(x-2)(x+3)(x-c) = x^3 + (1-c)x^2 - (c+6)x + 6c
So that gives you
a = 1
1-c = -2
b = -(c+6)
6c = 24
solve that and you have
a = 1
b = -10
c = 4
check: (x-2)(x+3)(x-4) = x^3 - 3x^2 - 10x + 24
It appears you have a typo in your polynomial.
the function f(x) = ax^3 - 2x^2 + bx + 24 has three factors. Two of the factors are x-2 and x+3. Determine the value of a and b, and then determine the other factor
4 answers
alternate approach , assuming there is no typo.
f(x) = ax^3 - 2x^2 + bx + 24
x-2 is a factor, so
f(2) = 8a - 8 + 2b + 24 = 0
4a + b = -8
x+3 is a factor, so
f(-3) = -27a - 18 - 3b + 24 = 0
9a + b = 2
subtract them
5a = 10
a = 2 , then b = -16
f(x) = 2x^3 - 2x^2 - 16x + 24
= (x-2)(x+3)(2x - 4) , the third factor would have to be (2x - 4)
f(x) = ax^3 - 2x^2 + bx + 24
x-2 is a factor, so
f(2) = 8a - 8 + 2b + 24 = 0
4a + b = -8
x+3 is a factor, so
f(-3) = -27a - 18 - 3b + 24 = 0
9a + b = 2
subtract them
5a = 10
a = 2 , then b = -16
f(x) = 2x^3 - 2x^2 - 16x + 24
= (x-2)(x+3)(2x - 4) , the third factor would have to be (2x - 4)
ah, yes - My bad. I assumed a was 1.
SMH
SMH
thank you so much guys!!!