30*5=150
32*34=1088
15*19=285
15*3=45
We haven n! = (n-1)! x n and (n+1)! = n! x (n + 1);
Then, (n!)^2 = n! x n! = n! x (n-1)! x n;
And (n+1)!(n-1)! = n! x (n + 1) x (n-1)!;
Finally, [n! x (n-1)! x n] / [n! x (n + 1) x (n-1)!] = (n+1)/n;
Answer:
x³-9x²+14x+24
Step-by-step explanation:
here given : 4, -1, 6
so put negative value of each digit in the equation
(x-4)(x+1)(x-6)
(x²-4x+x-4)(x-6)
(x²-3x-4)(x-6)
(x³-3x²-4x-6x²+18x+24)
x³-9x²+14x+24
The answer to your question is YES!