Synthetic Division: A Shortcut for Polynomial Division
Core Concepts
Master the technique of synthetic division to quickly divide polynomials by linear expressions of the form \(x - k\). Apply it to find function values and test for roots.
Synthetic Division: A Shortcut for Polynomial Division
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What is synthetic division primarily used for?
It is a shortcut method for dividing a polynomial by a linear binomial of the form \((x - k)\). It is faster and more compact than polynomial long division.
To divide \(P(x)\) by \((x - k)\) using synthetic division, what number do you place in the 'box' to the left?
You use the value \(k\), which is the zero of the divisor \((x-k)\). For example, to divide by \((x - 5)\), you use \(5\). To divide by \((x + 3)\), you use \(-3\).
Set up the synthetic division problem for dividing \(x^3 - 5x^2 + 2x - 10\) by \((x - 3)\).
You use \(k=3\) in the box. The coefficients of the polynomial are \(1\), \(-5\), \(2\), and \(-10\). The setup is: \(3 \,|\, 1\, -5\, 2\, -10\), with a line drawn underneath.
Given the setup \(3 \,|\, 1\, -5\, 2\, -10\), describe the first two steps of synthetic division.
1. Bring down the first coefficient (\(1\)) below the line.
2. Multiply the number in the box (\(3\)) by the number you just brought down (\(1\)) and write the result (\(3\)) under the next coefficient (\(-5\)).
How do you interpret the last number in the bottom row of a completed synthetic division problem?
The last number is the remainder of the division. If the remainder is \(0\), it means the divisor is a factor of the polynomial.
How do you interpret the other numbers (not the last one) in the bottom row of a synthetic division problem?
These numbers are the coefficients of the quotient polynomial. The degree of the quotient is one less than the degree of the original polynomial (the dividend).
Divide \(2x^3 - 7x^2 - 5x + 4\) by \((x - 4)\) using synthetic division and state the quotient and remainder.
Using \(k=4\): \(4 \,|\, 2\, -7\, -5\, 4 \) gives a bottom row of \(2, 1, -1, 0\).
The quotient is \(2x^2 + x - 1\).
The remainder is \(0\).
How must you handle 'missing terms' when setting up a synthetic division problem? Example: \((x^4 - 3x^2 + 1) \div (x+2)\).
You must use a zero as a placeholder for the coefficient of any missing term. The coefficients for \(x^4 - 3x^2 + 1\) are \(1\) (for \(x^4\)), \(0\) (for \(x^3\)), \(-3\) (for \(x^2\)), \(0\) (for \(x\)), and \(1\).
Set up the synthetic division for \((x^4 - 16) \div (x - 2)\).
The dividend is \(x^4 + 0x^3 + 0x^2 + 0x - 16\). Use \(k=2\). The setup is: \(2 \,|\, 1\, 0\, 0\, 0\, -16\).
Use synthetic division and the Remainder Theorem to find \(P(-3)\) for \(P(x) = x^3 + 5x^2 + 5x - 3\).
Divide \(P(x)\) by \((x+3)\) using \(k=-3\). The synthetic division gives a remainder of \(0\). Therefore, \(P(-3) = 0\).
After finding that \(k\) is a root of \(P(x)\) using synthetic division, what is the resulting quotient called?
The resulting quotient is often called the 'depressed polynomial'. It has a degree one less than \(P(x)\) and can be used to find the remaining roots.
Is it possible to use synthetic division to divide by \(2x - 6\)?
Yes, but with an extra step. First, factor out the \(2\) to get \(2(x-3)\). Perform synthetic division with \(k=3\). Then, divide every coefficient in the resulting quotient by \(2\). The remainder is not affected.
