5.2 Q-6

Exercise 5.2 — Question 6

This question applies the Factor Theorem and Remainder Theorem to factorize or analyze a cubic polynomial.

Key Theorems

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

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

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.
2. Confirm using the Factor Theorem: If $P(k)=0$, then $(x-k)$ is a factor.
3. Divide $P(x)$ by $(x-k)$ using synthetic division or long division to obtain a quadratic quotient $Q(x)$.
4. Factorize the quadratic $Q(x)$ by factoring, completing the square, or the quadratic formula.
5. Write the complete factorization: $P(x)=(x-k)\cdot Q(x)$.

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Worked Example

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

Step 1 — Hit and trial. Factors of $-6$: $\pm 1,\pm 2,\pm 3,\pm 6$.

$P(1)=1-6+11-6=0✓$

So $(x-1)$ is a factor.

Step 2 — Synthetic division by $(x-1)$:

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

Quotient: $x^2-5x+6$.

Step 3 — Factorize the quadratic:

$x^2-5x+6=(x-2)(x-3)$

Step 4 — Complete factorization:

$P(x)=(x-1)(x-2)(x-3)$

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