Factor theorem

<h2 id="factorization-of-polynomials">Factorization of polynomials</h2>
<p class="note">Main article: <a href="/facts/Factorization_of_polynomials/1S9HRKcI">Factorization of polynomials</a></p>
<p>Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.
</p><p>The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a>
</p>
<ol><li>Deduce the candidate of zero 
  
    
      
        a
      
    
    {\displaystyle a}
  
 of the polynomial 
  
    
      
        f
      
    
    {\displaystyle f}
  
 from its leading coefficient 
  
    
      
        
          a
          
            n
          
        
      
    
    {\displaystyle a_{n}}
  
 and constant term 
  
    
      
        
          a
          
            0
          
        
      
    
    {\displaystyle a_{0}}
  
. (See <a href="/facts/Rational_root_theorem/U97kHx9n">Rational Root Theorem</a>.)</li>
<li>Use the factor theorem to conclude that 
  
    
      
        (
        x
        −
        a
        )
      
    
    {\displaystyle (x-a)}
  
 is a factor of 
  
    
      
        f
        (
        x
        )
      
    
    {\displaystyle f(x)}
  
.</li>
<li>Compute the polynomial 
  
    
      
        g
        (
        x
        )
        =
        
          
            
              
                f
                (
                x
                )
              
              
                (
                x
                −
                a
                )
              
            
          
        
      
    
    {\textstyle g(x)={\dfrac {f(x)}{(x-a)}}}
  
, for example using <a href="/facts/Polynomial_long_division/MgJlJ2G7">polynomial long division</a> or <a href="/facts/Synthetic_division/meQEnhWh">synthetic division</a>.</li>
<li>Conclude that any root 
  
    
      
        x
        ≠
        a
      
    
    {\displaystyle x\neq a}
  
 of 
  
    
      
        f
        (
        x
        )
        =
        0
      
    
    {\displaystyle f(x)=0}
  
 is a root of 
  
    
      
        g
        (
        x
        )
        =
        0
      
    
    {\displaystyle g(x)=0}
  
. Since the <a href="/facts/Polynomial_degree/Bf8vEIhf">polynomial degree</a> of 
  
    
      
        g
      
    
    {\displaystyle g}
  
 is one less than that of 
  
    
      
        f
      
    
    {\displaystyle f}
  
, it is "simpler" to find the remaining zeros by studying 
  
    
      
        g
      
    
    {\displaystyle g}
  
.</li></ol>
<p>Continuing the process until the polynomial 
  
    
      
        f
      
    
    {\displaystyle f}
  
 is factored completely, which all its factors is irreducible on 
  
    
      
        
          R
        
        [
        x
        ]
      
    
    {\displaystyle \mathbb {R} [x]}
  
 or 
  
    
      
        
          C
        
        [
        x
        ]
      
    
    {\displaystyle \mathbb {C} [x]}
  
.
</p>
<h3>Example</h3>
<p>Find the factors of 
  
    
      
        
          x
          
            3
          
        
        +
        7
        
          x
          
            2
          
        
        +
        8
        x
        +
        2.
      
    
    {\displaystyle x^{3}+7x^{2}+8x+2.}

</p><p>Solution: Let 
  
    
      
        p
        (
        x
        )
      
    
    {\displaystyle p(x)}
  
 be the above polynomial
</p>
Constant term = 2
Coefficient of 
  
    
      
        
          x
          
            3
          
        
        =
        1
      
    
    {\displaystyle x^{3}=1}

<p>All possible factors of 2 are 
  
    
      
        ±
        1
      
    
    {\displaystyle \pm 1}
  
 and 
  
    
      
        ±
        2
      
    
    {\displaystyle \pm 2}
  
. Substituting 
  
    
      
        x
        =
        −
        1
      
    
    {\displaystyle x=-1}
  
, we get: 
</p>

(
        −
        1
        
          )
          
            3
          
        
        +
        7
        (
        −
        1
        
          )
          
            2
          
        
        +
        8
        (
        −
        1
        )
        +
        2
        =
        0
      
