Find the product of the ff expression
(2x+y)(3x²+2xy+5y²)
Answers
Kuch milte julte question:
Add the following:
(i) 3x and 7x
(ii) −5xy and 9xy
ANSWER:
We have
(i) 3x + 7x = (3 + 7)x = 10x
(ii) -5xy + 9xy = ( -5 + 9)xy = 4xy
Page No 7.13:
Question 2:
Simplify each of the following:
(i) 7x3y + 9yx3
(ii) 12a2b + 3ba2
ANSWER:
Simplifying the given expressions, we have
(i) 7x3y + 9yx3 = (7 + 9)x3y = 16x3y
(ii) 12a2b + 3ba2 = (12 + 3)a2b = 15a2b
Page No 7.13:
Question 3:
Add the following:
(i) 7abc, −5abc, 9abc, −8abc
(ii) 2x2y, − 4x2y, 6x2y, −5x2y
ANSWER:
Adding the given terms, we have
(i) 7abc + (- 5abc) + (9 abc) + (- 8abc)
= 7abc - 5abc + 9abc - 8abc
= (7 - 5 + 9 - 8)abc
= (16 - 13)abc
= 3abc
(ii) 2x2y + (- 4x2y) + 6x2y + (- 5x2y)
= 2x2y - 4x2y + 6x2y - 5x2y
= (2 - 4 + 6 - 5)x2y
= (8 - 9)x2y
= -x2y
Answer:
using 2x to multiply the second bracket
I.e (2x)(3x2 +2xy+5y2)
=(6x3 +4x2y +10xy2)
using (y) to multiply the second bracket also
= (3x2y +2xy2 +5y3)
bringing the two bracket together
= 6x3 +5y3 + (10xy2 + 2xy2) +(4x2y +3x2y)
= 6x3 + 5y3 + 12xy2 +7x2y