# HCF and LCM Polynomials

To obtain the HCF of algebraic expression, take the common of all the prime factors of two polynomials. To obtain the LCM of algebraic expression, take the product of all its prime factors.

### Relation Between HCF and LCM of Polynomials

$P(x)×Q(x)=$ HCF $(P(x)$

and

$Q(x)×$ LCM $(P(x)$ and $Q(x))$

### LCM and HCF of Polynomials

Factorise:

$x_{2}−12x=35$

$(x−7)(x−5)$

Factories:

$x_{2}−8x+7$

$(x−7)(x−1)$

So HCF $=(x−7)$

LCM $=(x−7)(x−5)(x−1)$

### HCF of polynomials

Solution: Factorise all:

$p(x)=x_{2}−2x+2$ = $(x−1_{2}$

$q(x)=x_{4}−1$ = $(x−1)(x+1)(x_{2}+1)$

$r(x)=x_{3}−2x_{2}−5x+6$ = $x_{3}−x_{2}−x_{2}+x−6x+6=x_{2}(x−1)−x(x−1)−6(x−1)=(x−1)(x_{2}−x−6)=(x−1)(x−3)(x+2)$

Observe HCF of $p(x),q(x)=(x−1)$, $(−x−1),r(x)=(x−1)$. hence HCF of

$p(x),q(x),r(x)$ is $x−1$

### LCM of polynomials

**Solution:**

Factorizing $x_{2}y_{2}−x_{2}$ by taking the common factor $x_{2}$ we get,

$x_{2}(y_{2}−1)$

Now by using the identity $a_{2}−b_{2}$.

$x_{2}(y_{2}−12)$

= $x_{2}(y+1)(y−1)$

Also, factorizing $xy_{2}−2xy−3x$ by taking the common factor $x$ we get,

$x(y_{2}−2y−3)$

$=x(y_{2}−3y+y−3)$

$=x[y(y−3)+1(y−3)]$

$=x(y−3)(y+1)$

Therefore, the L.C.M. of $x_{2}y_{2}−x_{2}$ and $xy_{2}−2xy−3x$ is $x_{2}(y+1)(y−1)(y−3)$.