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Factoring POlynomials and finding Zeros

  • Thread starter matt000
  • Start date
  • #1
5
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factoring=====================

1. 16x^3 - 54 This one, I've broken it to 2(2^3 x^3 - 3^2) but it's still wrong! I don't get how much further it can be broken down!
2. 3x^4 - 48 This one.. I've gotten far as 3x^2 (x^2 - 4x + 4) ! Still wrong.
=============================

Zeros... dang it I'm kinda embarrassed to even ask...

1. 9x^3 + 27x^2 = 4x - 12
I've put x = -3, but I know there's more. I don't know how to get more!

2. x^5 - 100x = 0
Dang it, I don't know the zero of this one >< !

THANK YOU!!
 

Answers and Replies

  • #2
G01
Homework Helper
Gold Member
2,665
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For the zeros, get x on one side of the equal side and zero on the other. Then solve for x. If you end up with x(Ax+b)=0 or something similar, for this equation to be zero, either x = 0 or Ax+b = 0. For these cases then, what can x be? This should be of help for number 2. For number one you need to factor a third order polynomial.

For the factoring, in number 1 you have a term containing the difference between two cubes, 8x^3-3^3, this can be factored further. There should be a formula you can follow in your text.In number 2, you have a quadratic,(x^2-4x-4) this can also be factored more.

Good Luck!

G01
 
  • #3
1
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3X^4-48 = 3(X^4-16)=3(X^2-4)(X^2+4)=3(X-2)(X+2)(X^2+4)

X^5-100X=0 -> X(X^4-100)=0 -> X=0 or (X^2-10)(X^2+10)=0 -> X = +- sqrt(10), X=+-i*sqrt(10)
 
  • #4
1,425
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Don't forget complex solutions.
 
  • #5
286
0
How far you can factor it depends on what you're factoring over.

ie. x^2 - 5 cannot be factored over the rational numbers,
however, it can be factored as [tex](x-\sqrt{5})(x+\sqrt{5})[/tex]
Similarly, x^2 + 1 cannot be factored over the reals, but it can be factored as [tex](x-i)(x+i)[/tex]

Note: For x^2 - 3x - 4 = 0, the roots are 4 and -1
The factors are (x-4) and (x+1)
And, for 2x^2 - 3x + 1, the roots are 1 and 1/2
Two of the factors are (x-1) and (x - 1/2) Note: multiplying these doesn't quite give you the original
2x^2 - 3x + 1. However, if you were to multiply:
(2)(x-1)(x - 1/2), you'd get that polynomial.
Almost interestingly, you can distribute that 2 into the 3rd factor and write it as (x-1)(2x-1)

In short, if you can find the roots of a polynomial, then it can be factored as (leading coefficient)(x-root1)(x-root2)...(x-rootn), n being the degree of the polynomial.
 
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