f(x)=(24x3+68x2+24x−32)/(4x3+56x2+236x+280)
I need help finding the roots and the hole for this problem.
well first of all divide top and bottom by 4
y=(24x3+68x2+24x−32)/(4x3+56x2+236x+280)
= (6x^3+17x^2+6 x-8)/(x^3+14x^2+59x+70)
try factoring the top. I went to an online graphing program and found roots of the top at
-2 , .5, - 1 1/3
so try
(x+2)(2x-1)(3x+4) that works
now the bottom same way
for the bottom roots are
-7, -.5 , -2
so
(x+7)(2x+1)(x+2)
well x+2 cancels so we really have
[ (2x-1)(3x+4) ] / [ (x+7)(2x+1) ]
that has zeros at 1/2 and -4/3
and is undefined at -7 and -1/2
To find the roots and the hole of the given rational function, let's start by simplifying the expression and determining its numerator and denominator.
Given Function: f(x) = (24x^3 + 68x^2 + 24x − 32) / (4x^3 + 56x^2 + 236x + 280)
Step 1: Simplify and factor the numerator and denominator.
The numerator of the function is already in simplified form, but we can factor the denominator for further analysis.
Denominator: 4x^3 + 56x^2 + 236x + 280
Start by finding the common factor, which is 4:
4(x^3 + 14x^2 + 59x + 70)
Since the coefficient of x^3 is 1, check for possible rational roots. The possible rational roots for this polynomial equation can be found using the Rational Root Theorem.
Step 2: Apply the Rational Root Theorem to find the possible rational roots.
The Rational Root Theorem states that if a polynomial equation has a rational root p/q, where p is the factors of the constant term (in this case, 70) and q is the factors of the leading coefficient (in this case, 1), then p/q is a possible rational root of the equation.
For this polynomial equation, p can be factors of 70, which are ±1, ±2, ±5, ±7, ±10, ±14, ±35, ±70.
q can be factors of 1, which are ±1.
Therefore, the possible rational roots can be calculated by taking each combination of p/q. In this case, we have 2 × 1 = 2 possible roots.
Step 3: Apply synthetic division to check for the actual rational roots.
Using synthetic division, we can check which of the possible roots from the previous step are actual roots of the polynomial equation.
Let's use the first possible root, p/q = 1.
Dividing the polynomial equation by (x − 1):
1 | 1 14 59 70
| -1 13 72 131
---------------------
| 1 13 72 201
The last entry in the bottom row represents the remainder, which is 201.
Since the remainder is not zero, we can conclude that x − 1 is not a factor of the polynomial equation. Therefore, 1 is not a root.
Now let's try the next possible root, p/q = 2.
Dividing the polynomial equation by (x − 2):
2 | 1 14 59 70
| -2 24 66 270
----------------------
| 1 12 83 340
Again, the last entry in the bottom row represents the remainder, which is 340.
Since the remainder is not zero, we can conclude that x − 2 is not a factor of the polynomial equation. Therefore, 2 is not a root either.
Now we have checked all the possible rational roots, and none of them are actual roots of the polynomial equation.
Step 4: Identifying the hole
Since we did not find any rational roots, there is no hole in the rational function.
To summarize:
- The given rational function does not have any real roots because it did not factor into linear factors with real coefficients.
- There is no hole in the rational function.
Please note that there might be complex roots that could not be determined using the Rational Root Theorem.