A and b are the variables cause they represent a number
5a2 + 6ab=a(5a+6b)
b(6a-8) if you are factoring
2a x 3b = 6ab
To factorize the expression 4ab - 6ab, you first need to identify the common factor between the two terms, which is 2ab. You can then factor out this common factor to rewrite the expression as 2ab(2 - 3). Therefore, the fully factorized form of 4ab - 6ab is 2ab(2 - 3) or simply -2ab.
6ab inches
The expression (5a^2 - 6ab) is a polynomial in two variables, (a) and (b). It consists of two terms: (5a^2), which is quadratic in (a), and (-6ab), which is a product of (a) and (b) with a coefficient of (-6). This polynomial cannot be simplified further without additional information about (a) and (b).
To multiply the expressions (4ab) and (6ab), you first multiply the coefficients (the numerical parts) and then multiply the variables. So, (4 \times 6 = 24), and for the variables, (a \times a = a^2) and (b \times b = b^2). Combining these, the product is (24a^2b^2).
2a x 3b = 6ab
The factors of 6Ab^2 are the numbers or variables that can be multiplied together to result in 6Ab^2. In this case, the factors of 6Ab^2 are 1, 2, 3, 6, A, B, A^2, B^2, AB, 2A, 3A, 6A, 2B, and 3B. These factors can be combined in various ways to represent the original expression 6Ab^2.
The expression ((6ab)^2) can be simplified by squaring both the coefficient and the variables inside the parentheses. Thus, ((6ab)^2 = 6^2 \cdot a^2 \cdot b^2 = 36a^2b^2).
5a2 + 6ab=a(5a+6b)
6ab-3b factorize = 3
3a2b is the simplest formImproved Answer:-3a x 2b = 6ab when simplified
0.3333
The expression (3a \times 2b) can be simplified by multiplying the coefficients and the variables separately. The coefficients 3 and 2 multiply to give 6, while the variables (a) and (b) remain as they are. Therefore, the simplified expression is (6ab).
No, 6ab and 4ba are not like terms. Like terms are terms that have the same variables raised to the same powers. In this case, the terms have the same variables, 'a' and 'b', but the order in which they appear is different. Therefore, they are not considered like terms in algebraic expressions.
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