Math, asked by rexvargas, 4 months ago

A. find the LCM of the following expressions.
1. \: 12 {}^{2}y {}^{3} and \: 15 {x}^{3} y
2. \:  \:  {x}^{2} - 7x + 6 \: and \:  {x}^{2}  - 1
Questions:
1. How did you get the lcm of the given?
2.What factoring techniques did you apply in item 2? ​

Answers

Answered by bson
0

Step-by-step explanation:

1. 12²y³ and 15x³y

2 x²-7x+6 and x²-1

1. 12²y³ = 3y×4×12× y²

15 x³y = 3y× 5x³

lcm = 3y ×4×12×y²×5x³ = 720 x³y³

2.x²-7x+6 = (x-1)(x-6)

x²-1 = (x+1)(x-1)

lcm = (x-1)(x-6)(x+1) = (x-6)(x²-1)

= x³-x-6x²+6=x³-6x²-x+6

Answered by Anonymous
4

Solution :

12 {y}^{3}  \:  \: and \:  \: 15 {x}^{3} y

Let's Split the Terms to prime factors

12 {y}^{3}  = 2 \times 3 \times 2 \times y \times y \times y \\ 15 {x}^{3}  = 5 \times 3 \times x \times x \times x \times y

Lets have the Common terms once and left terms in product form

 = 2 \times 3 \times 2 \times 5 \times x \times x \times x  \times y \times y \times y

 = 60 {x}^{3}  {y}^{3}

1. How did you get the LCM of the given ?

By splitting the terms into least prime factors and taking the common terms once and taking left terms as they were.

Solution :

 {x}^{2}  - 7x + 6 \:  \: and \:  \:  {x}^{2}  - 1

 {x}^{2}  - 7x + 6 =  {x}^{2}  - x - 6x - 6 \\ =  x(x - 1) - 6(x - 1) \\  = (x - 1)(x - 6)

 {x}^{2}  - 1 = ( {x}^{2}  -  {1}^{2} ) = ( {a }^{2} -  {b}^{2}   )

( {a}^{2}  -  {b}^{2} ) = (a + b)(a - b) = (x + 1)(x - 1)

 {x}^{2}  - 7x + 6 = (x - 6)(x - 1) \\  {x}^{2}  - 1 = (x + 1)(x - 1)

So the LCM is

(x - 1)(x + 1)(x + 6)

2. What factorising techniques you applied in terms 2 ?

For the first vale we applied quadratic equation technique and for the second value we have applied algebraic identity

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