If x=1 and x=2 are zeros of f(x)=3X^3+Ax^2+Bx+2, then the value of A+B is a)1, b)-1, c)9, d)-9, e) cannot be determined
I think it can't be determined but i don't know!
Find A+B in
f(x)=3X^3+Ax^2+Bx+2,
Hints:
1. two of the roots are 1 and 2, then two of the factors are (x-1) and (x-2)
2. the third root must be of the form
(3x±1) to satisfy the coefficient 3x^3 and the constant term.
3. Determine the sign of the factor in (2).
4. Expand
(x-1)(x-2)(3x±1)
to find A and B, and hence A+B.
To find the value of A+B, we can use the fact that if x=1 and x=2 are zeros of the polynomial f(x), then it means that when we substitute these values into f(x), we get a result of zero.
Substituting x=1 into f(x), we get:
f(1) = 3(1)^3 + A(1)^2 + B(1) + 2
= 3 + A + B + 2
= A + B + 5
Similarly, substituting x=2 into f(x), we get:
f(2) = 3(2)^3 + A(2)^2 + B(2) + 2
= 24 + 4A + 2B + 2
= 4A + 2B + 26
Since both values of x=1 and x=2 yield zeros, we can set the results equal to zero:
A + B + 5 = 0 ------(1)
4A + 2B + 26 = 0 ------(2)
To find the value of A+B, we need to eliminate one of the variables (A or B) and solve for the other. Let's eliminate B by multiplying equation (1) by 2 and equation (2) by -1:
2(A + B + 5) = 0 (equation (1) multiplied by 2)
-1(4A + 2B + 26) = 0 (equation (2) multiplied by -1)
Simplifying these equations, we get:
2A + 2B + 10 = 0
-4A - 2B - 26 = 0
Adding these two equations eliminates the B variable:
-2A - 16 = 0
Solving for A, we get:
-2A = 16
A = -8
Substituting the value of A into equation (1), we can solve for B:
-8 + B + 5 = 0
B = 3
Therefore, the value of A+B is -8+3 = -5.
So, the correct answer is not e) cannot be determined. The correct answer is b) -1.