5.1 Q-6

Exercise 5.1 — Question 6

This question covers polynomial division, the Remainder Theorem, and the Factor Theorem.

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Key Concepts

Polynomial Long Division / Synthetic Division (SLO M-11-A-42)

To divide a polynomial $P(x)$ by a linear factor $(x-k)$:

Synthetic Division Steps:

1. Write the root $k$ and the coefficients of $P(x)$.
2. Bring down the leading coefficient.
3. Multiply by $k$, add to the next coefficient.
4. Repeat until all coefficients are processed.
5. The last number is the remainder; the rest form the quotient.

Example: Divide $P(x)=2x^3-3x^2+x-5$ by $(x-2)$:

$\begin{array}{rrrrr}2& 2& -3& 1& -5\\ & & 4& 2& 6\\ & 2& 1& 3& 1\end{array}$

Quotient: $2x^2+x+3$, Remainder: $1$

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Remainder Theorem (SLO M-11-A-43)

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

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

Remainder $=5$.

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Factor Theorem (SLO M-11-A-44)

This is a special case of the Remainder Theorem — when the remainder is zero, the divisor is a factor.

Steps to Factorize a Cubic Polynomial $P(x)$:

1. Find a value $a$ such that $P(a)=0$ (try $\pm 1,\pm 2,\dots$).
2. Divide $P(x)$ by $(x-a)$ using synthetic or long division.
3. Factor the resulting quadratic quotient.
4. Write $P(x)$ as a product of all linear factors.

Example: Factorize $P(x)=x^3-6x^2+11x-6$.

• Test $x=1$: $P(1)=1-6+11-6=0$ ✓ → $(x-1)$ is a factor.
• Divide: $P(x)=(x-1)(x^2-5x+6)$
• Factor quadratic: $x^2-5x+6=(x-2)(x-3)$
• Result: $P(x)=(x-1)(x-2)(x-3)$