Chapter 5: Problem 17
For the following exercises, use the Factor Theorem to find all real zeros for the given polynomial function and one factor. $$ f(x)=2 x^{3}+3 x^{2}+x+6 ; x+2 $$
Short Answer
Expert verified
The real zero of the polynomial is \( x = -2 \).
Step by step solution
01
Application of Factor Theorem
According to the Factor Theorem, if \( (x + a) \) is a factor of the polynomial \( f(x) \), then \( f(-a) = 0 \). Given the factor \( (x + 2) \), we substitute \( x = -2 \) into \( f(x) = 2x^3 + 3x^2 + x + 6 \).
02
Evaluate the Polynomial at Specific Point
Substitute \( x = -2 \) into the polynomial: \[ f(-2) = 2(-2)^3 + 3(-2)^2 + (-2) + 6 \].
03
Simplify the Expression
Calculate each term: \[ 2(-2)^3 = 2(-8) = -16 \], \[ 3(-2)^2 = 3(4) = 12 \], \[ -2 = -2 \], and \[ +6 = 6 \]. Sum these to get \( f(-2) = -16 + 12 - 2 + 6 = 0 \). This confirms \( x + 2 \) is a factor because \( f(-2) = 0 \).
04
Perform Polynomial Division
Divide the polynomial \( f(x) = 2x^3 + 3x^2 + x + 6 \) by \( x + 2 \) using synthetic or long division. When performing synthetic division, the coefficients \( 2, 3, 1, 6 \) are used alongside dividing value \(-2\).
05
Process of Synthetic Division
1. Bring down the 2; 2. Multiply \( 2 \), our result, by \( -2 \) (value of the zero of \( x + 2 \)) to get \(-4\); 3. Add \(-4\) to the next coefficient 3, resulting in \(-1\); 4. Repeat these steps to the end yielding the quotient \( 2x^2 - x + 3 \).
06
Find Additional Factors
The quotient \( 2x^2 - x + 3 \) must be further factored. This quadratic does not factor nicely, so use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], with \( a = 2, b = -1, c = 3 \) to find real zeros.
07
Solve the Quadratic Equation
Calculate the discriminant \( b^2 - 4ac = (-1)^2 - 4(2)(3) = 1 - 24 = -23 \). Since it is negative, there are no real solutions. Thus, \( x + 2 \) is the only real factor and gives the only real zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Division
Polynomial division is a method used to divide one polynomial by another, similar to how we divide numbers. In the case of the polynomial \( f(x) = 2x^3 + 3x^2 + x + 6 \) divided by the factor \( x + 2 \), we use either synthetic division or long division. These methods help simplify the original polynomial into a form where we can easily identify factors and zeros.
- In our problem, synthetic division was chosen because it is faster and more straightforward for polynomials with linear divisors.
- We started the process by converting the divisor \( x + 2 \) into the value \( x = -2 \) because we solve at \( x = -a \) where \( a \) is the number attached to \( x \) in the divisor.
- We lined up the coefficients of the polynomial: \( 2, 3, 1, 6 \).
- Brought down the first coefficient, \( 2 \), then multiplied it by \( -2 \), resulting in \( -4 \).
- Added \( -4 \) to the next coefficient \( 3 \) to get \( -1 \), and repeated this process until reaching the end.
- The result was the quadratic \( 2x^2 - x + 3 \), which had to be factored further or solved using another method.
Quadratic Formula
When a quadratic expression does not easily factor, like \( 2x^2 - x + 3 \) from our polynomial division, the quadratic formula becomes our powerful tool. It helps find the roots or zeros of a quadratic equation \( ax^2 + bx + c = 0 \). The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation from our example, \( a = 2 \), \( b = -1 \), and \( c = 3 \). Plugging these values into the formula, we start by calculating the discriminant:
\( b^2 - 4ac = (-1)^2 - 4 \times 2 \times 3 = 1 - 24 = -23 \)
A negative discriminant indicates that the quadratic equation has no real solutions, only complex ones. Therefore, the quadratic part \( 2x^2 - x + 3 \) does not contribute additional real zeros to our original polynomial. In real-world applications, this negative discriminant tells us the curve does not cross the x-axis at any other point except those found from other factors.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation from our example, \( a = 2 \), \( b = -1 \), and \( c = 3 \). Plugging these values into the formula, we start by calculating the discriminant:
A negative discriminant indicates that the quadratic equation has no real solutions, only complex ones. Therefore, the quadratic part \( 2x^2 - x + 3 \) does not contribute additional real zeros to our original polynomial. In real-world applications, this negative discriminant tells us the curve does not cross the x-axis at any other point except those found from other factors.
Real Zeros
Real zeros of a polynomial are the \( x \)-values for which the polynomial equals zero. They correspond to the points where the graph intersects the x-axis. Finding real zeros is fundamental in polynomial graphing and helps us understand the behavior of functions in calculus and algebra.
- The Factor Theorem states that for \( f(x) = 2x^3 + 3x^2 + x + 6 \), \( x+2 \) is a factor if \( f(-2) = 0 \).
- Substituting \( x = -2 \) provides \( f(-2) = 0 \), which confirms that \( x = -2 \) is a real zero of the polynomial.
- The only real zero is found from this factorization method, which ends at \( x=-2 \) due to no other real solutions found by the quadratic part \( 2x^2 - x + 3 \).
In this exercise: