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1. HCF(GCD), LCM, LCD of Polynomials examples
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1. Examples
1. Find HCF(GCD), LCM of `(x^2-y^2),(x^3-y^3)`
Solution:
1. Factor of `(x^2-y^2)`
`=(x^2-y^2)`
`=x^2-y^2`
Rewrite `x^2=(x)^2` and `y^2=(y)^2`
`=(x)^2-(y)^2`
Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=x` and `b=y`
`=(x-y)(x+y)`
= ( x - y)( x + y)
2. Factor of `(x^3-y^3)`
`=(x^3-y^3)`
`=x^3-y^3`
Rewrite `x^3=(x)^3` and `y^3=(y)^3`
`=(x)^3-(y)^3`
Here both terms are perfect cubes, so factor using the difference of cubes formula, `a^3-b^3=(a-b)(a^2+ab+b^2)` where `a=x` and `b=y`
`=(x-y)(x^2+xy+y^2)`
= ( x - y)( x2 + xy + y2)
GCD = ( x - y)
LCM = ( x - y)( x + y)( x2 + xy + y2)
2. Find HCF(GCD), LCM of `4x^2-25,8x^3-125`
Solution:
1. Factor of `4x^2-25`
`=4x^2-25`
Rewrite `4x^2=(2x)^2` and `25=(5)^2`
`=(2x)^2-(5)^2`
Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=2x` and `b=5`
`=(2x-5)(2x+5)`
= ( 2 x - 5)( 2 x + 5)
2. Factor of `8x^3-125`
`=8x^3-125`
Rewrite `8x^3=(2x)^3` and `125=(5)^3`
`=(2x)^3-(5)^3`
Here both terms are perfect cubes, so factor using the difference of cubes formula, `a^3-b^3=(a-b)(a^2+ab+b^2)` where `a=2x` and `b=5`
`=(2x-5)(4x^2+10x+25)`
= ( 2 x - 5)( 4 x2 + 10 x + 25)
GCD = ( 2 x - 5)
LCM = ( 2 x - 5)( 2 x + 5)( 4 x2 + 10 x + 25)
3. Find HCF(GCD), LCM of `(x^2-4),(x^2-5x+6)`
Solution:
1. Factor of `(x^2-4)`
`=(x^2-4)`
`=x^2-4`
Rewrite `x^2=(x)^2` and `4=(2)^2`
`=(x)^2-(2)^2`
Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=x` and `b=2`
`=(x-2)(x+2)`
= ( x - 2)( x + 2)
2. Factor of `(x^2-5x+6)`
`=(x^2-5x+6)`
`=x^2-5x+6`
Hint for pair of factors
The coefficients are `a=1,b=-5,c=6`
Find pair of factors whose product is `a*c=1 * 6=6` and sum is `b=-5`
The pairs of factors for 6 and their sums are
`1,6 : 1+6=7`
`2,3 : 2+3=5`
Factors are `-2,-3` whose product is `-2 * -3=6` and sum is `(-2)+(-3)=-5`
So use `-2` and `-3` to split the `-5` coefficient on the middle term
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)`
`=(x-2)(x-3)`
= ( x - 2)( x - 3)
GCD = ( x - 2)
LCM = ( x - 2)(x + 2)(x - 3)
4. Find HCF(GCD), LCM of `(x+3)(x+5)^2,(x+5)(x+7)^2,(x+5)(x+3)^2`
Solution:
1. Factor of `(x+3)(x+5)^2`
`=(x+3)(x+5)^2`
= ( x + 3)( x + 5)2
2. Factor of `(x+5)(x+7)^2`
`=(x+5)(x+7)^2`
= ( x + 5)( x + 7)2
3. Factor of `(x+5)(x+3)^2`
`=(x+5)(x+3)^2`
= ( x + 5)( x + 3)2
GCD = ( x + 5)
LCM = ( x + 3)2( x + 5)2( x + 7)2
5. Find HCF(GCD), LCM of `(x^2-4),(x^2-5x+6),(x^2+x-6)`
Solution:
1. Factor of `(x^2-4)`
`=(x^2-4)`
`=x^2-4`
Rewrite `x^2=(x)^2` and `4=(2)^2`
`=(x)^2-(2)^2`
Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=x` and `b=2`
`=(x-2)(x+2)`
= ( x - 2)( x + 2)
2. Factor of `(x^2-5x+6)`
`=(x^2-5x+6)`
`=x^2-5x+6`
Hint for pair of factors
The coefficients are `a=1,b=-5,c=6`
Find pair of factors whose product is `a*c=1 * 6=6` and sum is `b=-5`
The pairs of factors for 6 and their sums are
`1,6 : 1+6=7`
`2,3 : 2+3=5`
Factors are `-2,-3` whose product is `-2 * -3=6` and sum is `(-2)+(-3)=-5`
So use `-2` and `-3` to split the `-5` coefficient on the middle term
`=x^2-2x-3x+6`
`=x(x-2)-3(x-2)`
`=(x-2)(x-3)`
= ( x - 2)( x - 3)
3. Factor of `(x^2+x-6)`
`=(x^2+x-6)`
`=x^2+x-6`
Hint for pair of factors
The coefficients are `a=1,b=1,c=-6`
Find pair of factors whose product is `a*c=1 * -6=-6` and sum is `b=1`
The pairs of factors for 6 and their differences are
`1,6 : 6-1=5`
`2,3 : 3-2=1`
Factors are `-2,3` whose product is `-2 * 3=-6` and sum is `(-2)+3=1`
So use `-2` and `3` to split the `1` coefficient on the middle term
`=x^2-2x+3x-6`
`=x(x-2)+3(x-2)`
`=(x-2)(x+3)`
= ( x - 2)( x + 3)
GCD = ( x - 2)
LCM = ( x - 2)( x + 2)( x - 3)( x + 3)
6. Find HCF(GCD), LCM of `(2x^2-4x),(3x^4-12x^2),(2x^5-2x^4-4x^3)`
Solution:
1. Factor of `(2x^2-4x)`
`=(2x^2-4x)`
`=2x(x-2)`
= 2 x( x - 2)
2. Factor of `(3x^4-12x^2)`
`=(3x^4-12x^2)`
`=3x^2(x^2-4)`
Rewrite `x^2=(x)^2` and `4=(2)^2`
`=3x^2((x)^2-(2)^2)`
Here both terms are perfect squares, so factor using the difference of squares formula, `a^2-b^2=(a-b)(a+b)` where `a=x` and `b=2`
`=3x^2(x-2)(x+2)`
= 3 x2( x - 2)( x + 2)
3. Factor of `(2x^5-2x^4-4x^3)`
`=(2x^5-2x^4-4x^3)`
`=2x^3(x^2-x-2)`
Hint for pair of factors
The coefficients are `a=1,b=-1,c=-2`
Find pair of factors whose product is `a*c=1 * -2=-2` and sum is `b=-1`
The pairs of factors for 2 and their differences are
`1,2 : 2-1=1`
Factors are `1,-2` whose product is `1 * -2=-2` and sum is `1+(-2)=-1`
So use `1` and `-2` to split the `-1` coefficient on the middle term
`=2x^3(x^2+x-2x-2)`
`=2x^3(x(x+1)-2(x+1))`
`=2x^3(x+1)(x-2)`
= 2 x3( x + 1)( x - 2)
GCD = x( x - 2)
LCM = 2 × 3 x3( x - 2)( x + 2)( x + 1)
LCM `= 6x^3(x-2)(x+2)(x+1)`
This material is intended as a summary. Use your textbook for detail explanation.
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