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Difference between revisions of "Factor Theorem"

(→‎Theorem and Proof: theorem)
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==Theorem==  
 
==Theorem==  
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If <math>P(x)</math> is a polynomial, then <math>x-a</math> is a factor <math>P(x)</math> iff <math>P(a)=0</math>.
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If <math>P(x)</math> is a polynomial, then <math>x-a</math> is a [[factor]] <math>P(x)</math> iff <math>P(a)=0</math>.
  
 
==Proof==
 
==Proof==
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Now suppose that <math>P(a) = 0</math>.
 
Now suppose that <math>P(a) = 0</math>.
  
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Apply division algorithm to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the remainder polynomial such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>.
+
Apply division [[algorithm]] to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the [[remainder polynomial]] such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>.
  
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This means that <math>R(x)</math> can be at most a constant polynomial.
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This means that <math>R(x)</math> can be at most a [[constant polynomial]].
  
 
Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>.
 
Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>.
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== See also ==
 
== See also ==
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 +
  
 
{{stub}}
 
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Revision as of 15:28, 9 February 2008

The Factor Theorem is a theorem relating to polynomials

Theorem

If $P(x)$ is a polynomial, then $x-a$ is a factor $P(x)$ iff $P(a)=0$.

Proof

If $x - a$ is a factor of $P(x)$, then $P(x) = (x - a)Q(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$. Then $P(a) = (a - a)Q(a) = 0$.

Now suppose that $P(a) = 0$.

Apply division algorithm to get $P(x) = (x - a)Q(x) + R(x)$, where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ and $R(x)$ is the remainder polynomial such that $0\le\deg(R(x)) < \deg(x - a) = 1$.

This means that $R(x)$ can be at most a constant polynomial.

Substitute $x = a$ and get $P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0$.

But $R(x)$ is a constant polynomial and so $R(x) = 0$ for all $x$.

Therefore, $P(x) = (x - a)Q(x)$, which shows that $x - a$ is a factor of $P(x)$.

Problems

See also

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