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Q: How do you find the zeros of cubic polynomial equation?

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Solve the equation - by whatever means available to you: factorising, graphical, numerical approximations, etc.

graph apex xD

Add two zeros and a percentage sign.For example, 23 = 2300%.

The answer depends on where, in the sequence, the missing number is meant to go.Furthermore, whatever number you choose and wherever in the sequence it is meant to be, it is always possible to find a polynomial of degree 5 that will go through all five points given in the question and your chosen one.Using a polynomial of degree 4, the next number is -218.The answer depends on where, in the sequence, the missing number is meant to go.Furthermore, whatever number you choose and wherever in the sequence it is meant to be, it is always possible to find a polynomial of degree 5 that will go through all five points given in the question and your chosen one.Using a polynomial of degree 4, the next number is -218.The answer depends on where, in the sequence, the missing number is meant to go.Furthermore, whatever number you choose and wherever in the sequence it is meant to be, it is always possible to find a polynomial of degree 5 that will go through all five points given in the question and your chosen one.Using a polynomial of degree 4, the next number is -218.The answer depends on where, in the sequence, the missing number is meant to go.Furthermore, whatever number you choose and wherever in the sequence it is meant to be, it is always possible to find a polynomial of degree 5 that will go through all five points given in the question and your chosen one.Using a polynomial of degree 4, the next number is -218.

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when the equation is equal to zero. . .:)

by synthetic division and quadratic equation

If there is one variable. Then put each variable equal to zero and then solve for the other variable.

Find All Possible Roots/Zeros Using the Rational Roots Test f(x)=x^4-81 ... If a polynomial function has integer coefficients, then every rational zero will ...

Multiply x3 - 2x2 - 13x - 10

In general, there is no simple method.

2x^3 - 5x^2 - 14x + 8 Let P(x) represents the cubic polynomial. We can find the sum of x-values which make P(x) = 0, (the sum of the roots of the equation) P(x) = 2x^3 - 5x^2 - 14x + 8 P(x) = 0 2x^3 - 5x^2 - 14x + 8 = 0 Since the degree of this polynomial is odd, then the sum of the roots is -[a(n - 1)/an], where a(n-1) is -5 and an is 2. So we have, -[a(n - 1)/an] = -(-5/2) = 5/2 Thus the sum of the roots is 5/2.

If you have the zeros of a polynomial, it is easy, almost trivial, to find an expression with those zeros. I am not sure I understood the question correctly, but let's assume you have the zero 2 with multiplicity 2, and other zeros at 3 and 5. Just write the expression: (x-2)(x-2)(x-3)(x-5). (Example with a negative zero: if there is a zero at "-5", the factor becomes (x- -5) = (x + 5).) You can multiply this out to get the polynomial if you like. For example, if you multiply every term in the first factor with every term in the second factor, you get x2 -2x -2x + 4 = x2 -4x + 4. Next, multiply each term of this polynomial with each term of the next factor, etc.

If the discriminant is positive, then the function has two real zeros. If it is zero, then the function has one real zero. If it is negative, then it has two complex conjugate zeros.This assumes that we are talking about a standard second order polynomial equation, i.e. quadratic equation, in the form Ax2 + Bx + C = 0, and that the discriminant is B2 - 4AC, which is a part of the standard solution of these kind of equations.

Solve the equation - by whatever means available to you: factorising, graphical, numerical approximations, etc.

They are all the points where the graph crosses (or touches) the x-axis.

The quadratic formula can be used to find the solutions of a quadratic equation - not a linear or cubic, or non-polynomial equation. The quadratic formula will always provide the solutions to a quadratic equation - whether the solutions are rational, real or complex numbers.

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