How do you know when to use each method and how do you do those methods

Question

1. 2x^3+128=0
2. x^4-5x^2+4=0
3. x^4-9x^2+14=0
4. x^4+13x^2+42=0
5. x^3+2x+9x+18=0
6. 6x^3+10x^2+5x=0
7. 3x^4-13x^3-10x^2=0

Answer 1

Hi Ann,

Mr. Jacques did a good job of succinctly answering the specific polynomials presented. If you were looking for a list of general steps to take on factoring polynomials, here is one for you:

To solve lower order polynomials of degree two or less:

1. See if you can isolate variables (letters) from constants (solitary numbers) and just "solve the equation normally."
2. See if you can factor out a common number from all terms and then solve the equation regularly.
3. See if you can simplify the factoring process by factoring out an xⁿ from all terms before solving normally.
4. Try to apply factoring by the FOIL method.
5. Try to factor by grouping.
6. At any point, you can apply the sum or differences of squares formulas if you recognize one of these useful patterns in your given polynomial.
7. Know that if you need to solve for quadratic roots, the quadratic formula will always have your back.

For higher-order polynomials:

1. Apply the Rational Roots Thoerem and synthetic division to factor out a second-degree polynomial.
2. Then, use the strategies above to find your zeroes (real roots only).
3. Apply the Complex Roots Thoerem and/or Complex Conjugates Theorem if you need to find imaginary roots.

Trying to find the single best way to factor a polynomial is like trying to find only one place to watch a sunset for the rest of your life....Not really possible or necessarily desirable!

Answer 2

Re: #5

Can also use 1st and 2nd terms; 3rd and 4th terms to factor.

x³ + 2x² + 9x + 18 = 0

x²(x + 2) + 9(x + 2) = 0

(x + 2)(x² + 9) = 0

Answer 3

1) Divide by 2 and solve x³ = -64

2-4) recognize the structure ax²ⁿ + bxⁿ + c = 0 which can be solved by the quadratic formula for xⁿ, then solved for x

5) I believe that it's supposed to be 2x² not 2x, in which case you can see that the first and third terms and the 2nd and 4th terms have the same factor: (x²+9)(x+2) = 0 which is now easily solvable

6 and 7) Factor out power of x in common (x for 6, x² for 7 knowing x=0 is a solution) and use quadratic equation on term left after dividing.

Please consider a tutor. Take care.