Problem 1
Answers:
The possible rational roots are:
-10, -5, -2, -1, -2/5, -1/5,
10, 5, 2, 1, 2/5, 1/5
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Explanation:
The first term is 5x^3 with the coefficient 5
The last term is 10
Factors of 10 (last term): 1, 2, 5, 10
Factors of 5 (first coefficient): 1, 5
Divide the factors of 10 over the factors of 5
10/1 = 10
10/5 = 2
5/1 = 5
5/5 = 1
2/1 = 2
2/5 = 2/5
1/1 = 1
1/5 = 1/5
And throw in the minus versions of those results as well: -10, -2, -5, -1, -2, -2/5, -1, -1/5
Combine the two sets and toss out the duplicates
-10, -5, -2, -1, -2/5, -1/5,
10, 5, 2, 1, 2/5, 1/5
So those are the possible rational roots
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Problem 2
Answer: Choice D) (5x+2)(x+6)(3x-4)
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Explanation:
I'm going to use the rule
if x = a/b is a root, then (bx-a) is a factor
If we set the factor bx-a equal to zero and solve, then we get
bx-a = 0
bx = a
x = a/b
where b is nonzero
So if x = -2/5 is a root, then
x = -2/5
5x = -2
5x+2 = 0
making (5x+2) a factor
If x = -6 is a root then
x = -6
x+6 = 0
making (x+6) a factor
And if x = 4/3 is a root, then
x = 4/3
3x = 4
3x-4 = 0
making (3x-4) a factor
The three factors found were: (5x+2) and (x+6) and (3x-4)
So one factorization could be (5x+2)(x+6)(3x-4)
which is what choice D is showing
-(-5)+7-18+4 first yo distribute the negative so it would be 5+7-18+4=-2
Answer:
2.75
Step-by-step explanation:
first convert it to a improper fraction
2 3/4 = 11/4
then dividing the numerator by the denominator.
11/4 = 2.75
hope this helped!
