Algebra · real student question

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

Question

Use synthetic division to divide 2t2−t−152t^2-t-15 by t−1t-1. State the quotient and the remainder.

Step-by-step solution

  1. Confirm the divisor is in the right shape. Synthetic division needs the divisor monic and linear; t−1t-1 is both, so c=1c=1. The dividend may have any leading coefficient — the 22 in 2t22t^2 causes no trouble at all, it simply becomes the leading coefficient of the quotient. The coefficient row is

    2,  −1,  −152,\;-1,\;-15

  2. Predict the remainder. By the remainder theorem the last column must come out as the value of the polynomial at t=1t=1:

    f(1)=2(1)2−1−15=2−16=−14f(1)=2(1)^2-1-15=2-16=-14

    Doing this first means the table is a confirmation rather than a guess.

  3. Run the multiply-and-add cycle with c=1c=1.

    12−1−152121−14\begin{array}{r|rrr}1 & 2 & -1 & -15\\ & & 2 & 1\\ \hline & 2 & 1 & -14\end{array}

    Bring down the 22; 2⋅1=22\cdot 1=2 and −1+2=1-1+2=1; 1⋅1=11\cdot 1=1 and −15+1=−14-15+1=-14.

  4. Read off a linear quotient. A quadratic divided by a linear divisor leaves a linear quotient, so the first two entries of the bottom row are the quotient and the last is the remainder:

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

  5. Verify and interpret. Written without fractions,

    2t2−t−15=(t−1)(2t+1)−142t^2-t-15=(t-1)(2t+1)-14

    and expanding gives (t−1)(2t+1)=2t2+t−2t−1=2t2−t−1(t-1)(2t+1)=2t^2+t-2t-1=2t^2-t-1, so subtracting 1414 yields 2t2−t−15  ✓2t^2-t-15\;\checkmark. Since the remainder is nonzero, t=1t=1 is not a root; the actual factorization is 2t2−t−15=(2t+5)(t−3)2t^2-t-15=(2t+5)(t-3), with roots t=3t=3 and t=−52t=-\tfrac52.

Answer

2t2−t−15t−1=2t+1−14t−1,quotient 2t+1, remainder −14\frac{2t^2-t-15}{t-1}=2t+1-\frac{14}{t-1},\qquad \text{quotient }2t+1,\ \text{remainder }-14

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