Find the products
2a3(3a + 5b)
To find the product, we will use distributive law as follows:
Find the products
-11a (3a + 2b)
To find the product, we will use distributive law as follows:.
Find the products
- 5a(7a – 2b)
To find the product, we will use distributive law as follows:
Find the products
-11y2(3y + 7)
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
xy (x3- y3)
To find the product, we will use distributive law as follows:
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0.1y (0.1 x5 + 0.1y)
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
1.5 x(10x2y – 100xy2)
To find the product, we will use distributive law as follows:
Find the products
4.1 xy(1.1x – y)
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the products
To find the product, we will use distributive law as follows:
Find the product 24x2(1-2x) and evaluate its value for x = 3.
To find the product, we will use distributive law as follows:
Find the product -3y(xy + y2) and find its value for x = 4 and y = 5.
To find the product, we will use distributive law as follows:
Multiply – (3/2)x2y3 by (2x – y) and verify the answer for x = 1 and y = 2.
To find the product, we will use distributive law as follows:
Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.05:
(i) 15y2 (2 – 3x)
(ii) −3x (y2 + z2)
(iii) Z2(x – y)
(iv) xz(x2 + y2)
Simplify:
(i) 2x2(x3 – x) –3x(x4+ 2x) -2(x4-3x2)
(ii) x3y (x2 – 2x) + 2xy(x3– x4)
(iii) 3a2 + 2(a + 2) -3a(2a + 1)
(iv) x(x + 4) + 3x(2x2–1) + 4x2 + 4
(v) a(b – c) – b(c – a) – c(a – b)
(vi) a(b – c) + b(c – a) + c(a – b)
(vii) 4ab(a–b) – 6a2(b – b2) –3b2(2a2 – a) + 2ab(b – a)
(viii) x2(x2 + 1) –x3(x + 1) –x(x3– x)
(ix) 2a2+ 3a(1- 2a3) + a(a + 1)
(x) a2(2a – 1) + 3a + a3 – 8
(xii) a2b(a – b2) + ab2(4ab – 2a2) – a3b(1 – 2b)
(xiii) a2b(a3 – a + 1) – ab(a4 –2a2 + 2a) – b(a3 – a2 – 1)