Use the rational root theorem to determine all possible rational roots of 5x^3 + 4x^2 + 4x + 10 = 0. do not find the actual roots. the roots of a certain polynomial are -2/5, -6, and 4/3. what might the linear factors of the polynomial be? a) (5x + 2)(x + 6)(3x + 4) b) (5x + 2)(x – 6)(3x – 4) c) (5x – 2)(x + 6)(3x – 4) d) (5x + 2)(x + 6)(3x – 4) answer both and show of explain answers : )

Problem 1

Answers: 
The possible rational roots are: 
-10, -5, -2, -1, -2/5, -1/5, 
10, 5, 2, 1, 2/5, 1/5



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



Problem 2

Choice D) (5x+2)(x+6)(3x-4)



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

Cd is the perpendicular bisector of both xy and st, and cy=20.find xy

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