Question

Find the product of
(7y + 4y+) (3y2 + &r?) and verify the result for x =
x = 1, y = 2.​

Answer 1

Answer:

y=2 given

(7×2+4×2)(3×2×2+r)

(14+8)(12+r)

22(12+r)

264+22r

hope so it helps u mate

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Answer 2

Step-by-step explanation:

solution-

( 7xy + 4y^2) ( 3y^2+8x^2)

7xy ( 3y^2+8x^2) +4y^2( 3 y^2+ 8x^2)

= 21xy^3+ 56x^3 y+12y^4 +32x^2 y^2

now putting x= 1 , y= 2 in

( 7xy + 4y^2) ( 3y^2+8x^2)

( 7×1 ×2+4×2^2) ( 3×2^2+8×1^2)

=21 ×1×2^3+56×1^3×2+12×2^4+32×1^2×2^2

=30×20=168+112+192+128

=600=600

hence ,the product is correct .