Hi Larry
x^3 - 168 = [ 24(x-2)] / 2
x^3 - 168 = (24x - 48)/2
x^3 - 168 = 12x - 24
Subtract 12x - 24 to both sides
x^3 - 168 - (12x - 24) = 12xx - 24 - (12x - 24)
x^3 - 12x - 144 = 0
Now, factor the left sides
(x - 6)(x^2 + 6x + 24) = 0
Set factors equal to 0
x - 6 = 0 or x^2 + 6x + 24 = 0
x = 0 + 6 or x^2 + 6x + 24 - 0
x = 6
Answer : X = 6
Good luck !
If the roots of the equation f(x)=0 are -4,-1, 2 and 5, then binomials (x+4), (x+1), (x-2) and (x-5) are factors of needed polynomial.
Thus, the polynomial will have form
where a is real coefficient (positive or negative). This polynomial has degree 4 as needed.
The form of the graph depends on the sign of coefficient a. Attached diagrams show two different cases of possible forms of graphs (first one for positive coefficient a, second one for negative coefficient a).
So distribute
2x+2+8=6
add like terms
2x+10=6
-10 -10
2x=-4
---------
2. 2
x=-2
Answer:
48
Step-by-step explanation:
n/-3=-16
n=-16*-3
n=48