Long Division of Polynomials with Missing Terms
2 3
3 2
2
2
x +5x -3 x 3x 2
x +5x 3x
-5x 6x 2
-5x 25x 15
31x- 17
5x 2
31 17
5 3
x
x x
You need to leave a hole when you have
missing terms. This technique will help
you line up like terms. See the dividend
above.
2 5 + 7 - 1
Using synthetic division instead of long division.
Notice that the divisor has to be a binomial of
degree 1 with no coefficients.
5
10
3
6
5
2
55 3
2
2 5 7 1
xx
x x x
Thus:
If you are given the function f(x)=x3- 4x2+5x+3 and
you want to find f(2), then the remainder of this
function when divided by x-2 will give you f(2)
f(2)=5
Example
Use synthetic division and the remainder
theorem to find the indicated function value. 3 2( ) 3 5 1; f(2)f x x x
Solve the equation 2x3-3x2-11x+6=0 given that 3 is
a zero of f(x)=2x3-3x2-11x+6. The factor theorem
tells us that x-3 is a factor of f(x). So we will use
both synthetic division and long division to show
this and to find another factor.
Another factor
Example
Solve the equation x3- 5x2 + 9x - 45 = 0 given
that 5 is a zero of f(x)= x3- 5x2 + 9x – 45.
Consider all complex number solutions.