Use synthetic division to show that x is a solution of the third-degree polynomial equation, and use the result to factor the polynomial completely. List all the real zeros of the function.
x^3 - 28x - 48 = 0
Value of x = -4
Please help!!Thank you
To determine if x is a solution of the polynomial equation, we can use synthetic division. Synthetic division is a method used to divide a polynomial by a linear binomial of the form (x - a). Here's how we can perform synthetic division:
1. Set up the synthetic division table. Write the coefficients of the polynomial equation in descending order, including any missing terms. In this case, the coefficients are 1, 0, -28, -48. Place them in the table as follows:
| 1 | 0 | -28 | -48 |
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-4 |
2. Bring down the leading coefficient, which is 1, into the division bar.
| 1 | 0 | -28 | -48 |
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-4 | 1 |
3. Multiply the value in the division bar by -4, and write the result below the second coefficient.
| 1 | 0 | -28 | -48 |
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-4 | 1 | -4 |
4. Add the numbers in the second and third columns and write the sum below the third coefficient.
| 1 | 0 | -28 | -48 |
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-4 | 1 | -4 | 112 |
5. Repeat the process by multiplying the new value in the division bar, which is 112, by -4 and writing the result below the fourth coefficient.
| 1 | 0 | -28 | -48 |
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-4 | 1 | -4 | 112 | -448 |
6. Add the numbers in the third and fourth columns, and write the sum below the fourth coefficient.
| 1 | 0 | -28 | -48 |
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-4 | 1 | -4 | 112 | -448 |
7. The result of synthetic division is in the last column. The coefficients of the quotient are 1, -4, 112, and -448.
If the remainder is zero, then x is a solution to the equation. In this case, the remainder is -448, which is not zero. Therefore, x = -4 is not a solution to the polynomial equation x^3 - 28x - 48 = 0.
To factor the polynomial completely and determine all real zeros of the function, we can use the Rational Root Theorem. According to the theorem, the possible rational roots of the polynomial equation are factors of the constant term divided by factors of the leading coefficient. In this case, the constant term is -48 and the leading coefficient is 1.
The factors of -48 are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, and ±48.
The factors of 1 are ±1.
Combining the possible factors, we get the following potential rational roots:
± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1, ± 8/1, ± 12/1, ± 16/1, ± 24/1, and ± 48/1.
By testing these values using synthetic division or polynomial long division, we can find the real roots of the polynomial equation.
To use synthetic division, we will divide the given polynomial by x + 4 because we know that x = -4 is a solution of the equation.
Performing the synthetic division:
-4 | 1 0 0 -28 -48
-4 16 -64 192
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1 -4 16 -64 144
The result of the synthetic division is 1 - 4 16 - 64 144.
So, the factored form of the third-degree polynomial is (x + 4)(x^2 - 4x + 16).
To find the real zeros of the function, we can set each factor equal to zero:
x + 4 = 0 --> x = -4
x^2 - 4x + 16 = 0
Using the quadratic formula, we can find the roots of the quadratic equation:
x = (-(-4) ± √((-4)^2 - 4(1)(16))) / (2(1))
= (4 ± √(16 - 64)) / 2
= (4 ± √(-48)) / 2
Since the discriminant (√(-48)) is negative, the quadratic equation has no real roots.
Therefore, the real zeros of the function x^3 - 28x - 48 = 0 are x = -4.