You didn't state the value of x which you want to be shown to be a solution of the given polynomial
3x³ - 19x² + 30x - 8 = 0.
This polynomial of degree three obviously, has three solutions, and 2 is one of the solutions.
I will assume 2 to be the value of x you didn't state, and work with it. But the number is not 2, then it must be one of the remaining two solutions which would follow the same procedure that I will show here.
Answer:
The solutions to the equation
3x³ - 19x² + 30x - 8 = 0
are x = (2, 4, 1/3).
Step-by-step explanation:
Given the polynomial
3x³ - 19x² + 30x - 8 = 0
We need to show that x = 2 is a solution using synthetic division. If x = 2 is a solution, the remainder obtained from the synthetic division
(3x³ - 19x² + 30x - 8) ÷ (x - 2)
is zero.
SYNTHETIC DIVISION
2 | 3 ..... -19 ..... 30 ..... -8
................. 6 .... -26 ...... 8
.... 3........ -13 ........4 .......0->Remainder
Explanation:
The first row:
2 is from the divisor (x - 2). Always take the opposite of what you have. If the divisor is (x + 5), take -5.
(3, -19, 30, -8) are the coefficients of the divisor, the polynomial of degree three.
The coefficient of x³ is dropped to the third column.
2 is multiplied by 3 in the third column to obtain 6 in the second column.
6 is added to -19 to obtain -13.
Again, multiply 2 by -13 in the third column, to obtain -26 in the second column. Add this to 30 to obtain 4
Repeat this process, 2 × 4 = 8
8 + (-8) = 0.
Since 0 is the remainder, x = 2 is a solution to the polynomial.
The remaining factors can be found from the resulting polynomial of degree two whose coefficients were obtained from the synthetic division.
The coefficients of the polynomial are: 3, -13, 4
So, the polynomial is: 3x² - 13x + 4.
The zeros of this polynomial are the remaining factors we are looking for.
We solve
3x² - 13x + 4 = 0
(x - 4)(3x - 1) = 0
x - 4 = 0
=> x = 4
3x - 1 = 0
=> x = 1/3
x = 4, 1/3
The solutions to the equation are x = (2, 4, 1/3).
) we can show:
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Checks true.