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Senin, 25 Juni 2018

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The Factor Theorem and The Remainder Theorem - YouTube
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In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.

The factor theorem states that a polynomial f ( x ) {\displaystyle f(x)} has a factor ( x - k ) {\displaystyle (x-k)} if and only if f ( k ) = 0 {\displaystyle f(k)=0} (i.e. k {\displaystyle k} is a root).


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Factorization of polynomials

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.

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:

  1. "Guess" a zero a {\displaystyle a} of the polynomial f {\displaystyle f} . (In general, this can be very hard, but maths textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
  2. Use the factor theorem to conclude that ( x - a ) {\displaystyle (x-a)} is a factor of f ( x ) {\displaystyle f(x)} .
  3. Compute the polynomial g ( x ) = f ( x ) / ( x - a ) {\displaystyle g(x)=f(x){\big /}(x-a)} , for example using polynomial long division or synthetic division.
  4. 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 polynomial degree 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} .

Example

Find the factors of

x 3 + 7 x 2 + 8 x + 2. {\displaystyle x^{3}+7x^{2}+8x+2.}

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if ( x - 1 ) {\displaystyle (x-1)} is a factor, substitute x = 1 {\displaystyle x=1} into the polynomial above:

x 3 + 7 x 2 + 8 x + 2 = ( 1 ) 3 + 7 ( 1 ) 2 + 8 ( 1 ) + 2 {\displaystyle x^{3}+7x^{2}+8x+2=(1)^{3}+7(1)^{2}+8(1)+2}
= 1 + 7 + 8 + 2 {\displaystyle =1+7+8+2}
= 18. {\displaystyle =18.}

As this is equal to 18 and not 0 this means ( x - 1 ) {\displaystyle (x-1)} is not a factor of x 3 + 7 x 2 + 8 x + 2 {\displaystyle x^{3}+7x^{2}+8x+2} . So, we next try ( x + 1 ) {\displaystyle (x+1)} (substituting x = - 1 {\displaystyle x=-1} into the polynomial):

( - 1 ) 3 + 7 ( - 1 ) 2 + 8 ( - 1 ) + 2. {\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2.}

This is equal to 0 {\displaystyle 0} . Therefore x - ( - 1 ) {\displaystyle x-(-1)} , which is to say x + 1 {\displaystyle x+1} , is a factor, and - 1 {\displaystyle -1} is a root of x 3 + 7 x 2 + 8 x + 2. {\displaystyle x^{3}+7x^{2}+8x+2.}

The next two roots can be found by algebraically dividing x 3 + 7 x 2 + 8 x + 2 {\displaystyle x^{3}+7x^{2}+8x+2} by ( x + 1 ) {\displaystyle (x+1)} to get a quadratic:

x 3 + 7 x 2 + 8 x + 2 x + 1 = x 2 + 6 x + 2 , {\displaystyle {x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2,}

and therefore ( x + 1 ) {\displaystyle (x+1)} and x 2 + 6 x + 2 {\displaystyle x^{2}+6x+2} are factors of x 3 + 7 x 2 + 8 x + 2. {\displaystyle x^{3}+7x^{2}+8x+2.} Of these the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic - 3 ± 7 . {\displaystyle -3\pm {\sqrt {7}}.} Thus the three irreducible factors 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}}).}


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References

Source of the article : Wikipedia

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