Question

14.
Factorize 2x² – 5 xy + 3y²} and use your result to
factorize 2(3a- b)²– 5 (3a-b) (2a-b)+3(2a-b)².
The factors are​

Answer 1

Answer:

ab

Step-by-step explanation:

Given an algebraic expression such that,

$2 {x}^{2} - 5xy + 3 {y}^{2}$

To factorise it,

We will use middle term splitting method.

Therefore, we will get,

$= 2 {x}^{2} - 2xy - 3xy + 3 {y}^{2} \\ \\ = 2x(x - y) - 3y(x - y) \\ \\ = (x - y)(2x - 3y)$

Now, another expression is such that,

$2 {(3a - b)}^{2} - 5(3a - b)(2a - b) + 3 {(2a - b)}^{2}$

Let's assume that,

• x = (3a-b)
• y = (2a-b)

Therefore, we have,

$= 2 {x}^{2} - 5xy + 3 {y}^{2}$

Factorising this as same as above is done,

Therefore, we will get,

= {(3a-b)-(2a-b)}{2(3a-b)-3(2a-b)}

= (3a-b-2a+b)(6a-2b-6a+3b)

= ab

Hence, the required value is ab.

Answer 2

$\bigstar$ Solution :

$\bigstar$ Question :

• Factorize 2x² – 5 xy + 3y² and use your result to  factorize

2(3a- b)²– 5 (3a-b) (2a-b)+3(2a-b)²

$\bigstar$ Explanation :

$\star$ Given algebraic expression is ,

$2x^2-5xy+3y^2$

$\star$ Now factorize the expression using the middle term splitting method.

$\implies 2x^2-2xy-3xy+3y^2\\\\\implies 2x(x-y)-3y(x-y)\\\\\implies (x-y)(2x-3y)$

$\star$ Now, another expression given is ,

$2(3a-b)^2-5(3a-b)(2a-b)+3(2a-b)^2$

$\star$ Let's assume that,

→ x = (3a-b)

→ y = (2a-b)

$\star$ Now sub. these in expression , we get ,

$\implies 2x^2-5xy+3y^2$

$\star$ We  already factorize the same expression above , so we get ,

$\\ \implies [(3a-b)-(2a-b)][{2(3a-b)-3(2a-b)}]\\\\ \implies (3a-b-2a+b)(6a-2b-6a+3b)\\\\ \implies (a).(b)\\\\\implies ab$

∴  The value is ab.

                                                                                         

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