Problem 26 - Algebra 2

Question

Find all of the zeros of each function. \(h(x)=10 x^{3}-17 x^{2}-7 x+2\)

Step-by-Step Solution

Answer

The zeros of the function are \(-\frac{1}{2}\), \(\frac{1}{5}\), and \(2\).

Step 1: Identify Possible Rational Zeros

According to the Rational Root Theorem, the possible rational zeros of a polynomial are the factors of the constant term divided by the factors of the leading coefficient. For the function \(h(x)=10x^3 - 17x^2 - 7x + 2\), the constant term is 2 and the leading coefficient is 10. The factors of 2 are \(\pm 1, \pm 2\), and the factors of 10 are \(\pm 1, \pm 2, \pm 5, \pm 10\). The possible rational zeros are \(\pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 2\).

Step 2: Use Synthetic Division

Test each possible rational zero using synthetic division. First, use \(+1\) as a test zero. Set up synthetic division for \(10x^3 - 17x^2 - 7x + 2\):\[\begin{array}{r|rrrr}1 & 10 & -17 & -7 & 2 \ & & 10 & -7 & -14 \\hline & 10 & -7 & -14 & -12\end{array}\]The remainder is not zero, so \(+1\) is not a zero. Repeat this process for each possible zero.

Step 3: Find a Zero Using Synthetic Division

After testing, \(-\frac{1}{2}\) is found to be a zero using synthetic division:\[\begin{array}{r|rrrr}-\frac{1}{2} & 10 & -17 & -7 & 2 \ & & -5 & 11 & -2 \\hline & 10 & -22 & 4 & 0\end{array}\]The remainder is zero, so \(-\frac{1}{2}\) is a zero.

Step 4: Factor the Polynomial

Use the zero found to factor \(h(x)\). We have \(h(x) = (x + \frac{1}{2})(10x^2 - 22x + 4)\). Simplify the quadratic by factoring out the greatest common divisor:\(10x^2 - 22x + 4 = 2(5x^2 - 11x + 2)\)Thus, \(h(x) = (x + \frac{1}{2})(2)(5x^2 - 11x + 2)\).

Step 5: Factor the Quadratic Equation

Now, solve \(5x^2 - 11x + 2 = 0\) by factoring further. The quadratic factors as:\((5x - 1)(x - 2) = 0\)The solutions to this equation are \(x = \frac{1}{5}\) and \(x = 2\).

Step 6: List All Zeros

The zeros of the original polynomial \(h(x)=10x^3 - 17x^2 - 7x + 2\) are: \(-\frac{1}{2}\), \(\frac{1}{5}\), and \(2\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a valuable tool when finding the zeros of a polynomial function. In this case, for the polynomial \( h(x) = 10x^3 - 17x^2 - 7x + 2 \), the theorem helps us identify potential rational zeros. To apply the Rational Root Theorem, one needs to consider two things:

• The factors of the constant term of the polynomial, which is 2 in this example. The factors of 2 are \( \pm 1 \) and \( \pm 2 \).
• The factors of the leading coefficient, which is 10. The factors of 10 are \( \pm 1, \pm 2, \pm 5, \pm 10 \).

Next, form all possible fractions where numerators are factors of the constant term and denominators are factors of the leading coefficient. Therefore, the list of all possible rational zeros is \( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 2 \). While this theorem only provides possible zeros, it significantly narrows down the candidates, saving analysis time.

Synthetic Division

Synthetic division offers a simplified way to divide a polynomial by a linear factor of the form \( x - c \). When searching for zeros of \( h(x) = 10x^3 - 17x^2 - 7x + 2 \), one tries each potential zero identified by the Rational Root Theorem to see if it results in a remainder of zero, thereby confirming it as an actual zero.
To perform synthetic division:

• Write down the coefficients of the polynomial: 10, -17, -7, and 2.
• Select a potential zero, for example, -\(\frac{1}{2} \), and place it at the start.
• Carry out the synthetic division process by bringing down the first coefficient, multiplying and adding as per the structure of synthetic division.
• If the remainder is zero, \( c \) is a root. Otherwise, it is not.

When \( -\frac{1}{2} \) was tested, it yielded a zero remainder, confirming it as a zero of \( h(x) \).

Factorization

Factorization involves expressing a polynomial as a product of its factors, and it becomes easier once you have identified at least one zero using synthetic division. After identifying \( -\frac{1}{2} \) as a zero, \( h(x) \) could be factored into \( (x + \frac{1}{2})(10x^2 - 22x + 4) \).

• The factor \( x + \frac{1}{2} \) comes from the zero \( -\frac{1}{2} \). When a zero is found, \( (x - c) \) becomes a factor, here adjusted for the negative zero.
• The remaining polynomial, \( 10x^2 - 22x + 4 \), can often be further factored. Simplify by factoring out any greatest common divisor, leading to \( 2(5x^2 - 11x + 2) \).
• This step reduces the polynomial, making it easier to solve or further analyze.

Quadratic Equation

Once the polynomial is reduced to a quadratic equation, solving it can provide any additional zeros. The remaining quadratic \( 5x^2 - 11x + 2 \) was solved through factorization. This process involved:

• Looking for two numbers that multiply to the product of 5 (leading coefficient) and 2 (constant term), in this case, 10, and adding to -11 (middle term).
• Finding that these numbers are -1 and -10, suggesting the factorization \( (5x - 1)(x - 2) = 0 \).
• Each factor provides a solution or zero: solving \( 5x - 1 = 0 \) yields \( x = \frac{1}{5} \), and solving \( x - 2 = 0 \) yields \( x = 2 \).

This process confirms all the zeros of the original polynomial: \( -\frac{1}{2}, \frac{1}{5}, \text{ and } 2 \). Understanding these steps in solving a quadratic can deeply cement how polynomials behave with respect to their zeros.