Q: When it is easier to solve a quadratic by factoring?

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Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.Yes and they do in factoring quadratic equations.

Solve by factoring. Solve by taking the square root of both sides.

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 means you are required to "solve" a quadratic equation by factorising the quadratic equation into two binomial expressions. Solving means to find the value(s) of the variable for which the expression equals zero.

y=b+x+x^2 This is a quadratic equation. The graph is a parabola. The quadratic equation formula or factoring can be used to solve this.

A quadratic equation.

The discriminant

The discriminant for the quadratic is b2-4ac = 302 - 4*4*45 = 900 - 720 = 180 Since 180 is not a perfect square, the roots of the equation are irrational and it is far from straightforward to solve such an equation by factoring.

(3x+4)(3x-4)=0 x=±4/3

Four? Factoring Graphing Quadratic Equation Completing the Square There may be more, but there's at least four.

using the quadratic formula or the graphics calculator. Yes, you can do it another way, by using a new method, called Diagonal Sum Method, that can quickly and directly give the 2 roots, without having to factor the equation. This method is fast, convenient and is applicable to any quadratic equation in standard form ax^2 +bx + c = 0, whenever it can be factored. It requires fewer permutations than the factoring method does, especially when the constants a, b, and c are large numbers. If this method fails to get answer, then consequently, the quadratic formula must be used to solve the given equation. It is a trial-and-error method, same as the factoring method, that usually takes fewer than 3 trials to solve any quadratic equation. See book titled:" New methods for solving quadratic equations and inequalities" (Trafford Publishing 2009)

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.