Problem 21
Question
Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=x^{3}+24 x^{2}+214 x+740$$
Step-by-Step Solution
Verified Answer
-4, -10 - \(\sqrt{15}\), and -10 + \(\sqrt{15}\) are the zeros of the function. The polynomial written as a product of linear factors is \(f(x) = (x + 4)(x - (-10 - \sqrt{15}))(x - (-10 + \sqrt{15})) \)
1Step 1: Finding the First Zero
Use the Rational Root Theorem to trial-and-error factors of 740 till finding a zero. After several attempts, you'll find that -4 is a zero of the function because \(f(-4) = 0\).
2Step 2: Simplify the Cubic Function
Perform synthetic division using -4 as the divisor. The result will be a quadratic function \(x^{2} + 20x + 185\).
3Step 3: Find Remaining Zeros
Use the quadratic formula, \(x = [-b \pm \sqrt{b^2 - 4ac}] / 2a\) to find the remaining roots of the quadratic equation. The roots are found to be -10 - \(\sqrt{15}\) and -10 + \(\sqrt{15}\).
4Step 4: Write the Polynomial as Product of Linear Factors
The original cubic function can be written as the product of linear factors: \(f(x) = (x + 4)(x - (-10 - \sqrt{15}))(x - (-10 + \sqrt{15})) \).
Key Concepts
Rational Root TheoremSynthetic DivisionQuadratic FormulaLinear Factors of a Polynomial
Rational Root Theorem
The Rational Root Theorem is a powerful tool for finding the possible zeros of a polynomial function. It states that any rational zero, or root, of a polynomial equation with integer coefficients will be a fraction where the numerator is a factor of the constant term and the denominator is a factor of the leading coefficient. In the exercise given, we are dealing with the polynomial function
\(f(x) = x^{3} + 24x^{2} + 214x + 740\).
To apply the Rational Root Theorem, we look at the factors of the constant term, 740, and the factors of the leading coefficient, which in this case is 1 (since the leading term is \(x^{3}\)). This means any rational root must be a factor of 740. By systematically trying these factors, we discovered that -4 is a zero of the function because \(f(-4) = 0\).
\(f(x) = x^{3} + 24x^{2} + 214x + 740\).
To apply the Rational Root Theorem, we look at the factors of the constant term, 740, and the factors of the leading coefficient, which in this case is 1 (since the leading term is \(x^{3}\)). This means any rational root must be a factor of 740. By systematically trying these factors, we discovered that -4 is a zero of the function because \(f(-4) = 0\).
Synthetic Division
Synthetic division is an efficient method for dividing a polynomial by a binomial of the form \(x - c\) and is particularly useful when we know one zero of the polynomial function. Once we have a zero, such as -4 in our exercise,
\(f(x) = x^{3} + 24x^{2} + 214x + 740\),
we can use synthetic division to simplify our cubic function to a quadratic one. It is done by placing the coefficients of \(f(x)\) in a row and using the zero (-4) to perform the division operation. After completing the synthetic division, we found a simplified quadratic function \(x^{2} + 20x + 185\). Synthetic division provides a simpler equation to work with and brings us a step closer to finding all zeros of the polynomial.
\(f(x) = x^{3} + 24x^{2} + 214x + 740\),
we can use synthetic division to simplify our cubic function to a quadratic one. It is done by placing the coefficients of \(f(x)\) in a row and using the zero (-4) to perform the division operation. After completing the synthetic division, we found a simplified quadratic function \(x^{2} + 20x + 185\). Synthetic division provides a simpler equation to work with and brings us a step closer to finding all zeros of the polynomial.
Quadratic Formula
When we are left with a quadratic equation after simplifying a polynomial, as we were with \(x^{2} + 20x + 185\) from synthetic division, the quadratic formula comes into play. This formula,
\(x = [-b \pm \sqrt{b^2 - 4ac}] / 2a\),
gives us the solutions for any quadratic equation \(ax^{2} + bx + c = 0\). By substituting the coefficients of our quadratic function into this formula, \(a=1\), \(b=20\), and \(c=185\), we can calculate the remaining zeros of the original cubic function. The quadratic formula yields two complex solutions, -10 - \(\sqrt{15}\) and -10 + \(\sqrt{15}\), which are the roots of our simplified quadratic function and, thus, also zeros of the original polynomial.
\(x = [-b \pm \sqrt{b^2 - 4ac}] / 2a\),
gives us the solutions for any quadratic equation \(ax^{2} + bx + c = 0\). By substituting the coefficients of our quadratic function into this formula, \(a=1\), \(b=20\), and \(c=185\), we can calculate the remaining zeros of the original cubic function. The quadratic formula yields two complex solutions, -10 - \(\sqrt{15}\) and -10 + \(\sqrt{15}\), which are the roots of our simplified quadratic function and, thus, also zeros of the original polynomial.
Linear Factors of a Polynomial
Writing a polynomial as a product of linear factors is essentially expressing the polynomial in its fully factored form, with each factor being a first-degree polynomial \(x - r\), where \(r\) is a zero of the function. In our original cubic function,
\(f(x) = x^{3} + 24x^{2} + 214x + 740\),
we found the zeros to be -4, -10 - \(\sqrt{15}\), and -10 + \(\sqrt{15}\). Therefore, we can write the function as a product of linear factors:
\(f(x) = (x + 4)(x + 10 + \sqrt{15})(x + 10 - \sqrt{15})\).
Each factor \((x - r)\) corresponds to a zero of the polynomial. This form is particularly insightful as it directly reveals the x-values where the function crosses the x-axis and enables us to understand the behavior of the polynomial around its zeros.
\(f(x) = x^{3} + 24x^{2} + 214x + 740\),
we found the zeros to be -4, -10 - \(\sqrt{15}\), and -10 + \(\sqrt{15}\). Therefore, we can write the function as a product of linear factors:
\(f(x) = (x + 4)(x + 10 + \sqrt{15})(x + 10 - \sqrt{15})\).
Each factor \((x - r)\) corresponds to a zero of the polynomial. This form is particularly insightful as it directly reveals the x-values where the function crosses the x-axis and enables us to understand the behavior of the polynomial around its zeros.
Other exercises in this chapter
Problem 21
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