Use synthetic division to divide 2t^3 - t - 15 by t - 1. Give the quotient and the remainder.

Step-by-step solution

1. 1

   Pad the missing $t^2$ term. The dividend is a cubic, so it needs four coefficients:

   $2t^3+0\cdot t^2-t-15\;\Longrightarrow\; 2,\;0,\;-1,\;-15$

   Leaving the $0$ out would turn this into a division of $2t^2-t-15$ instead, a genuinely different problem with a different answer.

2. 2

   Take $c=1$ and predict the remainder. From $t-1=t-(1)$ we use $c=1$, and the remainder theorem gives

   $f(1)=2-1-15=-14$

3. 3

   Multiply and add through the four columns.

   $\begin{array}{r|rrrr}1 & 2 & 0 & -1 & -15\\ & & 2 & 2 & 1\\ \hline & 2 & 2 & 1 & -14\end{array}$

   Bring down $2$; $2\cdot 1=2$ and $0+2=2$; $2\cdot 1=2$ and $-1+2=1$; $1\cdot 1=1$ and $-15+1=-14$, matching the prediction.

4. 4

   Read the quadratic quotient. Dropping one degree from the cubic:

   $\text{quotient}=2t^2+2t+1,\qquad \text{remainder}=-14$

5. 5

   Verify the division identity.

   $2t^3-t-15=(t-1)\left(2t^2+2t+1\right)-14$

   Expanding: $(t-1)(2t^2+2t+1)=2t^3+2t^2+t-2t^2-2t-1=2t^3-t-1$, and subtracting $14$ gives $2t^3-t-15\;\checkmark$. The nonzero remainder shows $t-1$ is not a factor, and since $2t^2+2t+1$ has discriminant $4-8=-4<0$, the cubic $2t^3-t-15$ has exactly one real root.