MATH SOLVE

9 months ago

Q:
# Use synthetic division and factor theorem to solve the below equation:
〖-3z〗^3+140-139z+38z^2=0

Accepted Solution

A:

Answer
To solve the equation -3z^3 + 38z^2 - 139z + 140 = 0 using synthetic division and the factor theorem, we need to first find a factor of the polynomial. One way to do this is to try different values of z until we find one that makes the polynomial equal to zero. However, since the polynomial has degree three, we know that it has at most three roots, so we can use the rational root theorem to narrow down the possible values of z.
The rational root theorem states that if a polynomial with integer coefficients has a rational root p/q, where p and q are integers with no common factors, then p must be a factor of the constant term of the polynomial, and q must be a factor of the leading coefficient of the polynomial. In this case, the constant term is 140, which has factors of ±1, ±2, ±4, ±5, ±7, ±10, ±14, ±20, ±28, ±35, ±70, and ±140, and the leading coefficient is -3, which has factors of ±1 and ±3. Therefore, the possible rational roots of the polynomial are:
±1/1, ±2/1, ±4/1, ±5/1, ±7/1, ±10/1, ±14/1, ±20/1, ±28/1, ±35/1, ±70/1, and ±140/1
±1/3, ±2/3, ±4/3, ±5/3, ±7/3, ±10/3, ±14/3, ±20/3, ±28/3, ±35/3, ±70/3, and ±140/3
We can use synthetic division to test each of these possible roots. For example, if we test z = 1, we get:
1 | -3 38 -139 140
| -3 35 -104
| -3 35 -104 36
Since the remainder is not zero, z = 1 is not a root of the polynomial. We can continue testing the other possible roots until we find one that works. In this case, we find that z = 4 is a root of the polynomial, so we can use synthetic division to factor the polynomial as:
(z - 4)(-3z^2 + 26z - 35) = 0
Therefore, the solutions of the equation are z = 4, z = (13 + √229)/3, and z = (13 - √229)/3.
In summary, we can use synthetic division and the factor theorem to solve the equation -3z^3 + 38z^2 - 139z + 140 = 0 by first finding a factor of the polynomial using the rational root theorem, and then using synthetic division to factor the polynomial and find the solutions of the equation.