Answer:
[tex]x^5+4x^3+x^3-4x+3+\dfrac{10x+5}{6x^2+5x+3}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{6x^7 + 29x^6 + 29x^5 + 17x^4 - 21x^3 - 2x^2 + 13x + 14}{6x^2 + 5x + 3}[/tex]
As the leading coefficient of the divisor is not 1, divide the dividend and divisor by the leading coefficient of the divisor (6):
[tex]\implies \dfrac{\dfrac{6}{6}x^7+\dfrac{29}{6}x^6+\dfrac{29}{6}x^5+\dfrac{17}{6}x^4-\dfrac{21}{6}x^3-\dfrac{2}{6}x^2+\dfrac{13}{6}x+\dfrac{14}{6}}{\dfrac{6}{6}x^2+\dfrac{5}{6}x+\dfrac{3}{6}}[/tex]
[tex]\implies \dfrac{x^7+\dfrac{29}{6}x^6+\dfrac{29}{6}x^5+\dfrac{17}{6}x^4-\dfrac{7}{2}x^3-\dfrac{1}{3}x^2+\dfrac{13}{6}x+\dfrac{7}{3}}{x^2+\dfrac{5}{6}x+\dfrac{1}{2}}[/tex]
Write the opposite sign of the coefficient of the x term and the constant term of the divisor on the outside of the synthetic division box. Write the coefficients of all the terms of the dividend inside the synthetic division box in descending order. As there are no missing terms, there is no need to use any zeros.
[tex]\begin{array}{rr|rrrrrrrr}-\dfrac{5}{6}&-\dfrac{1}{2}&1&\dfrac{29}{6}&\dfrac{29}{6}&\dfrac{17}{6}&-\dfrac{7}{2}&-\dfrac{1}{3}&\dfrac{13}{6}&\dfrac{7}{3}\\&&&\phantom{\dfrac12}&&&&&&\\\cline{3-10} &\phantom{\dfrac12}&&&&&&&&\\\phantom{\dfrac12}&&&&&&&&&\\ \cline{3-10} \end{array}[/tex]
Perform synthetic division:
[tex]\begin{array}{rr|rrrrrrrr}-\dfrac{5}{6}&-\dfrac{1}{2}&1&\dfrac{29}{6}&\dfrac{29}{6}&\dfrac{17}{6}&-\dfrac{7}{2}&-\dfrac{1}{3}&\dfrac{13}{6}&\dfrac{7}{3}\\&&&\phantom{\dfrac12}&-\frac{1}{2}&-2&-\frac{1}{2}&0&2&-\frac{3}{2}\\\cline{3-10} &\phantom{\dfrac12}&&\frac{29}{6} & \frac{13}{3}& \frac{5}{6}& -4&-\frac{1}{3}&\frac{25}{6}&\\\phantom{\dfrac12}&&&-\frac{5}{6}&-\frac{10}{3} & -\frac{5}{6}&0&\frac{10}{3} & -\frac{5}{2}&\\ \cline{3-10} \phantom{\dfrac12}&&1&4&1&0&-4&3&\frac{5}{3} & \frac{5}{6}\end{array}[/tex]
Since the leading coefficient of the dividend is 7 and the leading coefficient of the divisor is 2, the leading exponent of the quotient will be 7 - 2 = 5.
Therefore:
[tex]\textsf{Quotient}: \quad 1x^5+4x^4+1x^3+0x^2-4x+3[/tex]
[tex]\textsf{Remainder}: \quad \dfrac{5}{3}x+\dfrac{5}{6}[/tex]
[tex]\implies x^5+4x^3+x^3-4x+3+\dfrac{\frac{5}{3}x+\frac{5}{6}}{x^2+\frac{5}{6}x+\frac{1}{2}}[/tex]
Multiply the rational remainder by 6/6 so that its denominator is in the original form of the divisor:
[tex]\implies x^5+4x^3+x^3-4x+3+\left(\dfrac{\frac{5}{3}x+\frac{5}{6}}{x^2+\frac{5}{6}x+\frac{1}{2}}\right) \cdot \dfrac{6}{6}[/tex]
[tex]\implies x^5+4x^3+x^3-4x+3+\dfrac{10x+5}{6x^2+5x+3}[/tex]
