Class 7
nd
Semester: 2
Subject: Math
Chapter 11
Algebraic Products
Page: 197
Algebraic Products
Q. No: 25
(2x + 1)(1 + 3x)
= 2x(1 + 3x) + 1(1 + 3x)
= 2x + 6x² + 1
=
+ 3x
6x² + 5x + 1
Q. No: 26
(5x + 2)(2 - x)
= 5x(2 - x) + 2(2 - x)
10x - 5x² + 4 - 2x
= -5x² + 8x + 4
Q. No: 27
(6x - 1)(3 - x)
= 6x(3 - x) - 1(3 - x)
=
18x - 6x² -
3
+ x
= -6x² + 19x - 3
Q. No: 28
(5a - 2)(3 - 7a)
= 5a(3 - 7a) - 2(3 - 7a)
=
15a - 35a²
- 6 + 14a
= -35a² + 29a - 6
Q. No: 29
(3x + 2)(4 - x)
= 3x(4 - x) + 2(4 - x)
=
12x - 3x²
+ 8 - 2x
= -3x² + 10x + 8
Q. No: 30
(4x - 5)(3 + x)
=
4x(3 + x) - 5(3 + x)
= 12x + 4x²
=
- 15 -
4x² + 7x - 15
5x
Q. No: 31
(5x + 2)(4 + 3x)
=
5x(4 + 3x) + 2(4 + 3x)
= 20x + 15x² + 8 + 6x
= 15x² + 26x + 8
Q. No: 32
(7x + 4)(3 - 2x)
=
7x(3 - 2x) + 4(3 - 2x)
= 21x - 14x²
+ 12 - 8x
= -14x² + 13x + 12
Q. No: 33
(4x - 3)(3 - 5x)
=
4x(3 - 5x) - 3(3 - 5x)
= 12x - 20x² - 9 + 15x
= -20x² + 27x - 9
Q. No: 34
(3 - p)(4 + p)
=
3(4 + p) - p(4 + p)
= 12 + 3p - 4p - p²
= -p² - p + 12
Q. No: 35
(x - 5)(2 + x)
=
x(2 + x) - 5(2 + x)
= 2x + x² - 10 - 5x
= x² - 3x - 10
Solve
(z + 4)(z - 10)
= z(z - 10) + 4(z - 10)
=
z² - 10z + 4z - 40
= z² - 6z - 40