Math / Algebra

QuestionWatch viaeo
Question The function $f(x)=2 x^{4}+x^{3}-4 x^{2}-4 x-16$ has at least two rational roots. Use the ratio theorem to find those roots, then proceed to find all complex roots. (Note: roots may be integer, irrational, and/or complex.)
Answer Attempt 1 out of 2
There are $\square$ four roots :

Studdy Solution

STEP 1

What is this asking? We're on a root hunt!
We need to find all the roots (zeros) of a funky function, starting with at least two rational ones using the Rational Root Theorem, and then uncovering any hidden irrational or complex roots. Watch out! Don't forget that complex roots like to travel in pairs (conjugates)!
Also, the Rational Root Theorem gives us potential rational roots; we still need to test them!

STEP 2

1. Find Potential Rational Roots
2. Test Potential Roots
3. Factor the Polynomial
4. Find Remaining Roots

STEP 3

Let's define our function: $f(x) = 2x^4 + x^3 - 4x^2 - 4x - 16$ .
The Rational Root Theorem tells us that any rational root will be of the form $p/q$ , where $p$ is a factor of the constant term (-16) and $q$ is a factor of the leading coefficient (2).

STEP 4

The factors of -16 are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ .
The factors of 2 are: $\pm 1, \pm 2$ .

STEP 5

So, our potential rational roots are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8$ .
We can simplify this list by removing duplicates to: $\pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ .

STEP 6

Let's start testing!
We're looking for values of $x$ that make $f(x) = 0$ .
Let's try $x = 2$ : $f(2) = 2(2)^4 + (2)^3 - 4(2)^2 - 4(2) - 16 = 32 + 8 - 16 - 8 - 16 = 0$ .
Woohoo! $x = 2$ is a root!

STEP 7

Let's try $x = -2$ : $f(-2) = 2(-2)^4 + (-2)^3 - 4(-2)^2 - 4(-2) - 16 = 32 - 8 - 16 + 8 - 16 = 0$ .
Double woohoo! $x = -2$ is also a root!

STEP 8

Since $x = 2$ and $x = -2$ are roots, we know $(x - 2)$ and $(x + 2)$ are factors.
We can multiply these together to get $(x - 2)(x + 2) = x^2 - 4$ .

STEP 9

Now, we can perform polynomial long division to divide $f(x)$ by $x^2 - 4$ : $\begin{array}{cccccc} & 2x^2 & +x & \\ x^2 - 4 & 2x^4 & +x^3 & -4x^2 & -4x & -16 \\ & 2x^4 & & -8x^2 & \\ & & x^3 & +4x^2 & -4x & \\ & & x^3 & & -4x & \\ & & & 4x^2 & & -16 \\ & & & 4x^2 & & -16 \\ & & & & & 0 \\ \end{array}$ So, $f(x) = (x^2 - 4)(2x^2 + x + 4)$ .

STEP 10

We need to find the roots of $2x^2 + x + 4 = 0$ .
We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$ , $b = 1$ , and $c = 4$ .

STEP 11

Plugging in the values, we get: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 - 32}}{4} = \frac{-1 \pm \sqrt{-31}}{4} = \frac{-1 \pm i\sqrt{31}}{4}$

STEP 12

There are four roots: $x = 2$ , $x = -2$ , $x = \frac{-1 + i\sqrt{31}}{4}$ , and $x = \frac{-1 - i\sqrt{31}}{4}$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Find Potential Rational Roots
2. Test Potential Roots
3. Factor the Polynomial
4. Find Remaining Roots

STEP 3

Let's define our function: $f(x) = 2x^4 + x^3 - 4x^2 - 4x - 16$ .
The Rational Root Theorem tells us that any rational root will be of the form $p/q$ , where $p$ is a factor of the constant term (-16) and $q$ is a factor of the leading coefficient (2).

STEP 4

The factors of -16 are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ .
The factors of 2 are: $\pm 1, \pm 2$ .

STEP 5

So, our potential rational roots are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8$ .
We can simplify this list by removing duplicates to: $\pm \frac{1}{2}, \pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ .

