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Q: What quadratic equation has a leading coefficient of 4 and solutions 12 and?
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How can you determine whether a polynomial equation has imaginary solutions?

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.


What is the highest exponent of the leading variable in a quadratic equation?

the highest exponent of quadratic equation is 2 good luck on NovaNet peoples


What is the leading coefficient of each fungtion?

what is the leading coefficient -3x+8


How do you find zeros when the leading coefficient is one?

The answer depends on the what the leading coefficient is of!


Can all quadratic equations be solved?

Well, that depends on what you mean "solve by factoring." For any quadratic equation, it is possible to factor the quadratic, and then the roots can be recovered from the factors. So in the very weak sense that every quadratic can be solved by a method that involves getting the factors and recovering the roots from them, all quadratic equations can be solved by factoring. However, in most cases, the only way of factoring the quadratic in the first place is to first find out what its roots are, and then use the roots to factor the quadratic (any quadratic polynomial can be factored as k(x - r)(x - s), where k is the leading coefficient of the polynomial and r and s are its two roots), in which case trying to recover the roots from the factors is redundant (since you had to know what the roots were to get the factors in the first place). So to really count as solving by factoring, it makes sense to require that the solution method obtains the factors by means that _don't_ require already knowing the roots of the polynomial. And in this sense, most quadratic equations are not solvable through factoring.