Question

multiply (2x^2+3y) by (2x^2 - 3y ) and verify the result for x = 2 and y=3
please answer fast ⏩​

Answer 1

Answer:

multiplying  $(2x^2+3y)$ with $(2x^2-3y)$ and substituting x=2 and y=3 ; the solution is $-17$

Step-by-step explanation:

given that ;

we need to multiply $(2x^2+3y)$ with $(2x^2-3y)$ and verify the obtained result for $x =2$ and y=3.

so let's do the multiplication first:

⇒  $(2x^2+3y)$ x $(2x^2-3y)$

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

⇒  $4x^4 - 6x^2y + 6yx^2 - 9y^2$

⇒  $4x^4 - 9y^2$

the obtained result after multiplying  $(2x^2+3y)$ with $(2x^2-3y)$ is $4x^4 - 9y^2$

⇒ now, substitute the value of x and y for x=2 and y=3

⇒ $4x^4 - 9y^2$

⇒ $4(2^4) - 9(3^2)$

⇒  (4)(16) - (9)(9)

⇒  64 - 81

⇒ -17