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Parin Sir > Quantitative Aptitude (Maths) > H.C.F. & L.C.M. Of Polynomials > Examples > Examples

Examples

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(1) Find  the  H.C.F. and L.C.M. of 

(i)    8x2 + 2x – 21   and  10x2 – 11x – 6     (ii)    8 ( x3 + x2 + x )  and  28 ( x5 – x2 )

Soln. (i)

8x2 + 2x – 21

=  8x2+14x – 12x– 21

=  (2x – 3) (4x + 7)

10x2 – 11x – 6

=  10x2  –15x +4x –6

=  (2x – 3) (5x + 2)

\therefore  H.C.F.  =   (2x – 3)       and      L.C.M.  =  (2x – 3) (4x + 7) (5x + 2)

(ii)

8 ( x3 + x2 + x )

=  23x ( x2 + x + 1 )

28 (x5 – x2) =   28 x2 (x3 – 1 )

=   22 \times 7 x2 (x – 1 ) ( x2 + x + 1 )

\therefore  H.C.F.  =   22 \times ( x2 + x + 1 )      and      L.C.M.  =  23 \times 7 x2 (x – 1 ) ( x2 + x + 1 )

(2) Find the H.C.F.  and   L.C.M.  of   m4 – n4   and   m6 – n6 .

Soln. m4 – n4

=  ( m2 – n2) ( m2 + n2)

=  (m – n) (m + n) (m2 + n2)

m6 – n6

=  ( m3 – n3) ( m3 + n3)

=  ( m – n ) (m2 + mn + n2) ( m + n ) (m2 – mn + n2)

\therefore H.C.F.   =    ( m – n ) ( m + n )     and

L.C.M.   =   (m – n) (m + n) (m2 + n2) (m2 + mn + n2) (m2 – mn + n2)

(3) If  ( x – 3 ) is the H.C.F. of  ( 2x2 – 3x – 9 )  &  ( 4x2 – kx – 3 ), then  find k.

Soln. Clearly, ( 2x2 – 3x – 9 )  &  ( 4x2 – kx – 3 )  are divisible by  ( x – 3 )

So, x  =  3  will make each polynomial zero.

\therefore  For  ( 4x2 – kx – 3 ),

4 ( 3 ) 2 – k ( 3 )  – 3  =  0

\therefore  36 – 3k – 3  =  0

\therefore  k  =  11

Two  Important  Properties  :

◙          L.C.M.  of   two  polynomials   =    \frac{\textup{ Product of polynomials}}{\textup{Their H.C.F.}}

◙          For any two polynomials  p ( x )  and  q ( x ),  we have :

             p ( x ) \times q ( x )  =  ( Their  H.C.F. ) y \times ( Their  L.C.M. ).

(5) The   G.C.D. (H.C.F.)  of   two polynomials   is  x + 2  and  their  L.C.M.  is   x3  + 10x2 + 31x + 30.  If  one of  the polynomials is  x2 + 7x + 10, find the other.

Soln.

Other  polynomial   =   \frac{\textup{H.C.F.}\times \textup{L.C.M.}}{\textup{Given Polynomial}}

=   \frac{(x+2) \left (x^{3}+10x^{2}+31x+30 \right )}{\left (x^{2}+7x+10 \right )}

=  \frac{(x+2) \left (x^{3}+10x^{2}+31x+30 \right )}{(x+2)(x+5)}

=  \frac{(x+2) (x+5)(x^{2}+5x+6)}{(x+2)(x+5)}

=    x^{2}+5x+6

(6) Find the L.C.M. of  ( 2x2 – 18 ),  ( 4x2 – 4x – 24 )  and  (x3 – 7x – 6 ).

Soln. 2x2 – 18

=   2 ( x2 – 9 )

=   2 ( x  – 3 ) ( x  + 3 )

4x2 –  4x – 24

=  4 ( x2 –  x – 6 )

=  2( x  – 3 )( x  + 2 )

x3 – 7x – 6

=  x3 – 3x2 + 3x2 – 9x + 2x – 6

=  x2 ( x – 3 ) + 3x ( x – 3 ) + 2 ( x – 3)

=  ( x – 3 ) (x2  + 3x + 2 )

=  ( x – 3 ) ( x + 2 ) ( x + 1 )

   H.C.F. =  ( x  – 3 )  and   L.C.M. =  22( x  – 3 ) ( x  + 3 ) ( x  + 2 ) ( x + 1 )

 

Best of Luck

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