5.1 Q-1

Exercise 5.1 — Polynomial Division and Remainder Theorem

Key Concepts

Division Algorithm

For any polynomials $P(x)$ and $D(x)$ with $D(x)\ne 0$: $P(x)=D(x)\cdot Q(x)+R(x)$ where $Q(x)$ is the quotient and $R(x)$ is the remainder, with $\mathrm{deg}(R)<\mathrm{deg}(D)$.

• If $D(x)$ is linear (degree 1), then $R$ is a constant.
• If $D(x)$ is quadratic (degree 2), then $R$ is at most linear (i.e., $R=ax+b$).

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Long Division of Polynomials

Steps:

1. Arrange both polynomials in descending powers of $x$. Insert $0$ coefficients for missing terms.
2. Divide the leading term of the dividend by the leading term of the divisor to get the first term of $Q(x)$.
3. Multiply the entire divisor by that term and subtract from the dividend.
4. Bring down the next term and repeat until the degree of the remainder is less than the degree of the divisor.

Example: Divide $P(x)=2x^3+3x^2-4x+1$ by $(x+2)$.

$2x^3+3x^2-4x+1=(x+2)(2x^2-x-2)+5$

So Quotient $=2x^2-x-2$ and Remainder $=5$.

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Synthetic Division (for linear divisors $x-k$)

A shorthand method using only coefficients:

Example: Divide $P(x)=2x^3+3x^2-4x+1$ by $(x+2)$, i.e., $k=-2$.

Quotient $=2x^2-x-2$, Remainder $=5$. ✓

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Remainder Theorem

$\text{Remainder}=P(a)$

Proof sketch: By the division algorithm, $P(x)=(x-a)Q(x)+R$. Setting $x=a$: $P(a)=0\cdot Q(a)+R=R$.

Example: Find the remainder when $P(x)=2x^3+3x^2-4x+1$ is divided by $(x+2)$.

$R=P(-2)=2(-2{)}^3+3(-2{)}^2-4(-2)+1=-16+12+8+1=5$

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