GCD and LCM with factorisation

Example — GCD and LCM with factorisation

Calculate the GCD and LCM of $P_1 = 2x^3+3x^2-2x-3$ and $P_2 = x^4+2x^2-3$.

Factorisation of $P_1$ (partial grouping):

2x^3+3x^2-2x-3 &= x^2(2x+3) - (2x+3) \\ &= (x^2-1)(2x+3) = \boxed{(x-1)(x+1)(2x+3)} \end{aligned}$$ **Factorisation of $P_2$** (substitution $t=x^2$): $$\begin{aligned} x^4+2x^2-3 &= t^2+2t-3 = (t-1)(t+3) \\ &= (x^2-1)(x^2+3) = \boxed{(x-1)(x+1)(x^2+3)} \end{aligned}$$

Observation

$x^2+3$ is irreducible in $\mathbb{R}$: it is always positive, so it has no real roots.

Comparing the factors of the two polynomials:

	$(x-1)$	$(x+1)$	$(2x+3)$	$(x^2+3)$
$P_1$	✓	✓	✓
$P_2$	✓	✓		✓
GCD	✓	✓
LCM	✓	✓	✓	✓

The GCD collects the common factors, the LCM all the factors: $\text{MCD} = \boxed{(x-1)(x+1)}, \qquad \text{mcm} = \boxed{(x-1)(x+1)(2x+3)(x^2+3)}$

Links

Topics: Algebraic calculation
Concepts: GCD · LCM · Irreducible polynomial · Factorisation
Skills: Calculating · Factorising