Polynomials-Types, Graphs, Factor and Remainder theorem

Polynomials-Types, Graphs, Factor and Remainder theorem

Synthetic division of Horner’s method
Horner’S Method of synthetic deivision :
We shall explain the method with the following examples : 
To divide x⁴ + 4x³ + 3x² - 4x - 4 by (x-1)
(Multiplier = )\( 1\left| \begin{gathered} 1\,\,\,\,\,\,4\,\,\,\,3\,\,\,\, - 4\,\,\,\, - 4 \hfill \\ 0\,\,\,\,\,1\,\,\,\,\,5\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,4 \hfill \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \hfill \\ 1\,\,\,\,\,5\,\,\,\,\,8\,\,\,\,\,\,\,4\,\,\,\,\,\left| \!{\underline {\, {\,0} \,}} \right. \, \hfill \\ \end{gathered} \right. \)
The quotient is x³ + 5x² + 8x +4    
Explanation :
        First horizontal row contains the multiplier which is obtained by the zero of  x - 1, which is 1.
        The remaining elements in the first horizontal row are the coefficients of descending power of x(The coefficinet of missing power of x, if any, that should be taken as zero). 
        To form the second horizontal row start with zero right under the second element of first row and add, the result is 1 which is first entry in the third row. Multiply this 1 with the multiplier 1 put the result under 4 (the third entry of the first row). Thus we get the second entry 1 in the second row. Then add 4 and 1 to get the second entry in the third row which is 5. Now multiply this 5 with the multiplier, put the product right under 3(the fourth entry of the first row). Add 3 and 5 to get third entry in the third row. Thus the third entry in the third row is 8. Now multiply 8 with the multiplier 1, put the product under - 4 (the fifth entry of th first row). Add 8 and -4 to get the fourth entry of the third row. Repeat the same procedure to get zero as the fifth entry of the third row. The last entry in the third row stands for the remainder, while the first four figures stand for the             coefficients of descending powers of x of quotient.
    Thus the quotient is x³ + 5x² + 8x + 4. We shall return to our problem. 
    f(x) = x⁴ + 4x³ + 3x² - 4x - 4 = (x-1) (x³ +5x² +_8x +4)
    Now if we write g(x) =x³ - 5x² +8x +4
    g(-1) = (-1)³ + 5(-1)² + 8(-1) +4 = 0
    \( \therefore \) (x+1) is a factor of g(x). Hence a factor of f(x). 
    Now to divide g(x) by (x+1), the multiplier is -1, the zero of x + 1.
    Let us once again apply synthetic division, to g(x).
\( - 1\left| \begin{gathered} 1\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,4 \hfill \\ 0\,\,\,\,\, - 1\,\,\,\, - 4\,\,\,\,\, - 4 \hfill \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \hfill \\ 1\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,0 \hfill \\ \end{gathered} \right. \)
The quotient is x² + 4x + 4 = 0
     f(x) = (x-1) g(x) = (x-1) (x+1) (x² +4x +4) = (x-1) (x+1) (x+2)²