    
    {\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0}

<p>So, 
  
    
      
        (
        x
        −
        (
        −
        1
        )
        )
      
    
    {\displaystyle (x-(-1))}
  
, i.e, 
  
    
      
        (
        x
        +
        1
        )
      
    
    {\displaystyle (x+1)}
  
 is a factor of 
  
    
      
        p
        (
        x
        )
      
    
    {\displaystyle p(x)}
  
. On dividing 
  
    
      
        p
        (
        x
        )
      
    
    {\displaystyle p(x)}
  
 by 
  
    
      
        (
        x
        +
        1
        )
      
    
    {\displaystyle (x+1)}
  
, we get 
</p>
Quotient = 
  
    
      
        
          x
          
            2
          
        
        +
        6
        x
        +
        2
      
    
    {\displaystyle x^{2}+6x+2}

<p>Hence, 
  
    
      
        p
        (
        x
        )
        =
        (
        
          x
          
            2
          
        
        +
        6
        x
        +
        2
        )
        (
        x
        +
        1
        )
      
    
    {\displaystyle p(x)=(x^{2}+6x+2)(x+1)}

</p><p>Out of these, the quadratic factor can be further factored using the <a href="/facts/Quadratic_formula/j5ftMowG">quadratic formula</a>, which gives as roots of the quadratic 
  
    
      
        −
        3
        ±
        
          
            7
          
        
        .
      
    
    {\displaystyle -3\pm {\sqrt {7}}.}
  
 Thus the three <a href="/facts/Polynomial_factorization/1S9HRKcI">irreducible factors</a> of the original polynomial are 
  
    
      
        x
        +
        1
        ,
      
    
    {\displaystyle x+1,}
  
 
  
    
      
        x
        −
        (
        −
        3
        +
        
          
            7
          
        
        )
        ,
      
    
    {\displaystyle x-(-3+{\sqrt {7}}),}
  
 and 
  
    
      
        x
        −
        (
        −
        3
        −
        
          
            7
          
        
        )
        .
      
    
    {\displaystyle x-(-3-{\sqrt {7}}).}

</p>
<h2 id="proofs">Proofs</h2>
<p>Several proofs of the theorem are presented here.
</p><p>If 
  
    
      
        x
        −
        a
      
    
    {\displaystyle x-a}
  
 is a factor of 
  
    
      
        f
        (
        x
        )
        ,
      
    
    {\displaystyle f(x),}
  
 it is immediate that 
  
    
      
        f
        (
        a
        )
        =
        0.
      
    
    {\displaystyle f(a)=0.}
  
 So, only the converse will be proved in the following.
</p>
<h3>Proof 1</h3>
<p>This proof begins by verifying the statement for 
  
    
      
        a
        =
        0
      
    
    {\displaystyle a=0}
  
. That is, it will show that for any polynomial 
  
    
      
        f
        (
        x
        )
      
    
    {\displaystyle f(x)}
  
 for which 
  
    
      
        f
        (
        0
        )
        =
        0
      
    
    {\displaystyle f(0)=0}
  
, there exists a polynomial 
  
    
      
        g
        (
        x
        )
      
    
    {\displaystyle g(x)}
  
 such that 
  
    
      
        f
        (
        x
        )
        =
        x
        ⋅
        g
        (
        x
        )
      
    
    {\displaystyle f(x)=x\cdot g(x)}
  
. To that end, write 
  
    
      
        f
        (
        x
        )
      
    
    {\displaystyle f(x)}
  
 explicitly as 
  
    
      
        
          c
          
            0
          
        
        +
        
          c
          
            1
          
        
        
          x
          
            1
          
        
        +
        …
        +
        
          c
          
            n
          
        
        
          x
          
            n
          
        
      
    
    {\displaystyle c_{0}+c_{1}x^{1}+\dotsc +c_{n}x^{n}}
  
. Now observe that 
  
    
      
        0
        =
        f
        (
        0
        )
        =
        
          c
          
            0
          
        
      
