Problem 49
Question
Use synthetic division to show that \(x\) is a solution of the thirddegree polynomial equation, and use the result to factor the polynomial completely. List all the real solutions of the equation. Value of \(x\) \(x=4\) \(x=-6\) \(x=-\frac{3}{2}\) \(x=\frac{1}{3}\) \(x=\sqrt{3}\) \(x=2-\sqrt{5}\) Polynomial Equation $$2 x^{3}-17 x^{2}+12 x+63=0$$
Step-by-Step Solution
Verified Answer
The real solutions of the equation will be determined by synthetic division and solving the quotient equation derived from it.
1Step 1: Synthetic division with \(x=4\)
Firstly, perform the synthetic division using \(x=4\). Checking the values at the bottom row, if the last number is zero, then \(x=4\) is a root of the polynomial. The numbers before the last number would represent the quotient.
2Step 2: Factorization of the polynomial
Assuming that \(x=4\) is determined a root, which means it produced a zero remainder, then the given polynomial can be written as \(2(x - 4)(Q(x)) = 0\), where \(Q(x)\) is the quotient polynomial. Expanding \(Q(x)\) would give the more detailed factor form of the given equation.
3Step 3: Finding the remaining roots
Solve the equation \(Q(x) = 0\), and obtain the values of \(x\). Together with \(x=4\), these values would provide the complete list of real solutions of the equation.
4Step 4: Repetition of steps
Repeat Steps 1 to 3 with the rest of the given values of \(x\) and find the roots and factors of the polynomial equation. If any of the additional values of \(x\) such as -6, -3/2, 1/3, \(\sqrt{3}\), and \(2 - \sqrt{5}\) is a root of the equation, it will result in a zero remainder during synthetic division and can be used in the factorization of the equation.
Key Concepts
Polynomial FactorizationFinding Polynomial RootsThird-Degree Polynomial
Polynomial Factorization
Polynomial factorization is a critical concept in algebra that involves breaking down a polynomial into a product of simpler polynomials that can not be factored any further. This is an essential skill for students as it allows them to simplify complex polynomial expressions and solve polynomial equations more efficiently.
For instance, when analyzing the third-degree polynomial equation given in the exercise, \(2x^3 - 17x^2 + 12x + 63 = 0\), one of the steps in factorization involves identifying its roots, in this case through synthetic division. If synthetic division by a certain value of \(x\) yields a zero remainder, we have found a root and a factor of the polynomial. The polynomial can then be expressed as a product of a linear factor from the root (such as \((x - 4)\) if \(4\) is a root) and a quotient polynomial, representing the remaining factors.
To ensure students grasp this concept, explaining the connection between roots and factors is vital – each real root corresponds to a linear factor of the polynomial, and the polynomial can be expressed as the product of all its factors. Utilizing bullet points for clarity:
For instance, when analyzing the third-degree polynomial equation given in the exercise, \(2x^3 - 17x^2 + 12x + 63 = 0\), one of the steps in factorization involves identifying its roots, in this case through synthetic division. If synthetic division by a certain value of \(x\) yields a zero remainder, we have found a root and a factor of the polynomial. The polynomial can then be expressed as a product of a linear factor from the root (such as \((x - 4)\) if \(4\) is a root) and a quotient polynomial, representing the remaining factors.
To ensure students grasp this concept, explaining the connection between roots and factors is vital – each real root corresponds to a linear factor of the polynomial, and the polynomial can be expressed as the product of all its factors. Utilizing bullet points for clarity:
- Perform synthetic division to check if a number is a root of the polynomial.
- If a zero remainder is obtained, that number is a root, and the polynomial has a corresponding linear factor.
- Repeat this process with different numbers until all factors are found, or use other methods to find the remaining factors.
Finding Polynomial Roots
Finding polynomial roots is like detective work in algebra, looking for the values of \(x\) that satisfy the equation \(P(x) = 0\), where \(P(x)\) is a polynomial. For the exercise provided, one method to unearth these roots is synthetic division. When a particular value for \(x\), say \(4\), results in a zero at the end of the synthetic division process, it indicates that \(x=4\) is indeed a root of the polynomial.
The significance of identifying a polynomial's roots cannot be overstated; roots provide insight into the behavior of the polynomial function and are instrumental in fully factoring a polynomial expression. Once a root is found, it reveals a linear factor of the polynomial. The remaining quotient can then be examined for further roots. Here are a few tips for finding roots effectively:
The significance of identifying a polynomial's roots cannot be overstated; roots provide insight into the behavior of the polynomial function and are instrumental in fully factoring a polynomial expression. Once a root is found, it reveals a linear factor of the polynomial. The remaining quotient can then be examined for further roots. Here are a few tips for finding roots effectively:
- Apply synthetic division with possible rational roots derived from the Rational Root Theorem.
- Examine the remainder; a zero remainder signals a root, whereas a non-zero remainder means it is not a root.
- When a root is found, proceed to factor the polynomial or use the reduced polynomial to find other roots.
Third-Degree Polynomial
Third-degree polynomials, also called cubic polynomials, are equations that have the highest exponent of \(x\) as 3. They take the general form \(ax^3 + bx^2 + cx + d = 0\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). The cubic polynomial given in the exercise is \(2x^3 - 17x^2 + 12x + 63 = 0\), a typical example of a third-degree polynomial.
Third-degree polynomials are particularly interesting because they can have up to three real roots, unlike quadratics, which have at most two. Solving these polynomials might involve a variety of methods such as factoring, synthetic division, the Rational Root Theorem, or numerical approximations when roots cannot be found algebraically. Steps one can follow include:
Third-degree polynomials are particularly interesting because they can have up to three real roots, unlike quadratics, which have at most two. Solving these polynomials might involve a variety of methods such as factoring, synthetic division, the Rational Root Theorem, or numerical approximations when roots cannot be found algebraically. Steps one can follow include:
- Check for obvious roots, such as integers that satisfy the polynomial.
- Utilize synthetic division to confirm the roots and divide out the corresponding linear factor.
- Analyze the reduced polynomial (now of lower degree) for any remaining roots.
Other exercises in this chapter
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