(Hi),

What is the method to find the LCM and HCF of the following expressions?

$\displaystyle x^3 - 2x^2 - 13x -10$ and $\displaystyle x^3 - x^2 -10x - 8$

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- Oct 6th 2009, 04:07 AMsaberteethLCM and HCF of Algebric Expressions
(Hi),

What is the method to find the LCM and HCF of the following expressions?

$\displaystyle x^3 - 2x^2 - 13x -10$ and $\displaystyle x^3 - x^2 -10x - 8$ - Oct 6th 2009, 05:01 AMramiee2010
$\displaystyle P(x)=x^3 - x^2 -10x - 8 \ and \ Q(x)=x^3 - 2x^2 - 13x -10$

$\displaystyle factors\ of\ P(x) =(x+1)(x+2)(x+4)$

$\displaystyle factors\ of\ Q(x) =(x+1)(x+2)(x-5)$

$\displaystyle common\ factors\ of\ P(x)\ and\ Q(x) = (x+1),(x+2),(x+1)(x+2)$

Highest Common Factors (HCF) of P(x) and Q(x)$\displaystyle =(x+1)(x+2)$

$\displaystyle since \ (LCM)\ of \ P(x) \ and \ Q(x)= \frac{P(x) \times Q(x)}{HCF}$

$\displaystyle \ LCM =\frac{(x+1)(x+2)(x+4) \times (x+1)(x+2)(x-5)}{(x+1)(x+2)}$

$\displaystyle Least\ common\ multiplier\ (LCM)\ of\ P(x)\ and\ Q(x)=(x+1)(x+2)(x+4)(x-5)$ - Oct 6th 2009, 06:39 AMstapel
Factor each, and then

**make a table**with the factors:

[HTML]P(x): (x + 1)(x + 2)(x + 4)

Q(x): (x + 1)(x + 2) (x - 5)

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LCM: (x + 1)(x + 2)(x + 4)(x - 5)

HCF: (x + 1)(x + 2)[/HTML]

The LCM is the product of*all*the factors; the HCF (or GCF, for Americans) is the product of only the common factors. (Wink) - Oct 7th 2009, 02:59 AMsaberteeth
- Oct 7th 2009, 07:47 AMstapel
To learn how to factor polynomials, try

**here**. (Wink) - Oct 7th 2009, 10:14 AMsaberteeth
Cool. Just what i needed.