    
    {\displaystyle 0=f(0)=c_{0}}
  
, so 
  
    
      
        
          c
          
            0
          
        
        =
        0
      
    
    {\displaystyle c_{0}=0}
  
. Thus, 
  
    
      
        f
        (
        x
        )
        =
        x
        (
        
          c
          
            1
          
        
        +
        
          c
          
            2
          
        
        
          x
          
            1
          
        
        +
        …
        +
        
          c
          
            n
          
        
        
          x
          
            n
            −
            1
          
        
        )
        =
        x
        ⋅
        g
        (
        x
        )
      
    
    {\displaystyle f(x)=x(c_{1}+c_{2}x^{1}+\dotsc +c_{n}x^{n-1})=x\cdot g(x)}
  
. This case is now proven.
</p><p>What remains is to prove the theorem for general 
  
    
      
        a
      
    
    {\displaystyle a}
  
 by reducing to the 
  
    
      
        a
        =
        0
      
    
    {\displaystyle a=0}
  
 case. To that end, observe that 
  
    
      
        f
        (
        x
        +
        a
        )
      
    
    {\displaystyle f(x+a)}
  
 is a polynomial with a root at 
  
    
      
        x
        =
        0
      
    
    {\displaystyle x=0}
  
. By what has been shown above, it follows that 
  
    
      
        f
        (
        x
        +
        a
        )
        =
        x
        ⋅
        g
        (
        x
        )
      
    
    {\displaystyle f(x+a)=x\cdot g(x)}
  
 for some polynomial 
  
    
      
        g
        (
        x
        )
      
    
    {\displaystyle g(x)}
  
. Finally, 
  
    
      
        f
        (
        x
        )
        =
        f
        (
        (
        x
        −
        a
        )
        +
        a
        )
        =
        (
        x
        −
        a
        )
        ⋅
        g
        (
        x
        −
        a
        )
      
    
    {\displaystyle f(x)=f((x-a)+a)=(x-a)\cdot g(x-a)}
  
.
</p>
<h3>Proof 2</h3>
<p>First, observe that whenever 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
 belong to any commutative ring (the same one) then the identity 
  
    
      
        
          x
          
            n
          
        
        −
        
          y
          
            n
          
        
        =
        (
        x
        −
        y
        )
        (
        
          y
          
            n
            −
            1
          
        
        +
        
          x
          
            1
          
        
        
          y
          
            n
            −
            2
          
        
        +
        …
        +
        
          x
          
            n
            −
            2
          
        
        
          y
          
            1
          
        
        +
        
          x
          
            n
            −
            1
          
        
        )
      
    
    {\displaystyle x^{n}-y^{n}=(x-y)(y^{n-1}+x^{1}y^{n-2}+\dotsc +x^{n-2}y^{1}+x^{n-1})}
  
 is true. This is shown by multiplying out the brackets.
</p><p>Let 
  
    
      
        f
        (
        X
        )
        ∈
        R
        
          [
          X
          ]
        
      
    
    {\displaystyle f(X)\in R\left[X\right]}
  
 where 
  
    
      
        R
      
    
    {\displaystyle R}
  
 is any commutative ring. Write 
  
    
      
        f
        (
        X
        )
        =
        
          ∑
          
            i
          
        
        
          c
          
            i
          
        
        
          X
          
            i
          
        
      
    
    {\displaystyle f(X)=\sum _{i}c_{i}X^{i}}
  
 for a sequence of coefficients 
  
    
      
        (
        
          c
          
            i
          
        
        
          )
          
            i
          
        
      
    
    {\displaystyle (c_{i})_{i}}
  
. Assume 
  
    
      
        f
        (
        a
        )
        =
        0
      
    
    {\displaystyle f(a)=0}
  
 for some 
  
    
      
        a
        ∈
        R
      
    
    {\displaystyle a\in R}
  
. Observe then that 
  
    
      
        f
        (
        X
        )
        =
        f
        (
        X
        )
        −
        f
        (
        a
        )
        =
        
