Pregunta
Given the equation
, complete the following.
a. List all possible rational roots.
b. Use synthetic division to test several possible rational roots in order to identify one actual root.
c. Use the root from part (b) to solve the equation.
a. List all rational roots that are possible according to the Rational Zero Theorem.
(Use commas to separate answers as needed.)
b. Use synthetic division to test several possible rational roots in order to identify one actual root.
One rational root of the given equation is
(Simplify your answer.)
a. List all possible rational roots.
b. Use synthetic division to test several possible rational roots in order to identify one actual root.
c. Use the root from part (b) to solve the equation.
a. List all rational roots that are possible according to the Rational Zero Theorem.
b. Use synthetic division to test several possible rational roots in order to identify one actual root.
One rational root of the given equation is
(Simplify your answer.)
Ask by Beck Wyatt. in the United States
Dec 08,2024
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
a.
b.
b.
Solución
To solve the equation
, we will follow the steps outlined in your request.
Part a: List all possible rational roots.
According to the Rational Root Theorem, the possible rational roots of a polynomial are given by the factors of the constant term divided by the factors of the leading coefficient.
For the polynomial
:
- The constant term is
. - The leading coefficient is
.
The factors of
are
. Since the leading coefficient is
, the possible rational roots are:
Part b: Use synthetic division to test several possible rational roots.
Let’s test the possible rational roots using synthetic division. We will start with
:
Testing
:
The remainder is
, so
is not a root.
Testing
:
The remainder is
, so
is not a root.
Testing
:
The remainder is
, so
is not a root.
Testing
:
The remainder is
, so
is a root.
Part c: Use the root from part (b) to solve the equation.
Since we found that
is a root, we can factor the polynomial as follows:
Now we need to solve the cubic polynomial
. We can use synthetic division again to find more roots or apply the Rational Root Theorem again.
Testing
on the cubic:
The remainder is
, so
is not a root.
Testing
on the cubic:
The remainder is
, so
is a root of the cubic polynomial.
Now we can factor the cubic polynomial as:
Now we need to solve
using the quadratic formula:
Final Roots
Thus, the roots of the original polynomial
are:
Summary
- Possible rational roots:
- One actual root:
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Bonus Knowledge
To tackle the exploration of potential rational roots, we can first apply the Rational Root Theorem. This theorem suggests that any rational solution of the polynomial equation
must have
as a factor of the constant term and
as a factor of the leading coefficient. In our case, the constant term is
and the leading coefficient is
. Hence, all possible rational roots are
.
Now, let’s test these candidates through synthetic division. If we test
:
The remainder is
, which shows
is not a root. Next, let’s try
:
The remainder is
, so
is not a root either. Testing
:
Wow, the remainder is
, which means
is indeed a root! Let’s summarize the findings:
a. The possible rational roots were
.
b. One rational root of the given equation is -1.
b. One rational root of the given equation is -1.
Now that we have identified one actual root, we can use it to further solve the polynomial equation! To do that, we’ll perform synthetic division with
. The reduced polynomial obtained would give us a cubic equation to solve next. Isn’t it fun how one discovered root leads us further down the rabbit hole of polynomial solutions? Keep that mathematical curiosity alive!

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