Remainder and Factor Theorem

The Remainder Theorem and Factor Theorem are essential tools in polynomial algebra. They allow us to evaluate polynomials quickly and determine whether a given binomial is a factor of the polynomial, without performing full division.

Remainder Theorem:

If a polynomial $f(x)$ is divided by a linear divisor of the form $(x - c)$, the remainder is simply:

$$ \text{Remainder} = f(c) $$

This means you can find the remainder by directly substituting $x = c$ into the polynomial, avoiding long division.

Factor Theorem:

A special case of the Remainder Theorem: If $f(c) = 0$, then $(x - c)$ is a factor of $f(x)$.

$$ f(c) = 0 \quad \Rightarrow \quad (x - c) \text{ is a factor of } f(x) $$

This theorem is commonly used to find roots or to start factoring higher-degree polynomials.

How They Work Together:

Use the Remainder Theorem to test potential roots quickly.
If the result is zero, you’ve also confirmed a factor via the Factor Theorem.
After identifying a factor, you can reduce the polynomial via division or synthetic division.

These theorems simplify root-finding and polynomial factoring, especially for higher-order polynomials in engineering equations or signal analysis.

Problem 1:

What is the remainder when $9-3x+5x^2+6x^3$ is divided by $3x-2$?

Using all methods: Long Division, Synthetic Division, and Remainder Theorem

Problem 2:

When the expression $x^4+ax^3+5x^2+bx+6$ is divided by $(x-2)$, the remainder is 16. When it is divided by $(x+1)$, the remainder is 10. Find the value of the constant b.

Given $f(x)=x^4+ax^3+5x^2+bx+6$. When divided by $(x-2)$ the remainder is $16$ $\Rightarrow$ $f(2)=16$. When divided by $(x+1)$ the remainder is $10$ $\Rightarrow$ $f(-1)=10$.

$f(2)=2^4+a(2)^3+5(2)^2+b(2)+6=16+8a+20+2b+6=42+8a+2b$. Set $42+8a+2b=16 \Rightarrow 8a+2b=-26 \Rightarrow 4a+b=-13$.
$f(-1)=(-1)^4+a(-1)^3+5(-1)^2+b(-1)+6=1-a+5-b+6=12-a-b$. Set $12-a-b=10 \Rightarrow a+b=2$.
Solve the system: $\begin{cases}4a+b=-13\\ a+b=2\end{cases}\Rightarrow 3a=-15\Rightarrow a=-5,\ b=7$.

Therefore, $\boxed{b=7}$.

Check: $f(2)=42+8(-5)+2(7)=16$, $f(-1)=12-(-5)-7=10$.

Problem 3:

Find k so that $2x^2-kx-9$ has x-k as a factor.

If $x-k$ is a factor, then by the Factor Theorem we must have $f(k)=0$ for $f(x)=2x^2-kx-9$. Note that "the divisor is a factor" essentially means we do not have a remainder.
Substitute $x=k$: $$f(k)=2k^2-k(k)-9=2k^2-k^2-9=k^2-9.$$
Set equal to zero: $$k^2-9=0 \;\;\Rightarrow\;\; k^2=9.$$
Solve for $k$: $$k=\pm 3.$$

Final Answer: $\boxed{k=3}$ or $\boxed{k=-3}$