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Difference between revisions of "Factor Theorem"

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The '''Factor Theorem''' is a theorem relating to [[polynomials]]
 
The '''Factor Theorem''' is a theorem relating to [[polynomials]]
  
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== Theorem and Proof ==
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==Theorem==  
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'''Theorem:''' 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>.
  
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*'''Proof:''' If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
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==Proof==
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If <math>x - a</math> is a factor of <math>P(x)</math>, then <math>P(x) = (x - a)Q(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math>. Then <math>P(a) = (a - a)Q(a) = 0</math>.
  
 
Now suppose that <math>P(a) = 0</math>.
 
Now suppose that <math>P(a) = 0</math>.
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Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
 
Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>.
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== Problems ==
 
== Problems ==

Revision as of 17:50, 15 November 2007

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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