# How do you use synthetic division to divide (6t^3+5t^2+9) by 2t+3?

##### 1 Answer
Nov 19, 2015

{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,-9,+6,-9), (,,"-----","-----","-----","-----"), (div(2),"|",6,-4,+6,color(white)("X")color(red)(0)), (xx(-3),"|",color(blue)(3),color(blue)(-2),color(blue)(+3),) :}

#### Explanation:

Write the divisor in canonical form (your instructor may call this something else) i.e. in descending order of variable exponents with all exponents included (being careful to include those whose coefficients are $0$
$\textcolor{w h i t e}{\text{XXX}} 6 {t}^{2} + 5 {t}^{2} + 0 {t}^{1} + 9 {t}^{0}$ in this case

Write the coefficients in a row for line 1.
Leave a blank line as line 2 (or prefix it with a plus sign as I have done); values entered on this line will be added to values from line 1.
Draw a line separating lines 2 and 3.
On line 3 make a note (as a prefix) that you will need to divide by coefficient of the leading term of the binomial divisor.
On line 4 make a note (as a prefix) that you will need to multiply by the negative of the constant term of the divisor.

You should have something that looks like:
{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,,,), (,,"-----","-----","-----","-----"), (div(2),"|",,,,), (xx(-3),"|",,,,) :}

Copy the first coefficient ($6$) from line 1 to line 3 (as if you had added 0)
{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,,,), (,,"-----","-----","-----","-----"), (div(2),"|",6,,,), (xx(-3),"|",,,,) :}
Repeat the following process until the final column has been added:

• Divide the last number entered on line 3 by the coefficient of the $t$ term of the divisor and write the quotient below it on line 4.
• Multiple the last number written on line 4 by the negative of the constant term of the divisor and write the product in the next column of line 2
• Add the next two numbers from lines 1 and 2 to get a sum in line 3.

{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,-9,,), (,,"-----","-----","-----","-----"), (div(2),"|",6,-4,,), (xx(-3),"|",color(blue)(3),,,) :}

{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,-9,+6,-9), (,,"-----","-----","-----","-----"), (div(2),"|",6,-4,+6,), (xx(-3),"|",color(blue)(3),color(blue)(-2),,) :}

{: (,,6,color(white)("X")5,color(white)("X")0,color(white)("X")9), (+,,,-9,+6,-9), (,,"-----","-----","-----","-----"), (div(2),"|",6,-4,+6,color(white)("X")color(red)(0)), (xx(-3),"|",color(blue)(3),color(blue)(-2),color(blue)(+3),) :}

The last sum (in this case $\textcolor{red}{0}$) will be the remainder;
the last row will be the coefficients of the quotient polynomial (in this case $\textcolor{b l u e}{3} {t}^{2} \textcolor{b l u e}{- 2} t \textcolor{b l u e}{+ 3}$