Exercise 5.2 — Question 4

Concepts Covered

This question applies polynomial division, the Remainder Theorem, and the Factor Theorem to factorize cubic polynomials.

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Key Theorems

Remainder Theorem

If a polynomial $P(x)$ is divided by a linear polynomial $(x-a)$, the remainder is $P(a)$.

$P(x)=(x-a)\cdot Q(x)+P(a)$

Factor Theorem

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

For a zero of the form $x=-\frac{p}{q}$, the corresponding factor is $(qx+p)$.

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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. Hit and Trial: Test factors of the constant term $d$ (divided by factors of leading coefficient $a$) as potential zeros.
2. Verify: Confirm $P(a)=0$ using the Remainder Theorem.
3. Synthetic/Long Division: Divide $P(x)$ by $(x-a)$ to obtain a quadratic quotient $Q(x)$.
4. Factorize the Quotient: Factor $Q(x)$ by inspection, splitting the middle term, or the quadratic formula.
5. Write the Full Factorization: $P(x)=(x-a)\cdot Q(x)$.

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Worked Example

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

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

Test $x=1$: $P(1)=1-6+11-6=0✓$

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

Step 4 — Full Factorization: $P(x)=(x-1)(x-2)(x-3)$

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Flashcards

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