Solved on Dec 16, 2023

Find the zeros of the polynomial $f(x) = x^{3} - x + 2x^{2} - 2$ given that $(x - 1)$ is a factor.

STEP 1

Assumptions
1. The polynomial function is $f(x) = x^3 - x + 2x^2 - 2$ .
2. We are given that $(x - 1)$ is a factor of the polynomial.
3. We need to find all zeros of the polynomial.

STEP 2

Since $(x - 1)$ is a factor of $f(x)$ , we can use polynomial division or synthetic division to divide $f(x)$ by $(x - 1)$ to find the other factors.

STEP 3

First, let's reorder the terms of the polynomial by decreasing powers of $x$ :
$f(x) = x^3 + 2x^2 - x - 2$

STEP 4

Now, perform the synthetic division of $f(x)$ by $(x - 1)$ . Set up the synthetic division by writing down the coefficients of $f(x)$ :
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ \end{array}$

STEP 5

Bring down the leading coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

STEP 6

Multiply the divisor root (1) by the leading coefficient (1) and write the result under the second coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \end{array}$

STEP 7

Add the numbers in the second column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \hline & 1 & 3 & & \\ \end{array}$

STEP 8

Repeat the process: multiply the divisor root (1) by the new number in the bottom row (3) and write the result under the third coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \end{array}$

STEP 9

Add the numbers in the third column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \hline & 1 & 3 & 2 & \\ \end{array}$

STEP 10

Repeat the process for the last column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

STEP 11

Since the remainder is 0, $(x - 1)$ is indeed a factor, and the quotient is $x^2 + 3x + 2$ .

STEP 12

Now, factor the quadratic polynomial $x^2 + 3x + 2$ :
$x^2 + 3x + 2 = (x + 1)(x + 2)$

STEP 13

We have factored $f(x)$ completely:
$f(x) = (x - 1)(x + 1)(x + 2)$

STEP 14

Set each factor equal to zero to find the zeros of the polynomial:
$x - 1 = 0$ $x + 1 = 0$ $x + 2 = 0$

STEP 15

Solve each equation for $x$ :
$x = 1$ $x = -1$ $x = -2$
The zeros of the polynomial $f(x)$ are $x = 1$ , $x = -1$ , and $x = -2$ .

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Find the zeros of the polynomial $f(x) = x^{3} - x + 2x^{2} - 2$ given that $(x - 1)$ is a factor.

STEP 1

Assumptions
1. The polynomial function is $f(x) = x^3 - x + 2x^2 - 2$ .
2. We are given that $(x - 1)$ is a factor of the polynomial.
3. We need to find all zeros of the polynomial.

STEP 2

Since $(x - 1)$ is a factor of $f(x)$ , we can use polynomial division or synthetic division to divide $f(x)$ by $(x - 1)$ to find the other factors.

STEP 3

First, let's reorder the terms of the polynomial by decreasing powers of $x$ :
$f(x) = x^3 + 2x^2 - x - 2$

STEP 4

Now, perform the synthetic division of $f(x)$ by $(x - 1)$ . Set up the synthetic division by writing down the coefficients of $f(x)$ :
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ \end{array}$

STEP 5

Bring down the leading coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

STEP 6

Multiply the divisor root (1) by the leading coefficient (1) and write the result under the second coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \end{array}$

STEP 7

Add the numbers in the second column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \hline & 1 & 3 & & \\ \end{array}$

STEP 8

Repeat the process: multiply the divisor root (1) by the new number in the bottom row (3) and write the result under the third coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \end{array}$

STEP 9

Add the numbers in the third column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \hline & 1 & 3 & 2 & \\ \end{array}$

STEP 10

Repeat the process for the last column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

STEP 11

Since the remainder is 0, $(x - 1)$ is indeed a factor, and the quotient is $x^2 + 3x + 2$ .

STEP 12

Now, factor the quadratic polynomial $x^2 + 3x + 2$ :
$x^2 + 3x + 2 = (x + 1)(x + 2)$

STEP 13

We have factored $f(x)$ completely:
$f(x) = (x - 1)(x + 1)(x + 2)$

STEP 14

Set each factor equal to zero to find the zeros of the polynomial:
$x - 1 = 0$ $x + 1 = 0$ $x + 2 = 0$

STEP 15

Solve each equation for $x$ :
$x = 1$ $x = -1$ $x = -2$
The zeros of the polynomial $f(x)$ are $x = 1$ , $x = -1$ , and $x = -2$ .

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Find the zeros of the polynomial f(x)=x3−x+2x2−2f(x) = x^{3} - x + 2x^{2} - 2f(x)=x3−x+2x2−2 given that (x−1)(x - 1)(x−1) is a factor.

STEP 1

Assumptions1. The polynomial function is f(x)=x3−x+2x2−2 f(x) = x^3 - x + 2x^2 - 2 f(x)=x3−x+2x2−2.2. We are given that (x−1) (x - 1) (x−1) is a factor of the polynomial.3. We need to find all zeros of the polynomial.

