Q: In what form must a polynomial equation occur in order to solve?

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When the equation is a polynomial whose highest order (power) is 2. Eg. y= x2 + 2x + 10. Then you can use quadratic formula to solve if factoring is not possible.

It depends on what you wanted to do - graph it, solve it, factorise it, etc.

Can be done.

To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.

y = 5x^3 - 45x is a polynomial equation that crosses the y axis at x=3.

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You can evaluate a polynomial, you can factorise a polynomial, you can solve a polynomial equation. But a polynomial is not a specific question so it cannot be answered.

When the equation is a polynomial whose highest order (power) is 2. Eg. y= x2 + 2x + 10. Then you can use quadratic formula to solve if factoring is not possible.

It depends on what you wanted to do - graph it, solve it, factorise it, etc.

A parabola is a graph of a 2nd degree polynomial function. Two graph a parabola, you must factor the polynomial equation and solve for the roots and the vertex. If factoring doesn't work, use the quadratic equation.

The graph of a polynomial in X crosses the X-axis at x-intercepts known as the roots of the polynomial, the values of x that solve the equation.(polynomial in X) = 0 or otherwise y=0

It's quite convenient, for it offers a general method to solve any equation that involves a polynomial of degree two (in one variable).

Can be done.

Whenever there are polynomials of the form aX2+bX+c=0 then this type of equation is know as a quadratic equation. to solve these we usually break b into two parts such that there product is equal to a*c and I hope you know how to factor polynomials.

To determine whether a polynomial equation has imaginary solutions, you must first identify what type of equation it is. If it is a quadratic equation, you can use the quadratic formula to solve for the solutions. If the equation is a cubic or higher order polynomial, you can use the Rational Root Theorem to determine if there are any imaginary solutions. The Rational Root Theorem states that if a polynomial equation has rational solutions, they must be a factor of the constant term divided by a factor of the leading coefficient. If there are no rational solutions, then the equation has imaginary solutions. To use the Rational Root Theorem, first list out all the possible rational solutions. Then, plug each possible rational solution into the equation and see if it is a solution. If there are any solutions, then the equation has imaginary solutions. If not, then there are no imaginary solutions.

y = 5x^3 - 45x is a polynomial equation that crosses the y axis at x=3.

Either graph the polynomial on graph paper manually or on a graphing calculator. If it is a "y=" polynomial, then the zeroes are the points or point where the polynomial touches the x-axis. If it is an "x=" polynomial, then the zeroes are the points or point where the polynomial touches the y-axis. If it touches neither, then it has no zeroes.

There are several ways to solve such equations: (1) Write the equation in the form polynomial = 0, and solve the left part (where I wrote "polynomial"). (2) Completing the square. (3) Use the quadratic formula. Method (3) is by far the most flexible, but in special cases methods (1) and (2) are faster to solve.