Question

(3x + 2y)(3x + 2y) – (4x – 3y)2 + (2x +
+ (2x + 3y)(2x - 3y)
​

Answer 1

Step-by-step explanation:

Given:

$(3x + 2y)(3x + 2y)-(4x-3y)2 +(2x+3y)(2x-3y)$

To find:

The value of $(3x + 2y)(3x + 2y)-(4x-3y)2 +(2x+3y)(2x-3y)$ .

Solution:

Given that,

⇒ $(3x + 2y)(3x + 2y)-(4x-3y)2 +(2x+3y)(2x-3y)$

Here, multiply $(3x+2y)$ and $(3x+2y)$ to get $(3x+2y)^2.$

⇒ $(3x + 2y)^2-(4x-3y)2 +(2x+3y)(2x-3y)$

Now, use the binomial theorem $(a+b)^2=a^2+2ab+b^2$ to expand $(3x+2y)^2.$

⇒ $9x^2+12xy+4y^2-(4x-3y)2 +(2x+3y)(2x-3y)$

Now, use the distributive property to multiply $(4x-3y)$ by $2.$

⇒ $9x^2+12xy+4y^2-(8x-6y) +(2x+3y)(2x-3y)$

To find the opposite of $8x-6y$, find the opposite of each term.

⇒ $9x^2+12xy+4y^2-8x-(-6y) +(2x+3y)(2x-3y)$

Now, to the opposite of $-6y$ is $6y.$

⇒ $9x^2+12xy+4y^2-8x+6y+(2x+3y)(2x-3y)$

Here, consider $(2x+3y)(2x-3y).$ Multiplication can be transformed into the difference of squares using the rule: $(a-b)(a+b)=a^2-b^2$.

⇒ $9x^2+12xy+4y^2-8x+6y+(2x)^2-(3y)^2$

Expand $(2x^2).$

⇒ $9x^2+12xy+4y^2-8x+6y+2^2x^2-(3y)^2$

Here, calculate $2$ to the power of $2$ and get $4.$

⇒ $9x^2+12xy+4y^2-8x+6y+4x^2-(3y)^2$

Now, expand $(3y)^2.$

⇒ $9x^2+12xy+4y^2-8x+6y+4x^2-3^2y^2$

Here, calculate $3$ to the power of $2$ and get $9.$

⇒ $9x^2+12xy+4y^2-8x+6y+4x^2-9y^2$

Now, combine $9x^2$ and $4x^2$ to get $13x^2.$

⇒ $13x^2+12xy+4y^2-8x+6y-9y^2$

Again combine $4y^2$ and $-9y^2$ to get $-5y^2.$

⇒ $13x^2+12xy-5y^2-8x+6y$

Hence the answer is $13x^2+12xy-8x-5y^2+6y$.

Answer:

The value of $(3x + 2y)(3x + 2y)-(4x-3y)2 +(2x+3y)(2x-3y)$ is $13x^2+12xy-8x-5y^2+6y.$