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Magazine Chapter 4: Problem 48
Use the Factor Theorem and a calculator to factor the polynomial, as in Example 7. $$g(x)=x^{3}-5 x^{2}-5 x-6$$
Short Answer
Expert verified
Answer: The factored form of the polynomial is \(g(x) = (x + 2)\left(x - \frac{5 + \sqrt{5}}{2}\right)\left(x - \frac{5 - \sqrt{5}}{2}\right)\).
Step by step solution
01
Understanding the Factor Theorem
The Factor Theorem states that if \(g(x)\) is a polynomial and \(g(a) = 0\), then \((x - a)\) is a factor of \(g(x)\). So our goal is to find an "a" such that when we replace \(x\) with \(a\) in the polynomial, the result becomes zero.
02
Finding a zero of the function
Since we are allowed to use a calculator to find a zero of the function, we can first try plugging in integer values of \(x\). Given that the polynomial is $$g(x) = x^3 - 5x^2 - 5x - 6$$ let's try using -2 as x,
$$g(-2) = (-2)^3 - 5(-2)^2 - 5(-2) - 6 = -8 - 20 + 10 - 6 = 0$$
Since \(g(-2) = 0\), we found a zero of the function at x = -2.
03
Using synthetic division to find remaining factors
Now that we found a zero of the function at x = -2, we can use synthetic division to divide the polynomial by \((x - (-2)) = (x + 2)\) to find the remaining factors. Synthetic division is done as follows:
```
-2 | 1 -5 -5 -6
0 10 -20
---------------
1 -5 5 -26
```
The result of the synthetic division is represented by the coefficients of the remaining quadratic polynomial: $$x^2 - 5x + 5$$.
04
Factoring the quadratic
The remaining polynomial is a quadratic polynomial, so we can now try to find the zeroes of this quadratic by using the quadratic formula. The quadratic formula states that for a quadratic equation of the form $$ax^2 + bx + c = 0$$, the zeroes can be found using: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. The values of a, b, and c for our quadratic are: a = 1, b = -5, and c = 5.
Plugging in these values, we get:
$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)}$$
$$x = \frac{5 \pm \sqrt{25 - 20}}{2}$$
$$x = \frac{5 \pm \sqrt{5}}{2}$$
So, the zeroes of the quadratic are: $$x = \frac{5 + \sqrt{5}}{2}$$ and $$x = \frac{5 - \sqrt{5}}{2}$$.
05
Writing the factored polynomial
Now that we have all the zeroes of the polynomial, we can express the polynomial as a factored form by writing the polynomial in terms of its zeroes and their corresponding factors. From our work above, our zeroes are \(-2\), \(\frac{5 + \sqrt{5}}{2}\), and \(\frac{5 - \sqrt{5}}{2}\). So the factored polynomial is:
$$g(x) = (x + 2)\left(x - \frac{5 + \sqrt{5}}{2}\right)\left(x - \frac{5 - \sqrt{5}}{2}\right)$$
Our factored polynomial is now: $$g(x) = (x + 2)\left(x - \frac{5 + \sqrt{5}}{2}\right)\left(x - \frac{5 - \sqrt{5}}{2}\right)$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Factorization
Polynomial factorization is a method used to express a polynomial as the product of its factors. These factors are simpler polynomials, often linear or quadratic, that multiply to give back the original polynomial. For example, when we have a polynomial like \(g(x) = x^3 - 5x^2 - 5x - 6\), we aim to break it down into more manageable parts.
This process is guided by the Factor Theorem, which helps us determine if a linear expression \((x - a)\) is indeed a factor. If substituting \(a\) into the polynomial results in zero, then \((x - a)\) is a factor of the polynomial.
This process is guided by the Factor Theorem, which helps us determine if a linear expression \((x - a)\) is indeed a factor. If substituting \(a\) into the polynomial results in zero, then \((x - a)\) is a factor of the polynomial.
- First, use integer trials to find potential zeroes.
- Use the Factor Theorem to confirm these zeroes.
Synthetic Division
Synthetic division is a shortcut method for dividing a polynomial by a linear factor such as \((x + 2)\). It's particularly useful after finding a zero via the Factor Theorem. This process simplifies the division without writing out the long division. To perform synthetic division:
- Write the coefficients of the polynomial in a row.
- Use the zero found (e.g., -2 for \(g(x)\)) as the divisor.
- Bring down the first coefficient, then multiply by the zero, and add to the next coefficient.
- Repeat the process until all coefficients are addressed.
Quadratic Formula
Once a polynomial is simplified to a quadratic form, like \(x^2 - 5x + 5\), we use the quadratic formula to find its roots. This formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), provides a way to solve any quadratic equation. Here's a breakdown of its use:
- Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation \(ax^2 + bx + c = 0\).
- Substitute these values into the formula ensuring careful calculation of the discriminant \(b^2 - 4ac\).
- Solve for \(x\) using the positive and negative square root options.
