Synthetic Division and the Remainder Theorem: Additional Practice (Set 2)

The Remainder Theorem converts each condition into an equation: $P(1) = 12$ and $P(-2) = 0$.

$P(1) = 1 + a + b + 6 = 12 \;\Longrightarrow\; a + b = 5$

$P(-2) = -8 + 4a - 2b + 6 = 0 \;\Longrightarrow\; 2a - b = 1$

Adding the two equations gives $3a = 6$, so $a = 2$ and then $b = 3$. Checking: $P(x) = x^3 + 2x^2 + 3x + 6$ has $P(1) = 12$ and $P(-2) = -8 + 8 - 6 + 6 = 0$.

The $x^2$ term is missing, so the top row is $1,\; 0,\; -13,\; 12$ with corner $3$. Bring down the $1$; then $0 + 3 = 3$; then $-13 + 9 = -4$; then $12 - 12 = 0$.

$\begin{array}{c|rrrr} 3 & 1 & 0 & -13 & 12 \\ & & 3 & 9 & -12 \\ \hline & 1 & 3 & -4 & 0 \end{array}$

The quotient is $x^2 + 3x - 4$ with remainder $0$. Skipping the placeholder zero collapses the row to $1,\; -13,\; 12$ and yields the wrong answer $x - 10$ with remainder $-18$.

By the Remainder Theorem the answer is $P(2)$, and the two identities let you evaluate it in stages. The second identity gives $Q(2) = (2 - 2)S(2) + 4 = 4$.

$P(2) = (2 + 1)\,Q(2) - 6 = 3 \cdot 4 - 6 = 6$

So dividing $P$ by $x - 2$ leaves remainder $6$. Neither $-6$ nor $4$ is the answer; those constants are remainders of different divisions, and they combine through the evaluation.

The remainder equals $P(3)$, so set $P(3) = 4$.

$P(3) = 27 - 36 + 3k + 10 = 3k + 1 = 4 \quad \Longrightarrow \quad k = 1$

Checking: $P(x) = x^3 - 4x^2 + x + 10$ gives $P(3) = 27 - 36 + 3 + 10 = 4$ as required.

Nothing in the proof of the Remainder Theorem required $c$ to be real, so the remainder is $P(i)$.

$P(i) = i^2 + 1 = -1 + 1 = 0$

The division is exact, which matches the factorization $x^2 + 1 = (x - i)(x + i)$ over the complex numbers: the polynomial with no real roots still has linear factors once $i$ is allowed.

Run the tableau with corner $7$ and top row $1,\; -5,\; -9,\; -7$: bring down $1$; then $-5 + 7 = 2$; then $-9 + 14 = 5$; then $-7 + 35 = 28$.

$\begin{array}{c|rrrr} 7 & 1 & -5 & -9 & -7 \\ & & 7 & 14 & 35 \\ \hline & 1 & 2 & 5 & 28 \end{array}$

By the Remainder Theorem $P(7) = 28$, and the direct check $343 - 245 - 63 - 7 = 28$ agrees. Three multiplications replaced computing $7^3$ and $5 \cdot 7^2$ separately.

Reconstruct the middle row from the rhythm. The bring-down gives the first bottom entry $1$; the product $1 \times 3 = 3$ sits under $a$, and the column must add to the next bottom entry $1$.

$a + 3 = 1 \quad \Longrightarrow \quad a = -2$

The rest of the tableau confirms the reading: $1 \times 3 = 3$ and $-5 + 3 = -2$; then $-2 \times 3 = -6$ and $3 - 6 = -3$. Every column checks out with $a = -2$.

Factor the divisor: $3x + 4 = 3\left(x + \tfrac{4}{3}\right)$, so run the tableau with corner $-\tfrac{4}{3}$: bring down $6$; then $5 - 8 = -3$; then $-1 + 4 = 3$; then $-2 - 4 = -6$.

$\begin{array}{c|rrrr} -\tfrac{4}{3} & 6 & 5 & -1 & -2 \\ & & -8 & 4 & -4 \\ \hline & 6 & -3 & 3 & -6 \end{array}$

So $P(x) = \left(x + \tfrac{4}{3}\right)(6x^2 - 3x + 3) - 6 = (3x + 4)(2x^2 - x + 1) - 6$. The quotient is $2x^2 - x + 1$ (the tableau's quotient divided by $3$) and the remainder stays $-6$.

The bottom row gives quotient $2x^2 - x + 5$, so rebuild the dividend from the division identity.

$P(x) = (x + 3)(2x^2 - x + 5) - 7 = 2x^3 - x^2 + 5x + 6x^2 - 3x + 15 - 7$

Collecting like terms gives $P(x) = 2x^3 + 5x^2 + 2x + 8$. The remainder is added with its own sign; adding $+7$ instead produces the wrong constant $22$.

By the Remainder Theorem, test each divisor with one evaluation.

$P(4) = 48, \qquad P(-1) = -12, \qquad P(2) = -24, \qquad P(3) = 27 + 27 - 30 - 24 = 0$

Only $x = 3$ makes the polynomial vanish, so only $x - 3$ divides it exactly. (The full factorization is $(x - 3)(x + 2)(x + 4)$, but the other two factors are not among the choices.)

Exact division by $x + 1$ means $P(-1) = 0$ by the Remainder Theorem.

$P(-1) = -k + 4 - k + 4 = 8 - 2k = 0 \quad \Longrightarrow \quad k = 4$

Checking: $P(x) = 4x^3 + 4x^2 + 4x + 4 = 4(x + 1)(x^2 + 1)$, so the division by $x + 1$ is indeed exact.

Since $2x - 3 = 2\left(x - \tfrac{3}{2}\right)$, the remainder equals $P\left(\tfrac{3}{2}\right)$, and only the quotient (not the remainder) absorbs the factor $2$.

$P\left(\tfrac{3}{2}\right) = 2 \cdot \tfrac{27}{8} - \tfrac{9}{4} + \tfrac{9}{2} - 5 = \tfrac{27}{4} - \tfrac{9}{4} + \tfrac{18}{4} - \tfrac{20}{4} = \tfrac{16}{4} = 4$

The remainder is $4$. Halving it to $2$ is the classic non-monic mistake: the constant remainder never gets divided by the leading coefficient.