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Title: Algebra _ G.C.D and L.C.M of Polynomials
Description: Algebra _ G.C.D and L.C.M of Polynomials 1. To find GCD or LCM of polynomials, first we have to factor the given polynomials. 2. The coefficients of the variables like x as much as possible. 3. If there is quadratic or cubic polynomial, then it has to be factored suitable algebraic identities. 4. To find GCD, multiply the common factors 5. To find LCM, multiply the factors with highest exponents.
Description: Algebra _ G.C.D and L.C.M of Polynomials 1. To find GCD or LCM of polynomials, first we have to factor the given polynomials. 2. The coefficients of the variables like x as much as possible. 3. If there is quadratic or cubic polynomial, then it has to be factored suitable algebraic identities. 4. To find GCD, multiply the common factors 5. To find LCM, multiply the factors with highest exponents.
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Algebra
G
...
D and L
...
M of Polynomials
I
...
C
...
4 ab2 , 2 a2b
Solution:
2
4,
2
2,
1
G
...
D = 2 ab
2
...
C
...
5 a3 b3, 15 a b c2
Solution:
5 5, 15
1, 3
G
...
D = 5ab
4
...
C
...
8 a2 x, 6 ab xy, 10 a b x3 y2
Solution:
2
8, 6, 10
4,
3,
G
...
D = 2 ax
5
6
...
C
...
Find the H
...
F of the following:
1
...
C
...
(2x 7) (3x + 4), (2x 7)2( x + 3)
Solution:
G
...
D = (2x 7)
3
...
C
...
(x + 4)2 (x 3)3, (x 1) (x + 4) (x 3)2
Solution:
G
...
D = (x + 4) (x 3)2
5
...
C
...
a2 + ab, a2 – b2
Solution:
a2 + ab
= a (a + b)
a2 – b2
= (a + b) (a b)
G
...
D
= (a + b)
7
...
C
...
x2 + 3x + 2, x2 – 4
Solution:
x2 + 3x + 2
= (x + 1) (x )
x2 – 4
= (x + 2) (x 2)
G
...
D
= (x + 2)
9
...
C
...
x2 + 6x + 5, x2 + 8x + 15
Solution:
x2 + 6x + 5 = (x + 1) (x + 5)
x2 + 8x + 15 = (x + 3)(x + 5)
G
...
D = x + 5
11
...
C
...
x2 17x + 66, x2 + 5x – 66
Solution:
x2 – 17x + 66 = (x – 11 ) (x – 6)
x2 + 5x 66 = (x + 11) ( x – 6)
G
...
D = ( x – 6)
13
...
C
...
12x2 + x – 1, 15x2 + 8x + 1
Solution:
12x2 + x – 1 = 12x2 + 4x – 3x 1
= 4x (3x + 1) – 1 (3x + 1) = (3x + 1) (4x – 1)
2
15x + 8x + 1 = 15x2 + 5x + 3x + 1
= 5x(3x + 1) + 1(3x + 1) = (3x + 1) (5x + 1)
G
...
D = (3x + 1)
15
...
C
...
(x – 4) (x2 11x + 30), (x6 5x5 + x – 5)
Solution:
(x – 4) (x2 11x + 30)
x6 5x5 + x – 5
G
...
D
=
=
=
=
(x – 4) (x – 5) (x 6)
x5 (x 5) + 1 (x – 5)
(x – 5) (x5 + 1)
x–5
17
...
C
...
x3 + 8x2 – x – 8, x3 + x2 – x 1
Solution:
x3 + 8x2 – x – 8
= x2(x + 8) – 1 (x + 8)
= (x + 8) (x2 – 1) = (x + 8) (x –1) (x + 1)
x3 + x2 – x 1
= x2 (x + 1) – (x + 1)
= (x + 1) (x2 – 1)
= (x + 1) (x + 1) (x – 1)
G
...
D
...
(x – 3)2, x2 – 9, x2 – x 6
Solution
(x – 3)2
= (x – 3 ) (x – 3)
x2 – 9
= (x + 3) (x 3)
x2 – x 6
= (x 3) ( x + 2)
G
...
D = (x – 3)
20
...
C
Title: Algebra _ G.C.D and L.C.M of Polynomials
Description: Algebra _ G.C.D and L.C.M of Polynomials 1. To find GCD or LCM of polynomials, first we have to factor the given polynomials. 2. The coefficients of the variables like x as much as possible. 3. If there is quadratic or cubic polynomial, then it has to be factored suitable algebraic identities. 4. To find GCD, multiply the common factors 5. To find LCM, multiply the factors with highest exponents.
Description: Algebra _ G.C.D and L.C.M of Polynomials 1. To find GCD or LCM of polynomials, first we have to factor the given polynomials. 2. The coefficients of the variables like x as much as possible. 3. If there is quadratic or cubic polynomial, then it has to be factored suitable algebraic identities. 4. To find GCD, multiply the common factors 5. To find LCM, multiply the factors with highest exponents.