Long Division and Synthetic Division Dividing Polynomials Long Division and Synthetic Division
Example 1: x + 5 x − 3 x2 + 2x − 15 -x2 + 3x x2 – 3x 5x – 15 -5x + 15 x + 5 x − 3 x2 + 2x − 15 -x2 + 3x x2 – 3x 5x – 15 -5x + 15 5x – 15
Example 2: 2x + 5 x − 3 2x2 − x − 4 -2x2 + 6x 2x2 – 6x 5x – 4 -5x + 15 2x + 5 x − 3 2x2 − x − 4 -2x2 + 6x 2x2 – 6x 5x – 4 -5x + 15 5x – 15 11
Example 3: x2 – x – 3 x4 − x3 + 4x + 2 x2 + 3 –x4 – 3x2 x4 + 3x2 x2 – x – 3 x4 − x3 + 4x + 2 x2 + 3 –x4 – 3x2 x4 + 3x2 − x3 – 3x2 + 4x + 2 + x3 + 3x – x3 – 3x – 3x2 + 7x + 2 +3x2 + 9 – 3x2 – 9 7x + 11
There is a shorthcut method… … called Synthetic Division a shorthand, or shortcut, method of polynomial division Uses just the Coefficients Only when dividing by a linear factor. For example (x+2) or (x-5) Synthetic division is generally used for finding zeros (or roots) of polynomial functions
Synthetic Division Change the sign If the remainder is 0, then Change the sign If the remainder is 0, then (x-3) is a factor x=3 is a zero. remainder The first term’s exponent is 1 less than the original.
Synthetic Division: Change the sign
Synthetic Division: Change the sign
f(x) = 3x3 + 2x2 − x + 3 is divided by g(x) = x − 4. Example 4: Use synthetic division to find the quotient Q(x) and the remainder R(x) when: f(x) = 3x3 + 2x2 − x + 3 is divided by g(x) = x − 4.
Example 5: Put in missing exponents and insert 0’s if needed.
Remainder Theorem: If the polynomial P(x) is divided by D(x) = x − c, then becomes Plugging in x=c to the above equation one sees that
Example 6: Use the Remainder Theorem to find f(−2) when f(x) = x3 + 2x2 − 7 Same!
Factor Theorem: The number c is a zero of P(x) if and only if (x−c) is a factor of P(x); that is, P(x) = Q(x) · (x − c) for some polynomial Q(x). In other words, when using Synthetic division if the remainder = 0, you found a factor that “goes in evenly”
Example 7: Use the Factor Theorem to determine whether x + 2 is a factor of f(x) = 3x6 + 2x3 − 176. YES!!!!!!!!!!