According to the given cubic equation 8x³-10x² - 21x + 18 = 0 the other value of x will be 4.
<h3>How does one solve a three-degree equation?</h3>
Traditionally, a cubic problem can be solved by converting it to a quadratic equation and then solving it using factoring or the quadratic formula. A cubic equation, like a quadratic equation, has three roots.
<h3>What is the simplest method for solving a cubic equation?</h3>
The typical strategy for solving a cubic problem is to convert it to a quadratic equation and then solve the quadratic using the standard methods, either factoring or using the formula. All of them are cubic equations. A cubic equation, like a quadratic equation, may have two or three real roots.
<h3>According to the given data:</h3>
8x³-10x² - 21x + 18 = 0
x= 2
substituting the value in the equation we get:
8x³-10x² - 21x + 18 = 0
8(2)³-10(2)² - 21(2) + 18 = 0
68 - 40 - 42 + 18 = 0
4 = 0
According to the given equation 8x³-10x² - 21x + 18 = 0 the other value of x will be 4.
To know more about cubic equation visit:
brainly.com/question/13730904
#SPJ9
Using the Factor Theorem, it is found that yes, it is possible for a sixth degree polynomial function with integer coefficients to have no real zeroes, as they can have three complex-conjugate pairs.
<h3>What is the Factor Theorem?</h3>
The Factor Theorem states that a polynomial function with roots
is given by:
![f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20a%28x%20-%20x_1%29%28x%20-%20x_2%29%20%5Ccdots%20%28x%20-%20x_n%29)
In which a is the leading coefficient.
If a complex number is a root of a function, it's conjugate will also be a root. Thus, with three pairs of complex-conjugate roots, for example,
, a sixth degree function with no real zeros is formed, so the answer is Yes.
More can be learned about the Factor Theorem at brainly.com/question/24380382
He will travel 55km In 2.5 hours at a rate of 22kmph
Answer:
27
Step-by-step explanation:
So I dont really get the equation because its written confusingly but I will try my best.
1. √(2x+10) -6=2
2.√{2x+10}=8
3.√{2x+10}^2=8^2
4.2x+10=64
5.x=27