STEP 6

Let's start testing!
We're looking for values of $x$ that make $f(x) = 0$ .
Let's try $x = 2$ : $f(2) = 2(2)^4 + (2)^3 - 4(2)^2 - 4(2) - 16 = 32 + 8 - 16 - 8 - 16 = 0$ .
Woohoo! $x = 2$ is a root!

STEP 7

Let's try $x = -2$ : $f(-2) = 2(-2)^4 + (-2)^3 - 4(-2)^2 - 4(-2) - 16 = 32 - 8 - 16 + 8 - 16 = 0$ .
Double woohoo! $x = -2$ is also a root!

STEP 8

Since $x = 2$ and $x = -2$ are roots, we know $(x - 2)$ and $(x + 2)$ are factors.
We can multiply these together to get $(x - 2)(x + 2) = x^2 - 4$ .

STEP 9

STEP 10

We need to find the roots of $2x^2 + x + 4 = 0$ .
We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$ , $b = 1$ , and $c = 4$ .

STEP 11

Plugging in the values, we get: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 - 32}}{4} = \frac{-1 \pm \sqrt{-31}}{4} = \frac{-1 \pm i\sqrt{31}}{4}$

STEP 12

There are four roots: $x = 2$ , $x = -2$ , $x = \frac{-1 + i\sqrt{31}}{4}$ , and $x = \frac{-1 - i\sqrt{31}}{4}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Find Potential Rational Roots2. Test Potential Roots3. Factor the Polynomial4. Find Remaining Roots

STEP 3

STEP 4

The **factors of -16** are: ±1,±2,±4,±8,±16\pm 1, \pm 2, \pm 4, \pm 8, \pm 16±1,±2,±4,±8,±16.The **factors of 2** are: ±1,±2\pm 1, \pm 2±1,±2.

STEP 5

STEP 6

STEP 7

Let's try x=−2x = -2x=−2: f(−2)=2(−2)4+(−2)3−4(−2)2−4(−2)−16=32−8−16+8−16=0f(-2) = 2(-2)^4 + (-2)^3 - 4(-2)^2 - 4(-2) - 16 = 32 - 8 - 16 + 8 - 16 = 0f(−2)=2(−2)4+(−2)3−4(−2)2−4(−2)−16=32−8−16+8−16=0.Double woohoo! x=−2x = -2x=−2 is also a root!

STEP 8

Since x=2x = 2x=2 and x=−2x = -2x=−2 are roots, we know (x−2)(x - 2)(x−2) and (x+2)(x + 2)(x+2) are factors.We can **multiply** these together to get (x−2)(x+2)=x2−4(x - 2)(x + 2) = x^2 - 4(x−2)(x+2)=x2−4.

STEP 9

STEP 10

We need to find the roots of 2x2+x+4=02x^2 + x + 4 = 02x2+x+4=0.We can use the **quadratic formula**: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​ where a=2a = 2a=2, b=1b = 1b=1, and c=4c = 4c=4.

STEP 11

STEP 12

There are **four** roots: x=2x = 2x=2, x=−2x = -2x=−2, x=−1+i314x = \frac{-1 + i\sqrt{31}}{4}x=4−1+i31​​, and x=−1−i314x = \frac{-1 - i\sqrt{31}}{4}x=4−1−i31​​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find Potential Rational Roots
2. Test Potential Roots
3. Factor the Polynomial
4. Find Remaining Roots

The factors of -16 are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ .
The factors of 2 are: $\pm 1, \pm 2$ .

Let's try $x = -2$ : $f(-2) = 2(-2)^4 + (-2)^3 - 4(-2)^2 - 4(-2) - 16 = 32 - 8 - 16 + 8 - 16 = 0$ .
Double woohoo! $x = -2$ is also a root!

Since $x = 2$ and $x = -2$ are roots, we know $(x - 2)$ and $(x + 2)$ are factors.
We can multiply these together to get $(x - 2)(x + 2) = x^2 - 4$ .

We need to find the roots of $2x^2 + x + 4 = 0$ .
We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$ , $b = 1$ , and $c = 4$ .

There are four roots: $x = 2$ , $x = -2$ , $x = \frac{-1 + i\sqrt{31}}{4}$ , and $x = \frac{-1 - i\sqrt{31}}{4}$ .