One zero for the polynomial function is given. Find all others.3f(x) = x° + 4x - 2x - 20; 2 is a zero.А. 3+ i, 3- iО В. - 3+ і, -3-iС.- 2+ i, -2- iD. - 3-i,3— і Use the discriminant to determine whether the equation has two rational solutions, one rational solution, two irrational solutions, or twononreal complex solutions. Do not actually solve.4x + 8x + 4= 0A. Two irrational solutionsB. Two rational solutionsO C. One rational solutionO D. Two nonreal complex solutions

Answered: One zero for the polynomial function is…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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One zero for the polynomial function is given. Find all others.
3
f(x) = x° + 4x - 2x - 20; 2 is a zero.
А. 3+ i, 3- i
О В. - 3+ і, -3-i
С.
- 2+ i, -2- i
D. - 3-i,3— і

Use the discriminant to determine whether the equation has two rational solutions, one rational solution, two irrational solutions, or two
nonreal complex solutions. Do not actually solve.
4x + 8x + 4= 0
A. Two irrational solutions
B. Two rational solutions
O C. One rational solution
O D. Two nonreal complex solutions

Expert Solution

Step 1

Consider the polynomial $f (x) = x^{3} + 4 x^{2} - 2 x - 20$ .

Given that 2 is a zero of the polynomial. So, $(x - 2)$ is a factor of the polynomial.

Since the given polynomial is of degree 3, there are 3 zeros.

Divide the polynomial $f (x) = x^{3} + 4 x^{2} - 2 x - 20$ by $(x - 2)$ .

Step 2

The given polynomial can be written as,

$f (x) = x^{3} + 4 x^{2} - 2 x - 20 = (x - 2) (x^{2} + 6 x + 10)$

To find the zeroes equate $f (x)$ to zero.

$f (x) = 0 (x - 2) (x^{2} + 6 x + 10) = 0 x - 2 = 0 o r x^{2} + 6 x + 10 = 0$

Solve the equation $x^{2} + 6 x + 10 = 0$ using the formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .

Here a=1,b=6,c=10.

$x = \frac{- 6 \pm \sqrt{6^{2} - 4 (1) (10)}}{2 (1)} = \frac{- 6 \pm \sqrt{36 - 40}}{2} = \frac{- 6 \pm \sqrt{- 4}}{2} = \frac{- 6 \pm \sqrt{{(2 i)}^{2}}}{2} \sin c e i^{2} = - 1 = \frac{- 6 \pm 2 i}{2} = - 3 \pm i$

Therefore, the zeroes of the polynomial are 2, $- 3 - i a n d - 3 + i$ .

So, option B is correct.

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