STEP 2

Since (x−1) (x - 1) (x−1) is a factor of f(x) f(x) f(x), we can use polynomial division or synthetic division to divide f(x) f(x) f(x) by (x−1) (x - 1) (x−1) to find the other factors.

STEP 3

First, let's reorder the terms of the polynomial by decreasing powers of x x x:f(x)=x3+2x2−x−2 f(x) = x^3 + 2x^2 - x - 2 f(x)=x3+2x2−x−2

STEP 4

Now, perform the synthetic division of f(x) f(x) f(x) by (x−1) (x - 1) (x−1). Set up the synthetic division by writing down the coefficients of f(x) f(x) f(x):112−1−2 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ \end{array} 1​1​2​−1​−2​

STEP 5

Bring down the leading coefficient:112−1−21321320 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array} 1​11​213​−132​−220​​

STEP 6

Multiply the divisor root (1) by the leading coefficient (1) and write the result under the second coefficient:112−1−21 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \end{array} 1​1​21​−1​−2​

STEP 7

Add the numbers in the second column:112−1−2113 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \hline & 1 & 3 & & \\ \end{array} 1​11​213​−1​−2​​

STEP 8

Repeat the process: multiply the divisor root (1) by the new number in the bottom row (3) and write the result under the third coefficient:112−1−213 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \end{array} 1​1​21​−13​−2​

STEP 9

Add the numbers in the third column:112−1−213132 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \hline & 1 & 3 & 2 & \\ \end{array} 1​11​213​−132​−2​​

STEP 10

Repeat the process for the last column:112−1−21321320 \begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array} 1​11​213​−132​−220​​

STEP 11

Since the remainder is 0, (x−1) (x - 1) (x−1) is indeed a factor, and the quotient is x2+3x+2 x^2 + 3x + 2 x2+3x+2.

STEP 12

Now, factor the quadratic polynomial x2+3x+2 x^2 + 3x + 2 x2+3x+2:x2+3x+2=(x+1)(x+2) x^2 + 3x + 2 = (x + 1)(x + 2) x2+3x+2=(x+1)(x+2)

STEP 13

We have factored f(x) f(x) f(x) completely:f(x)=(x−1)(x+1)(x+2) f(x) = (x - 1)(x + 1)(x + 2) f(x)=(x−1)(x+1)(x+2)

STEP 14

Set each factor equal to zero to find the zeros of the polynomial:x−1=0 x - 1 = 0 x−1=0 x+1=0 x + 1 = 0 x+1=0 x+2=0 x + 2 = 0 x+2=0

STEP 15

Solve each equation for x x x:x=1 x = 1 x=1 x=−1 x = -1 x=−1 x=−2 x = -2 x=−2The zeros of the polynomial f(x) f(x) f(x) are x=1 x = 1 x=1, x=−1 x = -1 x=−1, and x=−2 x = -2 x=−2.

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Find the zeros of the polynomial $f(x) = x^{3} - x + 2x^{2} - 2$ given that $(x - 1)$ is a factor.

Assumptions
1. The polynomial function is $f(x) = x^3 - x + 2x^2 - 2$ .
2. We are given that $(x - 1)$ is a factor of the polynomial.
3. We need to find all zeros of the polynomial.

Since $(x - 1)$ is a factor of $f(x)$ , we can use polynomial division or synthetic division to divide $f(x)$ by $(x - 1)$ to find the other factors.

First, let's reorder the terms of the polynomial by decreasing powers of $x$ :
$f(x) = x^3 + 2x^2 - x - 2$

Now, perform the synthetic division of $f(x)$ by $(x - 1)$ . Set up the synthetic division by writing down the coefficients of $f(x)$ :
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ \end{array}$

Bring down the leading coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

Multiply the divisor root (1) by the leading coefficient (1) and write the result under the second coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \end{array}$

Add the numbers in the second column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & & \\ \hline & 1 & 3 & & \\ \end{array}$

Repeat the process: multiply the divisor root (1) by the new number in the bottom row (3) and write the result under the third coefficient:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \end{array}$

Add the numbers in the third column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & \\ \hline & 1 & 3 & 2 & \\ \end{array}$

Repeat the process for the last column:
$\begin{array}{c|cccc} 1 & 1 & 2 & -1 & -2 \\ & & 1 & 3 & 2 \\ \hline & 1 & 3 & 2 & 0 \\ \end{array}$

Since the remainder is 0, $(x - 1)$ is indeed a factor, and the quotient is $x^2 + 3x + 2$ .

Now, factor the quadratic polynomial $x^2 + 3x + 2$ :
$x^2 + 3x + 2 = (x + 1)(x + 2)$

We have factored $f(x)$ completely:
$f(x) = (x - 1)(x + 1)(x + 2)$

Set each factor equal to zero to find the zeros of the polynomial:
$x - 1 = 0$ $x + 1 = 0$ $x + 2 = 0$

Solve each equation for $x$ :
$x = 1$ $x = -1$ $x = -2$
The zeros of the polynomial $f(x)$ are $x = 1$ , $x = -1$ , and $x = -2$ .