I already know I can divide, but I have to use substitution. Do I just plug in the x-2 for x? Please HELPPP!!
2007-09-02 06:32:45 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
if x-2 is a factor to ax^4+bx^3 + cx^2 + ex + f = 0
then (x-2)(rest of factorization) = 0
so 2 is a solution to this equation...
Therefore if you substitute in 2, you better get 0, Otherwise it's not a factor
2007-09-02 06:42:12 · answer #1 · answered by radne0 5 · 1⤊ 0⤋
If x-2 is a factor, then x =2 is a root. So plug x =2 into the polynomial and if the result is zero, x-2 is a factor.
Synthetic division is a fast way to do this.
2007-09-02 06:43:09 · answer #2 · answered by ironduke8159 7 · 1⤊ 0⤋
Use remainder theory to check if f(2) = 0. If f(2) = 0, then x-2 is a factor. Otherwise, x-2 is not a factor of the polynomial.
2007-09-02 06:42:47 · answer #3 · answered by sahsjing 7 · 1⤊ 0⤋
I recall a theorem, namely, the factor theorem, which states:
Let p(x) be a polynomial of degree n>0. If p(a)=0 for a real number a, then (x-a) is a factor of p(x). Conversely, if (x-a) is factor of p(x), then p(a)=0.
In your problem a=2
2007-09-02 06:48:34 · answer #4 · answered by Christine P 5 · 0⤊ 0⤋
Hi Mark,
I'm not sure what you are asking. No, you wouldn't substitute x -2 for x. It would help immensely if you could post the polynomial here. Can you amend your question?
Radne0's advice is sound.
James :-)
2007-09-02 06:42:43 · answer #5 · answered by ? 3 · 0⤊ 0⤋
Simple answer - see if x=2 solves the equation.
If it does then x-2 is a factor.
2007-09-02 06:46:39 · answer #6 · answered by piscesgirl 3 · 0⤊ 0⤋