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Math  /  Algebra

QuestionAnswer the following questions about the equation below. 12x3+53x2−34x+5=012 x^{3}+53 x^{2}-34 x+5=0 (a) List all rational roots that are possible according to the Rational Zero Theorem. Choose the correct A. ±1,±2,±3,±4,±6,±12\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 B. ±1,±5,±12,±52,±13,±53,±14,±54,±16,±56,±112,±512\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{1}{4}, \pm \frac{5}{4}, \pm \frac{1}{6}, \pm \frac{5}{6}, \pm \frac{1}{12}, \pm \frac{5}{12} C. ±1,±2,±3,±4,±6,±12,±15,±25,±35,±45,±65,±125\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{5}, \pm \frac{2}{5}, \pm \frac{3}{5}, \pm \frac{4}{5}, \pm \frac{6}{5}, \pm \frac{12}{5} D. ±1,±5\pm 1, \pm 5 (b) Use synthetic division to test several possible rational roots in order to identify one actual root.
One rational root of the given equation is □\square (Simplify your answer.)

Studdy Solution

STEP 1

1. We are given a polynomial equation 12x3+53x2−34x+5=0 12x^3 + 53x^2 - 34x + 5 = 0 .
2. We need to use the Rational Root Theorem to list possible rational roots.
3. The Rational Root Theorem states that any rational solution, expressed as a fraction pq \frac{p}{q} , has p p as a factor of the constant term and q q as a factor of the leading coefficient.
4. We will use synthetic division to test possible rational roots.

STEP 2

1. Apply the Rational Root Theorem to list possible rational roots.
2. Use synthetic division to test possible rational roots.
3. Identify one actual rational root.

STEP 3

Identify the constant term and the leading coefficient of the polynomial:
The constant term is 5 5 and the leading coefficient is 12 12 .

STEP 4

List the factors of the constant term 5 5 :
The factors of 5 5 are ±1,±5 \pm 1, \pm 5 .

STEP 5

List the factors of the leading coefficient 12 12 :
The factors of 12 12 are ±1,±2,±3,±4,±6,±12 \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 .

STEP 6

According to the Rational Root Theorem, the possible rational roots are all combinations of the factors of the constant term divided by the factors of the leading coefficient:
±11,±51,±12,±52,±13,±53,±14,±54,±16,±56,±112,±512 \pm \frac{1}{1}, \pm \frac{5}{1}, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{1}{4}, \pm \frac{5}{4}, \pm \frac{1}{6}, \pm \frac{5}{6}, \pm \frac{1}{12}, \pm \frac{5}{12}
Thus, the correct choice is B.

STEP 7

Use synthetic division to test possible rational roots. We will test x=1 x = 1 as a potential root:
Set up synthetic division with coefficients 12,53,−34,5 12, 53, -34, 5 .
Perform synthetic division:
1. Bring down the 12.
2. Multiply 12 by 1, add to 53, giving 65.
3. Multiply 65 by 1, add to -34, giving 31.
4. Multiply 31 by 1, add to 5, giving 36.
Since the remainder is not zero, x=1 x = 1 is not a root.

STEP 8

Test another possible rational root, x=−12 x = -\frac{1}{2} :
Set up synthetic division with coefficients 12,53,−34,5 12, 53, -34, 5 .
Perform synthetic division:
1. Bring down the 12.
2. Multiply 12 by −12-\frac{1}{2}, add to 53, giving 47.
3. Multiply 47 by −12-\frac{1}{2}, add to -34, giving -57.5.
4. Multiply -57.5 by −12-\frac{1}{2}, add to 5, giving 33.75.
Since the remainder is not zero, x=−12 x = -\frac{1}{2} is not a root.

STEP 9

Test another possible rational root, x=13 x = \frac{1}{3} :
Set up synthetic division with coefficients 12,53,−34,5 12, 53, -34, 5 .
Perform synthetic division:
1. Bring down the 12.
2. Multiply 12 by 13\frac{1}{3}, add to 53, giving 57.
3. Multiply 57 by 13\frac{1}{3}, add to -34, giving -15.
4. Multiply -15 by 13\frac{1}{3}, add to 5, giving 0.
Since the remainder is zero, x=13 x = \frac{1}{3} is a root.
The rational root of the given equation is 13 \boxed{\frac{1}{3}} .

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