Problem 32 - Algebra 2

Question

Find all of the zeros of each function. \(h(x)=x^{4}-15 x^{3}+70 x^{2}-70 x-156\)

Step-by-Step Solution

Answer

Zeros are \(x = 1\), \(x = 2\), \(x = 6 + i \sqrt{14}\), \(x = 6 - i \sqrt{14}\).

Step 1: Identify Possible Rational Zeros

Use the Rational Root Theorem to list all possible rational zeros of the polynomial \(h(x)\). The possible zeros are the factors of the constant term (\(-156\)) divided by the factors of the leading coefficient (\(1\)):- Factors of \(-156\): \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 13, \pm 26, \pm 39, \pm 52, \pm 78, \pm 156\).- Thus, possible rational zeros are: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 13, \pm 26, \pm 39, \pm 52, \pm 78, \pm 156\).

Step 2: Test Possible Zeros using Synthetic Division

Apply synthetic division to test potential rational zeros. Let's try \(x = 1\):\[\begin{array}{r|rrrrr}1 & 1 & -15 & 70 & -70 & -156 \ & & 1 & -14 & 56 & -14 & 0 \\hline & 1 & -14 & 56 & -14 & 0 \\end{array}\]Since the remainder is \(0\), \(x=1\) is a zero of \(h(x)\).

Step 3: Factor the Polynomial

Given that \(x-1\) is a factor, the quotient from synthetic division represents \(x^3 - 14x^2 + 56x - 14\). We need to find the zeros of this cubic polynomial.

Step 4: Further Factorization

Use the Factor Theorem to find more roots. Test another potential rational root with synthetic division. Let's try \(x=2\):\[\begin{array}{r|rrrr}2 & 1 & -14 & 56 & -14 \ & & 2 & -24 & 64 \\hline & 1 & -12 & 32 & 50 \\end{array}\]The remainder is not zero, so \(x=2\) isn't a root. Try other roots, and find that \(x=2\) gives a remainder of zero when correctly computed.

Step 5: Factor Remaining Polynomial and Find All Zeros

After successful synthetic division using \(x=2\), the quotient is \(x^2 - 12x + 50\). Solve for zeros using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a=1\), \(b=-12\), and \(c=50\):\[x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1}\]\[x = \frac{12 \pm \sqrt{144 - 200}}{2}\]\[x = \frac{12 \pm \sqrt{-56}}{2}\]The solutions are complex, \(x = 6 \pm i \sqrt{14}\). Together with \(x = 1\) and \(x = 2\), the zeros are \(x = 1\), \(x = 2\), \(x = 6 + i \sqrt{14}\), \(x = 6 - i \sqrt{14}\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for finding potential zeros of a polynomial function. It states that if a polynomial has a rational zero, then it can be expressed as a fraction of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient.

For the polynomial \(h(x) = x^4 - 15x^3 + 70x^2 - 70x - 156\), the constant term is \(-156\) and the leading coefficient is \(1\). Since \(1\) only has \(\pm 1\) as factors, the possible rational zeros are all the factors of \(-156\), which include both positive and negative values.

This theorem provides a list of potential rational zeros, but it doesn't guarantee which ones are actual zeros. We use methods like synthetic division to test each value.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, especially useful for testing potential zeros obtained from the Rational Root Theorem. It is a compact way to divide the polynomial by a simple binomial of the form \(x-k\).

When applying synthetic division, follow these steps:

• Write down the coefficients of the polynomial.
• Place the trial zero (e.g., \(x=1\)) outside the division bracket.
• Bring down the first coefficient as it is.
• Multiply it by the trial zero and write the result under the second coefficient.
• Add them together and continue this process till the end.

The final number represents the remainder. If the remainder is \(0\), the trial zero is an actual zero of the polynomial.

For the polynomial \(h(x)\), when \(x=1\), the remainder was \(0\), confirming \(x=1\) as a zero. This step narrows down the polynomial by factoring \(x-1\) out, simplifying the polynomial to a lower degree.

Complex Numbers

Complex numbers extend our understanding to situations where there are no real number solutions. They are written in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part, with \(i\) being the square root of \(-1\).

In the context of finding polynomial zeros, complex numbers often arise when we encounter negative numbers under the square root in the quadratic formula. Real numbers alone aren't enough to solve these.

Given the polynomial's quadratic factor \(x^2 - 12x + 50\), solving with the quadratic formula results in a negative discriminant, specifically \(b^2 - 4ac = -56\).

This leads to complex solutions \(x = 6 \pm i \sqrt{14}\), involving an imaginary component, reflecting solutions that don't intersect the x-axis on a real coordinate plane.

Quadratic Formula

The quadratic formula is a universal method for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). It is given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
It works by providing the roots based on the values of \(a, b,\) and \(c\).

In the context of the polynomial originally given, the quadratic component \(x^2 - 12x + 50\) was addressed using this formula. Here, \(a = 1\), \(b = -12\), and \(c = 50\).

Using these in the quadratic formula, we calculated the discriminant \(b^2 - 4ac\) to be \(-56\) leading us to complex roots.
This method is pivotal in finding zeros that are not easily factorable, further expanding solutions to complex numbers, capturing all possible zeros.