Remainder Theorem for Polynomials Solutions

• The Remainder Theorem: If the polynomial P(x) is divided by (x – k), then the    remainder     is equal to the value, P(k).
• Rather than using    substitution     for all values of x, use synthetic division to find the value of the remainder.
• Synthetic division is more    efficient     than substitution because you do not need to raise any term to a power.

Example 6

Determine if P(5) and P(–5) are factors of  ${\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{–}\mathbf{17}\mathit{x}\mathbf{+}\mathbf{15}$.

   P(5) is not a factor    because the remainder is not zero.

The value –5 is a root of the polynomial because    the remainder is 0   ; therefore, (x + 5) is    a factor of P(x)   .

Example 7

Find the missing value when P(n) = –5 for $\mathit{P}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{–}\mathbf{6}\mathit{x}\mathbf{+}\mathbf{3}$.

Plan
Substitute the values into the synthetic division frame

Complete synthetic division

Solve for n

 

$\begin{array}{rcl}3+n\left(n–6\right)& =& –5\\ 3+n^2–6n& =& –5\\ n^2–6n+8& =& 0\\ \left(n–2\right)\left(n–4\right)& =& 0\\ n& =& 2, 4\\ & & \begin{array}{cc}P\left(2\right)=–5,& P\left(4\right)=–5\end{array}\end{array}$