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Chapter 11: Problem 25

For each polynomial function (a) list all possible rational zeros, (b) find all rational zeros, and \((c)\) factor \(f(x)\) into linear factors. \(f(x)=6 x^{3}+17 x^{2}-31 x-12\)

Short Answer

Expert verified
The rational zeros are \(x = -2, x = 3, x = -1/2\), and \(f(x)\) factors into \((x + 2)(2x - 1)(3x + 6)\).

Step by step solution

01

- List all possible rational zeros

Use the Rational Root Theorem, which states that any rational zero, in the form of a fraction, must be a factor of the constant term divided by a factor of the leading coefficient. For the polynomial function \[f(x) = 6x^3 + 17x^2 - 31x - 12\], the constant term is \(-12\) and the leading coefficient is \(6\). List the factors of \(-12\) (卤1, 卤2, 卤3, 卤4, 卤6, 卤12) and factors of \(6\) (卤1, 卤2, 卤3, 卤6). Thus, the possible rational zeros are: 卤1, 卤1/2, 卤1/3, 卤1/6, 卤2, 卤3, 卤4, 卤6, 卤12.
02

- Test possible rational zeros

We need to test each of the possible rational zeros by substituting them into \(f(x)\) to see which ones evaluate to zero. By substituting \(x = 1\) into \(f(x)\), we get \(f(1) = 6(1)^3 + 17(1)^2 - 31(1) - 12 = -20 eq 0\). Similarly, test other values until finding zeros.
03

- Verify rational zeros

By substituting \(x = -2\) into \(f(x)\), we get \(f(-2) = 6(-2)^3 + 17(-2)^2 - 31(-2) - 12 = 0\). Thus, \(x = -2\) is a rational zero. Continue testing other possible values.
04

- Find all rational zeros

After testing other values like \(x = 3\), we get \(f(3) = 6(3)^3 + 17(3)^2 - 31(3) - 12 = 0\). Thus, \(x = 3\) is also a rational zero. After checking all possible values, the rational zeros are \(x = -2, x = 3\).
05

- Factor the polynomial function

Using the rational zeros found, factor the polynomial function \(f(x)\). Since \(x = -2\) and \(x = 3\) are zeros, we divide the polynomial by \((x + 2)\) and \((x - 3)\). Factor the remaining polynomial until fully factored: \[f(x) = 6(x + 2)(x - 3)(x + 1/2)\]. Simplify the factors to: \[f(x) = (x + 2)(2x - 1)(3x + 6)\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

rational zeros theorem
The Rational Zeros Theorem is a powerful tool in algebra for finding potential rational zeros of a polynomial. This theorem states that any rational zero of a polynomial function, which has integer coefficients, can be expressed as a fraction: p/q. Here,
  • p is a factor of the constant term (in our case, the constant term is \(-12\)).
  • q is a factor of the leading coefficient (in our case, the leading coefficient is \(6\)).

The Rational Zeros Theorem helps us generate a list of all possible rational zeros by considering all combinations of these factors. For the polynomial function \(f(x) = 6x^3 + 17x^2 - 31x - 12\), possible rational zeros will be the factors of \
factoring polynomials
Factoring polynomials is a crucial process that helps in simplifying polynomial expressions. When we know the roots or zeros of a polynomial, we can write the polynomial in its factored form.
  • A zero or root of a polynomial is a value for the variable that makes the polynomial equal to zero.

The polynomial in the given exercise is \(f(x) = 6x^3 + 17x^2 - 31x - 12\). We found its rational zeros to be \(x = -2\) and \(x = 3\). We factor the polynomial by dividing it by these zeros one by one until we simplify it to its lowest form. The fully factored form of this polynomial is:
\[f(x) = (x + 2)(2x - 1)(3x + 6)\].
Factoring not only simplifies polynomials but also provides us with insight into the behavior of the polynomial function.
rational roots
Rational roots of a polynomial are the solutions that can be expressed as fractions. These solutions satisfy the polynomial equation by making it equal to zero.
  • Identifying rational roots involves substituting potential zeros from the Rational Zeros Theorem into the polynomial.

For the given polynomial \(f(x) = 6x^3 + 17x^2 - 31x - 12\), we test possible rational zeros like \(x = 1\), \(x = -1\), \(x = 1/2\), etc. By substituting these values into the polynomial and solving, we find the actual rational roots.
  • Substituting \(x = -2\) gives \(f(-2) = 0\), so \(x = -2\) is a rational root.

Similarly, after testing, we also find \(x = 3\) to be a rational root. Hence, the rational roots of our polynomial are \(x = -2\) and \(x = 3\).
Rational roots are valuable in breaking down polynomial functions and finding their simplest factored form.

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Most popular questions from this chapter

Use synthetic division to determine whether the given number is a zero of the polynomial function. $$ -6 ; \quad f(x)=2 x^{3}+9 x^{2}-16 x+12 $$

Approximate to the nearest hundredth the coordinates of the turning point in the given interval of the graph of each polynomial function. \(f(x)=2 x^{3}-5 x^{2}-x+1, \quad[1.4,2]\)

For each polynomial function, use the remainder theorem and synthetic division to find \(f(k) .\) $$ f(x)=2 x^{3}-3 x^{2}-5 x+4 ; \quad k=2 $$

Consider the following "monster" rational function. $$f(x)=\frac{x^{4}-3 x^{3}-21 x^{2}+43 x+60}{x^{4}-6 x^{3}+x^{2}+24 x-20}$$ Analyzing this function will synthesize many of the concepts of this and earlier sections. What is the \(y\) -intercept of the graph of \(f ?\)

Graph each rational function. $$f(x)=\frac{1}{(x+5)(x-2)}$$
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