Exercise 5.1 — Question 8

Key Concepts Covered

This question draws on three core skills:

1. Polynomial Division (long division or synthetic division)
2. Remainder Theorem
3. Factor Theorem (as a special case of the Remainder Theorem)

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Remainder Theorem

Why it works: By the division algorithm, $P(x)=(x-a)\cdot Q(x)+R$ Setting $x=a$ gives $P(a)=0\cdot Q(a)+R$, so $R=P(a)$.

Quick 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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Synthetic Division

Synthetic division is a shorthand method to divide $P(x)$ by $(x-k)$.

Steps:

1. Write the coefficients of $P(x)$ in order (use 0 for missing terms).
2. Write $k$ to the left.
3. Bring down the first coefficient.
4. Multiply by $k$, add to the next coefficient. Repeat.
5. The last number is the remainder; the others form the quotient coefficients.

Example

Divide $P(x)=x^3-3x^2+4x-2$ by $(x-1)$, so $k=1$:

$\begin{array}{rrrrr}1& 1& -3& 4& -2\\ & & 1& -2& 2\\ & 1& -2& 2& 0\end{array}$

Quotient: $x^2-2x+2$, Remainder: $0$

Since the remainder is 0, $(x-1)$ is a factor of $P(x)$.

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Factor Theorem

This is a direct consequence of the Remainder Theorem: if $P(a)=0$, the remainder is zero, so $(x-a)$ divides $P(x)$ exactly.

Factorizing a Cubic Polynomial

To fully factorize $P(x)=x^3-6x^2+11x-6$:

Step 1 — Find a root by trial (test factors of the constant term $\pm 1,\pm 2,\pm 3,\pm 6$): $P(1)=1-6+11-6=0✓$ So $(x-1)$ is a factor.

Step 2 — Divide $P(x)$ by $(x-1)$ using synthetic division: $\begin{array}{rrrrr}1& 1& -6& 11& -6\\ & & 1& -5& 6\\ & 1& -5& 6& 0\end{array}$ Quotient: $x^2-5x+6$

Step 3 — Factorize the quadratic: $x^2-5x+6=(x-2)(x-3)$

Full factorization: $P(x)=(x-1)(x-2)(x-3)$

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Summary Table