Ok this is a great problem to work with. Let me show you two types of examples here:

Example 1: x^{2}+3x+2

Step 1: Factor this problem out! --------> x^{2}+3x+2 =
(x+1)(x+2)

Step 2: Now you know from multiplication rule that if one of the sums within the parenthesis equals 0, then the whole product of the two parenthesis would be 0. So you equal each parenthsis to 0.

(x+1) = 0 -------> x=-1

(x+2) = 0 -------> x=-2

Notice that when x = -1 the expression

(x+1)(x+2)= (-1+1)(-1+2) = (0)(1)=0

that when x = -2 the expression

(x+1)(x+2)= (-2+1)(-2+2) = (-1)(0)=0

Tadaa! So when a problem asks to get zeros of a polynomial you are just
finding the x values where the polynomial expression would equal 0. The fastest way to do it is through factoring.

Example 2: x^{2}+2x+3

Sometimes you cannot factor. So you should use the quadratic formula for this. You can use the quadratic formula for example 1 but it's just faster to factor.

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...

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If a polynomial can be factorized easily, then factor it out. You have a lot of tools to do that such as factor theorem, rational root theorem, synthetic division etc. .

But if a polynomial is not factorable, then you have to use some numerical method to find its zeros. For example, you can use Newton's method.

Furthermore, if you want to find real and complex zeros, then it is way more complicated.

Specific question: Is f(x) = x^4 + 4a^4 factorable?

The first step is to factor the polynomial completely as lower degree polynomials are easier to solve. If it factors, you just set each factor to zero and solve.

Example: x^{2}-4x+3=0 factors to (x-1)(x-3)=0 so x=1 and x=3 are the two solutions.

Solving polynomials that don't factor:

Linear: ax+b=0

ax = -b

x= -b/a

Quadratic: ax^{2}+bx+c=0

Use the quadratic formula or complete the square.

For example of completing the square, consider x^{2}-4x+2=0. Then you get

x^{2}-4x+4 = 2

(x-2)^{2 }= 2

x-2 = ±√2

x = 2±√2

For Cubic and Quartic polynomials general formulas also exist in terms of radicals (Cardano's and Ferraria's formula respectively) and there are a number of websites that show how to solve the general cases but I will not bother to show them here.

Polynomials of degree 5 and higher are not solvable by radicals in the general case (Abel-Ruffini Theorem). However, advanced methods exist that solve in radicals, all solvable cases.

## Comments

Rizul -

For clarification, x

^{2}+2x+2 does not factor to (x+1)(x+2), but x^{2}+3x+2 does.