Synthetic Division and the Remainder Theorem: Additional Practice (Set 1)

The top row needs a placeholder for the missing $x^2$ term: $2,\; -3,\; 0,\; 1,\; -4$ with corner $-1$. Bring down the $2$; then $-3 - 2 = -5$; then $0 + 5 = 5$; then $1 - 5 = -4$; then $-4 + 4 = 0$.

$\begin{array}{c|rrrrr} -1 & 2 & -3 & 0 & 1 & -4 \\ & & -2 & 5 & -5 & 4 \\ \hline & 2 & -5 & 5 & -4 & 0 \end{array}$

The quotient is $2x^3 - 5x^2 + 5x - 4$ with remainder $0$, confirmed by $P(-1) = 2 + 3 - 1 - 4 = 0$.

The Remainder Theorem replaces a $51$-column division with one evaluation at $x = -1$, where the even power gives $+1$.

$P(-1) = (-1)^{50} + (-1) + 1 = 1 - 1 + 1 = 1$

The remainder is $1$. Evaluating at $+1$ by mistake gives $3$, and forgetting that an even power of $-1$ is positive leads to $-1$.

Run the tableau with corner $-2$ and top row $3,\; 7,\; -2,\; 1,\; -5$: bring down $3$; then $7 - 6 = 1$; then $-2 - 2 = -4$; then $1 + 8 = 9$; then $-5 - 18 = -23$.

$\begin{array}{c|rrrrr} -2 & 3 & 7 & -2 & 1 & -5 \\ & & -6 & -2 & 8 & -18 \\ \hline & 3 & 1 & -4 & 9 & -23 \end{array}$

So $P(-2) = -23$. The direct check agrees: $3(16) + 7(-8) - 2(4) + (-2) - 5 = 48 - 56 - 8 - 2 - 5 = -23$.

The division identity holds at every input, so evaluate both factors at $x = 5$.

$P(5) = (5 - 4)\left(25 + 5 + 1\right) + 11 = (1)(31) + 11 = 42$

So $P(5) = 42$. The remainder $11$ by itself only answers the question about $P(4)$, where the product term vanishes.

The Remainder Theorem turns the two conditions into $P(1) = 6$ and $P(-2) = 0$.

$P(1) = 1 + a + b + 2 = 6 \;\Longrightarrow\; a + b = 3$

$P(-2) = -8 + 4a - 2b + 2 = 0 \;\Longrightarrow\; 2a - b = 3$

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

Peel off one factor of $x$ at a time, keeping each coefficient's own sign.

$5x^3 - 4x^2 + 2x - 7 = \bigl((5x - 4)x + 2\bigr)x - 7$

Expanding back: $(5x - 4)x = 5x^2 - 4x$; adding $2$ and multiplying by $x$ gives $5x^3 - 4x^2 + 2x$; subtracting $7$ finishes it. This nesting is exactly the tableau's multiply-and-add rhythm, one multiplication per coefficient. The other expressions are products of factors, which change the polynomial entirely.

The remainder is $P(1)$, and evaluating any polynomial at $1$ simply adds up its coefficients.

$P(1) = 7 - 3 + 2 - 1 + 4 - 6 = 3$

The remainder is $3$. This special case is worth remembering: dividing by $x - 1$ always leaves the sum of the coefficients, so a polynomial's coefficients sum to zero exactly when $x - 1$ divides it.

Divide by $x - 5$ first: corner $5$, top row $1,\; -7,\; 7,\; 15$ gives bottom row $1,\; -2,\; -3$ with remainder $0$, so $P(x) = (x - 5)(x^2 - 2x - 3)$.

$x^2 - 2x - 3 = (x - 3)(x + 1)$

The second exact division confirms the factor $x - 3$ and leaves $x + 1$. Altogether $P(x) = (x - 5)(x - 3)(x + 1)$; expanding restores $x^3 - 7x^2 + 7x + 15$, since $(x - 5)(x - 3) = x^2 - 8x + 15$ and multiplying by $x + 1$ gives the original cubic.

The remainder is $P(c) = c^2 - 8c + 15$, so set it equal to $3$ and solve the quadratic.

$c^2 - 8c + 15 = 3 \quad \Longrightarrow \quad c^2 - 8c + 12 = 0 \quad \Longrightarrow \quad (c - 2)(c - 6) = 0$

So $c = 2$ or $c = 6$; both divisions leave remainder $3$. The values $3$ and $5$ answer a different question: they are where the remainder is $0$.

Since $2x + 3 = 2\left(x + \tfrac{3}{2}\right)$, the remainder equals $P\left(-\tfrac{3}{2}\right)$.

$P\left(-\tfrac{3}{2}\right) = 4 \cdot \tfrac{9}{4} - 9 = 9 - 9 = 0$

The remainder is $0$, matching the difference-of-squares factorization $4x^2 - 9 = (2x + 3)(2x - 3)$: the divisor is one of the factors, so nothing is left over.

The remainder is $P(-1)$; set it equal to $5$ and solve for $k$.

$P(-1) = -1 + k(-1)^2 + k = -1 + 2k = 5 \quad \Longrightarrow \quad k = 3$

Both appearances of $k$ survive at $x = -1$ because $(-1)^2 = 1$. Checking: $P(x) = x^3 + 3x^2 + 3$ gives $P(-1) = -1 + 3 + 3 = 5$ as required.

Evaluating the identity at $x = -2$ kills the product term, so $P(-2) = r$, which pins down $r = -5$. Now evaluate the same identity at $x = 0$.

$P(0) = (0 + 2)\left(0 - 0 + 4\right) + (-5) = 2 \cdot 4 - 5 = 3$

So $P(0) = 3$. The two evaluations use the same division identity at two different inputs: one to find the remainder, one to find the requested value.