Question

Factorise: (3x - 1/2 y + 1/3 z)^2​

Answer 1

Step-by-step explanation:

Given expression is {3x - (1/2)y + (1/3)z}²

⇛{3x + (-1/2)y + (1/3)z}²

Now,

This is in the form of (a+b+c)²

Here,

a = 3x, b = (-1/2)y and c = (1/3)z

So, using identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, we get

{3x + (-1/2)y + (1/3)z}²

= (3x)²+{-(1/2)y}² + {(1/3)z}² + 2(3x){-(1/2)y} + 2{-(1/2)y}{(1/3)z} + 2{(1/3)z}(3x)

= 3*3*x*x+ {(-1*-1/2*2)y*y} + {(1*1/3*3)z*z} + 2(3x){-(1/2)y} + 2{-(1/2)y}{(1/3)z} + 2{(1/3)z}(3x - (yz/3) + 2zx

= 9x² + (y²/4) + (z²/9) - 3xy - (yz/3) + 2zx

Take the LCM of the denominator 4, 3 and 9 is 36.

= {(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)/36}

= (1/35)(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)

Answer: Therefore, {3x - (1/2)y + (1/3)z}² = (1/35)(32x² + 9y² + 4z² - 108xy - 12yz + 72zx)

Please let me know if you have any other.

Answer 2

The expansion of the given expression is:

9x^2 + (1/4)y^2 + (1/9)z^2 - 3xy + 2xz - (1/3)yz

How to expand the expression?

Here we just need to use the simple rule:

(a + b + c)^2 = a^2 + b^2 + c^2 + 2*(ab + ac + bc).

(it is an expansion, the factorization is the inverse process actually)

Our expression is:

(3x - (1/2)y + (1/3)z)^2

Using the above rule we get:

= (3x)^2 + (-(1/2)y)^2 + ((1/3)z)^2 + 2*( (-3/2)xy + xz - (1/6)yz)

= 9x^2 + (1/4)y^2 + (1/9)z^2 - 3xy + 2xz - (1/3)yz

This is the expansion of the given expression.

If you want to learn more about factorization, you can read:

https://brainly.com/question/11579257