          ∑
          
            i
          
        
        
          c
          
            i
          
        
        (
        
          X
          
            i
          
        
        −
        
          a
          
            i
          
        
        )
      
    
    {\displaystyle f(X)=f(X)-f(a)=\sum _{i}c_{i}(X^{i}-a^{i})}
  
. Observe that each summand has 
  
    
      
        X
        −
        a
      
    
    {\displaystyle X-a}
  
 as a factor by the factorisation of expressions of the form 
  
    
      
        
          x
          
            n
          
        
        −
        
          y
          
            n
          
        
      
    
    {\displaystyle x^{n}-y^{n}}
  
 that was discussed above. Thus, conclude that 
  
    
      
        X
        −
        a
      
    
    {\displaystyle X-a}
  
 is a factor of 
  
    
      
        f
        (
        X
        )
      
    
    {\displaystyle f(X)}
  
.
</p>
<h3>Proof 3</h3>
<p>The theorem may be proved using <a href="/facts/Euclidean_division_of_polynomials/4c3dUbJr">Euclidean division of polynomials</a>: Perform a Euclidean division of 
  
    
      
        f
        (
        x
        )
      
    
    {\displaystyle f(x)}
  
 by 
  
    
      
        (
        x
        −
        a
        )
      
    
    {\displaystyle (x-a)}
  
 to obtain 
  
    
      
        f
        (
        x
        )
        =
        (
        x
        −
        a
        )
        Q
        (
        x
        )
        +
        R
        (
        x
        )
      
    
    {\displaystyle f(x)=(x-a)Q(x)+R(x)}
  
 where 
  
    
      
        deg
        ⁡
        (
        R
        )
        <
        deg
        ⁡
        (
        x
        −
        a
        )
      
    
    {\displaystyle \deg(R)<\deg(x-a)}
  
. Since 
  
    
      
        deg
        ⁡
        (
        R
        )
        <
        deg
        ⁡
        (
        x
        −
        a
        )
      
    
    {\displaystyle \deg(R)<\deg(x-a)}
  
, it follows that 
  
    
      
        R
      
    
    {\displaystyle R}
  
 is constant. Finally, observe that 
  
    
      
        0
        =
        f
        (
        a
        )
        =
        R
      
    
    {\displaystyle 0=f(a)=R}
  
. So 
  
    
      
        f
        (
        x
        )
        =
        (
        x
        −
        a
        )
        Q
        (
        x
        )
      
    
    {\displaystyle f(x)=(x-a)Q(x)}
  
.
</p><p>The Euclidean division above is possible in every commutative ring since 
  
    
      
        (
        x
        −
        a
        )
      
    
    {\displaystyle (x-a)}
  
 is a <a href="/facts/Monic_polynomial/4JzuIJ5R">monic polynomial</a>, and, therefore, the <a href="/facts/Polynomial_long_division/MgJlJ2G7">polynomial long division algorithm</a> does not involve any division of coefficients.
</p>
<h3>Corollary of other theorems</h3>
<p>It is also a <a href="/facts/Corollary/ntQ8zR9N">corollary</a> of the <a href="/facts/Polynomial_remainder_theorem/tTz57bPw">polynomial remainder theorem</a>, but conversely can be used to show it.
</p><p>When the polynomials are multivariate but the coefficients form an <a href="/facts/Algebraically_closed_field/HuVQClok">algebraically closed field</a>, the <a href="/facts/Hilbert%2527s_Nullstellensatz/TlgoRHuM">Nullstellensatz</a> is a significant and deep generalisation.
</p>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2 <a href="0-13-370149-2" target="_blank">0-13-370149-2</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
<li id="fn:2"><p>Sehgal, V K; Gupta, Sonal (September 2009), Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN 978-81-317-2816-1. <a href="978-81-317-2816-1" target="_blank">978-81-317-2816-1</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li>
<li id="fn:3"><p>Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN 81-7008-629-9. <a href="81-7008-629-9" target="_blank">81-7008-629-9</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li>
</ol>

Factor theorem open-in-new

Factor theorem