitute x = 1 into the polynomial, we get (1)^2 + 3(1) - 4 = 1 + 3 - 4 = 0. Therefore, D is a factor.
So, the factor that is not a factor of x^2 + 3x - 4 is option C, x - 4.