LCM and HCF of Algebric Expressions

**saberteeth** · October 6th 2009, 04:07 AM

,
What is the method to find the LCM and HCF of the following expressions?
$x^3 - 2x^2 - 13x -10$ and $x^3 - x^2 -10x - 8$

**ramiee2010** · October 6th 2009, 05:01 AM

$P(x)=x^3 - x^2 -10x - 8 \ and \ Q(x)=x^3 - 2x^2 - 13x -10$
$factors\ of\ P(x) =(x+1)(x+2)(x+4)$
$factors\ of\ Q(x) =(x+1)(x+2)(x-5)$
$common\ factors\ of\ P(x)\ and\ Q(x) = (x+1),(x+2),(x+1)(x+2)$

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

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

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

$Least\ common\ multiplier\ (LCM)\ of\ P(x)\ and\ Q(x)=(x+1)(x+2)(x+4)(x-5)$

**stapel** · October 6th 2009, 06:39 AM

Originally Posted by saberteeth

,
What is the method to find the LCM and HCF of the following expressions?
$x^3 - 2x^2 - 13x -10$ and $x^3 - x^2 -10x - 8$

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)
----------------------------------
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.

**saberteeth** · October 7th 2009, 02:59 AM

Originally Posted by ramiee2010

$P(x)=x^3 - x^2 -10x - 8 \ and \ Q(x)=x^3 - 2x^2 - 13x -10$
$factors\ of\ P(x) =(x+1)(x+2)(x+4)$
$factors\ of\ Q(x) =(x+1)(x+2)(x-5)$
$common\ factors\ of\ P(x)\ and\ Q(x) = (x+1),(x+2),(x+1)(x+2)$

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

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

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

$Least\ common\ multiplier\ (LCM)\ of\ P(x)\ and\ Q(x)=(x+1)(x+2)(x+4)(x-5)$

Thanks, i got it. But, how did you find the factors so accurately? What method did you use?

**stapel** · October 7th 2009, 07:47 AM

Originally Posted by saberteeth

...how did you find the factors so accurately? What method did you use?

To learn how to factor polynomials, try here.

**saberteeth** · October 7th 2009, 10:14 AM

Cool. Just what i needed.

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