5.2 Q-8

Exercise 5.2 — Question 8

Problem Statement

Factorize the cubic polynomial using the Factor Theorem and synthetic (or long) division.

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Key Theorems Recap

Remainder Theorem: When a polynomial $P(x)$ is divided by $(x-a)$, the remainder is $P(a)$.

Factor Theorem: $(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$.

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Method: Factorizing a Cubic Polynomial

Given a cubic polynomial $P(x)=a{x}^3+b{x}^2+cx+d$:

Step 1 — Find a zero by hit-and-trial. Test factors of the constant term $d$ (divided by factors of leading coefficient $a$). These are the only possible rational roots: $\text{Possible rational roots}=\pm \frac{\text{factors of }d}{\text{factors of }a}$

Step 2 — Verify using the Factor Theorem. If $P(k)=0$, then $(x-k)$ is a factor.

Step 3 — Perform synthetic division. Divide $P(x)$ by $(x-k)$ to obtain a quadratic quotient $Q(x)$: $P(x)=(x-k)\cdot Q(x)$

Step 4 — Factorize the quadratic. Factor $Q(x)$ by inspection, completing the square, or the quadratic formula: $Q(x)=(x-r)(x-s)$

Step 5 — Write the complete factorization. $P(x)=(x-k)(x-r)(x-s)$

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Worked Example

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

Step 1: Constant term $=-6$. Test $x=1$: $P(1)=1-6+11-6=0✓$

Step 2: By the Factor Theorem, $(x-1)$ is a factor.

Step 3: Synthetic division by $(x-1)$:

$\begin{array}{rrrrr}1& 1& -6& 11& -6\\ & & 1& -5& 6\\ & 1& -5& 6& 0\end{array}$

Quotient: $Q(x)=x^2-5x+6$

Step 4: Factorize $x^2-5x+6$: $x^2-5x+6=(x-2)(x-3)$

Step 5: Complete factorization: $P(x)=(x-1)(x-2)(x-3)$

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Synthetic Division — Quick Reference

To divide $P(x)=x^3+p{x}^2+qx+r$ by $(x-k)$:

1. Write coefficients: $1,p,q,r$
2. Bring down the leading coefficient.
3. Multiply by $k$, add to next coefficient; repeat.
4. Last entry = remainder (should be $0$ if $(x-k)$ is a factor).

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