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If f(x) = x³ 3x² - 22x + 24 and x - 6 is a factor of f(x), then find all of the zeros of f(x) algebraically.

User Genaro
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1 Answer

5 votes

Answer:

x = (3 ± sqrt(-47)) / 2.

Explanation:

If x - 6 is a factor of f(x), then we know that (x - 6) must divide evenly into f(x), which means that there is another factor of f(x) that we can find by polynomial long division or synthetic division.

Using synthetic division:

We start by writing the coefficients of f(x) in order:

1 3 -22 24

We then write the factor (x - 6) to the left of the coefficients, and draw a line:

6 | 1 3 -22 24

We bring down the first coefficient: 1

1 3 -22 24

1

We multiply 6 by the first coefficient and write the result under the second coefficient: 1

6 | 1 3 -22 24

-6

-3

If x - 6 is a factor of f(x), then we know that (x - 6) must divide evenly into f(x), which means that there is another factor of f(x) that we can find by polynomial long division or synthetic division.

Using synthetic division:

We start by writing the coefficients of f(x) in order:

1 3 -22 24

We then write the factor (x - 6) to the left of the coefficients, and draw a line:

6 | 1 3 -22 24

We bring down the first coefficient:

Copy code

1

6 | 1 3 -22 24

1

We multiply 6 by the first coefficient and write the result under the second coefficient:

Copy code

1

6 | 1 3 -22 24

-6

Copy code

-3

We add the second and third coefficients to get -19, then multiply by 6 and write the result under the third coefficient:

1

6 | 1 3 -22 24

-6

-3

42

66

We add the last two numbers to get 90. This means that we can write f(x) as:

f(x) = (x - 6)(x² - 3x + 14)

To find the zeros of f(x), we need to solve the equation (x - 6)(x² - 3x + 14) = 0.

The first factor gives us x = 6.

To solve the quadratic factor, we can use the quadratic formula:

x = (-b ± sqrt(b² - 4ac)) / 2a

In this case, a = 1, b = -3, and c = 14. Substituting these values into the formula, we get:

x = (3 ± sqrt(3² - 4(1)(14))) / 2(1)

x = (3 ± sqrt(-47)) / 2

Since the square root of a negative number is not a real number, the quadratic factor does not have any real zeros.

Therefore, the zeros of f(x) are x = 6, and the complex numbers x = (3 ± sqrt(-47)) / 2.

User Scabbiaza
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