Math: complete factor with Degree 3& zeros:-1, 4, 5

I dont know how to figure out these types of problems so it would helpful do have demenstration on how to figure out these problems.

1) Degree 3& zeros:-1, 4, 5 leading coefficient = 1

a) (x+1)(x-4)(x+5)
b) (x-1)(x+4)(x-5)
c) (x+1)(x+4)(x+5)
d) (x-1)(x-4)(x-5)


2) f(x)= x^3+5x^2-2x-24 -3 zero

a) f(x)=(x+4)(x-2)(x+3)
b) (x-4)(x-2)(x-3)
c) (x-4)(x+2)(x-3)
d) (x+4)(x+2)(x+3)

thank you!! =)

1)

Roots:
x = −1 -----> x + 1 = 0 -----> (x + 1) is a factor
x = 4 -------> x − 4 = 0 -----> (x − 4) is a factor
x = 5 -------> x − 5 = 0 -----> (x − 5) is a factor

f(x) = a (x + 1) (x − 4) (x − 5)

Leading coefficient = 1 -----> a = 1

f(x) = (x + 1) (x − 4) (x − 5)

This doesn't match any of the answers. Are you sure you typed everything in correctly?

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2)

Looking at possible answers, we see that zeros are ±4, ±2, ±3
When we try these, we find that f(−4), f(2), f(−3) are all zeros
Therefore, factors are (x + 4) (x − 2) (x + 3)

f(x) = (x + 4) (x − 2) (x + 3)

1) Since the zeros are -1,4,5 and the degree is 3.
that means the polynomial has 3 factors as
f(x)=a(x+1)(x-4)(x-5), where a is a constant. But
the leading coefficient=1=> a=1. So,
f(x)=(x+1)(x-4)(x-5)
Note that none of your answers is correct.

2) f(x)=x^3+5x^2-2x-24 & -3 is a zero=>
x+3 is a factor.
Using division:
1 +5 -2 -24 |-3
...- 3 -6+24 |
----------------|
1 +2 -8 + 0
Quotient=x^2+2x-8
Remainder=0
=>
f(x)=
(x+3)(x^2+2x-8)=
(x+3)(x+4)(x-2)
Ans.a)