Question

Use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor \(f\) over the real numbers. $$ f(x)=2 x^{4}+x^{3}-7 x^{2}-3 x+3 $$

Short answer

The real zeros are \-1, \frac{1}{2}, \sqrt{3}, \pm \sqrt{3}\. The polynomial factors as \(2(x + 1)(x - \frac{1}{2})(x - \sqrt{3})(x + \sqrt{3})\).

Step-by-step solution

State the Rational Zeros Theorem

The Rational Zeros Theorem states that any rational zero of a polynomial function, in the form \( a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \), will be of the form \frac{p}{q}\ where \p\ is a factor of the constant term \a_0\ and \q\ is a factor of the leading coefficient \a_n\.

Identify the factors of the constant term and leading coefficient

The constant term \a_0\ in \(f(x) = 2x^4 + x^3 - 7x^2 - 3x + 3\) is 3. The factors of 3 are \pm 1, \pm 3\.The leading coefficient \a_n\ is 2. The factors of 2 are \pm 1, \pm 2\.Thus, the possible rational zeros are \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\.

Test possible rational zeros using synthetic division

Use synthetic division to test possible rational zeros. Start with +1:\[ \begin{array}{r|rrrrr} \ 1 & 2 & 1 & -7 & -3 & 3 \ \ \ \ \hline \ & 2 & 3 & -4 & -7 & -4 \ \end{array} \] Remainder is -4, so +1 is not a zero.Now test -1:\[ \begin{array}{r|rrrrr} \ -1 & 2 & 1 & -7 & -3 & 3 \ \ \ \ \hline \ & 2 & -1 & -6 & 3 & 0 \ \end{array} \] Remainder is 0, so -1 is a zero.

Divide the polynomial by \(x+1\)

Since -1 is a zero, divide the polynomial \(2x^4 + x^3 - 7x^2 - 3x + 3\) by \(x+1\) using synthetic division:\[ \begin{array}{r|rrrrr} \ -1 & 2 & 1 & -7 & -3 & 3 \ \hline \ & 2 & -1 & -6 & 3 & 0 \ \end{array} \]So the quotient is \2x^3 - x^2 - 6x + 3\.

Find the zeros of the quotient polynomial using the Rational Zeros Theorem

Now apply the Rational Zeros Theorem to \(2x^3 - x^2 - 6x + 3\). The constant term 3 has factors \pm 1, \pm 3\, and the leading coefficient 2 has factors \pm 1, \pm 2\. Possible rational zeros are \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}\.Test +1:\[ \begin{array}{r|rrr} \ 1 & 2 & -1 & -6 & 3 \ \ \ \ \hline \ & 2 & 1 & -5 & -2 \ \end{array} \] Remainder is -2, so +1 is not a zero.Now test +3:\[ \begin{array}{r|rrr} \ 3 & 2 & -1 & -6 & 3 \ \ \ \ \hline \ & 2 & 5 & 9 & 30 \ \end{array} \] Remainder is 30, so +3 is not a zero.

Test remaining possible rational zeros using synthetic division

Now test -3:\[ \begin{array}{r|rrr} \ -3 & 2 & -1 & -6 & 3 \ \ \ \ \hline \ & 2 & -7 & 15 & -42 \ \end{array} \] Remainder is -42, so -3 is not a zero.Test \frac{1}{2}\:\[ \begin{array}{r|rrr} \ \frac{1}{2} & 2 & -1 & -6 & 3 \ \ \ \ \hline \ & 2 & 0 & -6 & 0 \ \end{array} \] Remainder is 0, so \frac{1}{2}\ is a zero.

Divide the quotient polynomial by \(x - \frac{1}{2}\)

Since \frac{1}{2}\ is a zero, divide the quotient polynomial \(2x^3 - x^2 - 6x + 3\) by \(x - \frac{1}{2}\) using synthetic division:\[ \begin{array}{r|rrr} \ \frac{1}{2} & 2 & -1 & -6 & 3 \ \hline \ & 2 & 0 & -6 & 0 \ \end{array} \]So the quotient is \(2x^2 - 6\).

Solve the remaining quadratic equation

Now solve the remaining quadratic equation \(2x^2 - 6 = 0\):\[2x^2 - 6 = 0 \] \[2x^2 = 6 \] \[x^2 = 3 \] \[x = \pm \sqrt{3}\]\.

List all real zeros and factor the polynomial

The real zeros are \x = -1, \frac{1}{2}, \sqrt{3}, \-\sqrt{3}\.Thus, \(f(x)\) factors over the real numbers as \(f(x) = 2(x + 1)(x - \frac{1}{2})(x - \sqrt{3})(x + \sqrt{3})\).

Key concepts

Polynomial Roots

Understanding polynomial roots is essential for solving polynomial equations. A root of a polynomial is a value for which the polynomial evaluates to zero. In simpler terms, if you substitute a root into the polynomial equation, the result will be zero. For instance, for a polynomial function \( f(x) = 2x^4 + x^3 - 7x^2 - 3x + 3 \), if \( x = -1 \) is a root, then \( f(-1) = 0 \).
Identifying the roots helps in breaking down and simplifying complex polynomial equations. Roots can be real or complex, but the Rational Zeros Theorem assists in recognizing potential rational (fractions or whole numbers) roots that are easier to manage.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear binomial of the form \( x - c \). This process streamlines the long polynomial division and is particularly useful when testing possible rational roots identified by the Rational Zeros Theorem.
For example, when testing the possible root \( x = -1 \) for the polynomial \( 2x^4 + x^3 - 7x^2 - 3x + 3 \), you arrange the coefficients in a row and perform operations based on the supposed root. The final row of numbers provides the coefficients of the resulting quotient polynomial and the remainder, helping you decide if the tested root is correct. If the remainder is zero, the test root is indeed a polynomial root.
Synthetic division is not only faster but also less error-prone, allowing for efficient root testing.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler terms, known as factors, that when multiplied together give the original polynomial. This is particularly useful in solving polynomial equations as it reveals the roots directly.
Using the Rational Zeros Theorem and synthetic division, you can systematically simplify the polynomial step by step. For instance, once you determine \( x = -1 \) is a root, you divide the polynomial by \( x + 1 \). This results in a quotient polynomial of a lower degree. Repeating this process with the new polynomial and additional roots eventually brings the polynomial down to a product of linear and possibly quadratic factors that are much easier to handle.
Factoring polynomials is fundamental in algebra and precalculus for understanding polynomial behavior and solving higher-degree equations.

Quadratic Equations

Quadratic equations are polynomials of degree two, usually expressed in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations is a crucial step when polynomial division eventually reduces the polynomial to a quadratic form.
For instance, in solving \( 2x^2 - 6 = 0 \) that results from continuous synthetic division, you simplify and solve it step by step:
\[ 2x^2 - 6 = 0 \] Divide both sides by 2:
\[ x^2 - 3 = 0 \] Isolate \( x^2 \) by adding 3 to both sides:
\[ x^2 = 3 \] Solve for \( x \) by taking the square root:
\[ x = \pm \sqrt{3} \].
Understanding how to solve quadratic equations is vital not only for polynomials but also for various other areas of mathematics, making this a foundational skill.