Below are 10 practice questions on synthetic division of polynomials. Each problem asks you to find the quotient \(Q\) and remainder \(R\), or to apply the Remainder Theorem. Problems include real, fractional, decimal, and complex divisors. Complete step-by-step synthetic division tables are provided in the solutions — ideal for algebra and precalculus students.
Use synthetic division to find the quotient \(Q\) and remainder \(R\) when dividing \( 2x^2 + x \) by \( x - 1 \).
Root \(c = 1\); coefficients \(2, 1, 0\) (include the placeholder for the constant):
\[ \begin{array}{c|rrr} 1 & 2 & 1 & 0 \\ & & 2 & 3 \\ \hline & 2 & 3 & \mathbf{3} \end{array} \]Answer: \( Q = 2x + 3,\ R = 3 \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( 4x^3 + x^2 - 2x + 3 \) by \( x + 2 \).
Root \(c = -2\); coefficients \(4, 1, -2, 3\):
\[ \begin{array}{c|rrrr} -2 & 4 & 1 & -2 & 3 \\ & & -8 & 14 & -24 \\ \hline & 4 & -7 & 12 & \mathbf{-21} \end{array} \]Answer: \( Q = 4x^2 - 7x + 12,\ R = -21 \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( 7x^5 - 2x^4 - 5x^3 + x^2 - 3x + 5 \) by \( x - 2 \).
Root \(c = 2\); coefficients \(7, -2, -5, 1, -3, 5\):
\[ \begin{array}{c|rrrrrr} 2 & 7 & -2 & -5 & 1 & -3 & 5 \\ & & 14 & 24 & 38 & 78 & 150 \\ \hline & 7 & 12 & 19 & 39 & 75 & \mathbf{155} \end{array} \]Answer: \( Q = 7x^4 + 12x^3 + 19x^2 + 39x + 75,\ R = 155 \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( \frac{1}{2}x^3 - \frac{1}{3}x^2 - \frac{3}{2}x + \frac{1}{3} \) by \( x - \frac{1}{2} \).
Root \(c = \tfrac{1}{2}\):
\[ \begin{array}{c|rrrr} \tfrac{1}{2} & \tfrac{1}{2} & -\tfrac{1}{3} & -\tfrac{3}{2} & \tfrac{1}{3} \\ & & \tfrac{1}{4} & -\tfrac{1}{24} & -\tfrac{37}{48} \\ \hline & \tfrac{1}{2} & -\tfrac{1}{12} & -\tfrac{37}{24} & \mathbf{-\tfrac{7}{16}} \end{array} \]Note: older answer keys listed the middle coefficient as \(-\tfrac{1}{2}\); the correct coefficient is \(-\tfrac{1}{12}\).
Answer: \( Q = \frac{1}{2}x^2 - \frac{1}{12}x - \frac{37}{24},\ R = -\frac{7}{16} \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( 0.3x^3 - 0.2x^2 - 0.1x + 0.3 \) by \( x - 0.2 \).
Root \(c = 0.2\):
\[ \begin{array}{c|rrrr} 0.2 & 0.3 & -0.2 & -0.1 & 0.3 \\ & & 0.06 & -0.028 & -0.0256 \\ \hline & 0.3 & -0.14 & -0.128 & \mathbf{0.2744} \end{array} \]Answer: \( Q = 0.3x^2 - 0.14x - 0.128,\ R = 0.2744 \)
Use synthetic division and the Remainder Theorem to find \( f(-2) \), given \( f(x) = 5x^4 - 4x^3 - x^2 + 5 \).
Root \(-2\); coefficients \(5, -4, -1, 0, 5\):
\[ \begin{array}{c|rrrrr} -2 & 5 & -4 & -1 & 0 & 5 \\ & & -10 & 28 & -54 & 108 \\ \hline & 5 & -14 & 27 & -54 & \mathbf{113} \end{array} \]Answer: \( f(-2) = 113 \)
Use synthetic division and the Remainder Theorem to find \( f\!\left(\tfrac{1}{2}\right) \), given \( f(x) = -3x^5 + x^3 - 2x^2 + x - 3 \).
Root \(\tfrac{1}{2}\); coefficients \(-3, 0, 1, -2, 1, -3\):
\[ \begin{array}{c|rrrrrr} \tfrac{1}{2} & -3 & 0 & 1 & -2 & 1 & -3 \\ & & -\tfrac{3}{2} & -\tfrac{3}{4} & \tfrac{1}{8} & -\tfrac{15}{16} & \tfrac{1}{32} \\ \hline & -3 & -\tfrac{3}{2} & \tfrac{1}{4} & -\tfrac{15}{8} & \tfrac{1}{16} & \mathbf{-\tfrac{95}{32}} \end{array} \]Note: older answer keys omitted the negative sign; the correct remainder is \(-\tfrac{95}{32}\).
Answer: \( f\!\left(\tfrac{1}{2}\right) = -\tfrac{95}{32} \)
Use synthetic division and the Remainder Theorem to find \( f(0.2) \), given \( f(x) = 0.1x^3 - 0.6x^2 - 0.6 \).
Root \(0.2\); coefficients \(0.1, -0.6, 0, -0.6\):
\[ \begin{array}{c|rrrr} 0.2 & 0.1 & -0.6 & 0 & -0.6 \\ & & 0.02 & -0.116 & -0.0232 \\ \hline & 0.1 & -0.58 & -0.116 & \mathbf{-0.6232} \end{array} \]Answer: \( f(0.2) = -0.6232 \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( x^2 + 2x + 3 \) by \( x - i \), where \( i = \sqrt{-1} \).
Root \(c = i\):
\[ \begin{array}{c|rrr} i & 1 & 2 & 3 \\ & & i & -1+2i \\ \hline & 1 & 2+i & \mathbf{2+2i} \end{array} \](Since \(i \cdot (2 + i) = 2i + i^2 = -1 + 2i\).)
Answer: \( Q = x + (2 + i),\ R = 2 + 2i \)
Use synthetic division to find \(Q\) and \(R\) when dividing \( x^5 + x^4 + x^3 + 2x - 5 \) by \( x + i \).
Root \(c = -i\); coefficients \(1, 1, 1, 0, 2, -5\) (include \(0\) for the missing \(x^2\) term):
\[ \begin{array}{c|rrrrrr} -i & 1 & 1 & 1 & 0 & 2 & -5 \\ & & -i & -1-i & -1 & i & 1-2i \\ \hline & 1 & 1-i & -i & -1 & 2+i & \mathbf{-4-2i} \end{array} \]Check each product:
Answer: \( Q = x^4 + (1 - i)x^3 - i x^2 - x + (2 + i),\ R = -4 - 2i \)
Synthetic division is a fast way to divide a polynomial by a linear divisor of the form \(x - c\). It's also used with the Remainder Theorem to evaluate \(f(c)\) quickly.
Synthetic division only works when the divisor is linear, \(x - c\). For divisors like \(x^2 + 1\) or \(2x - 3\) with a leading coefficient other than 1, use polynomial long division (or adjust coefficients carefully).
The final entry is the remainder \(R\). The other entries are the coefficients of the quotient polynomial, whose degree is one less than the dividend.
The Remainder Theorem says \(f(c)\) equals the remainder when \(f(x)\) is divided by \(x - c\). Synthetic division gives you that remainder directly.
Yes. Simply use the complex number \(c = a + bi\) as the root. Arithmetic works the same way; just remember \(i^2 = -1\).