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Factor 3ab plus 3ac plus 2b2 plus 2bc?
3a(b+c)+2b(b+c)
What is 2a divided by 6ab?
0.3333
How do you factorise 3ac-6ad-2bc plus 4bd?
To factorise the expression (3ac - 6ad - 2bc + 4bd), first group the terms: ((3ac - 6ad) + (-2bc + 4bd)). From the first group, factor out (3a), giving (3a(c - 2d)). From the second group, factor out (-2b), yielding (-2b(c - 2d)). Now, combine the factored terms: ((3a - 2b)(c - 2d)).
How do you factor a times a cubed plus 2 times a times b plus b squared?
the answer to factorising (a x a3 + 2ab + b2) would be (a4+2ab+b2)
What is 5a2-6ab?
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).
What is 3ac plus 2ca?
3ac + 2ca = 3ac + 2ac = (3 + 2)(ac) = 5ac
6ab plus 9b divided by 2a plus 3?
(6ab + 9b)/(2a + 3) = 3b(2a + 3)/(2a + 3) = 3b
5 plus 7 ac plus 3b plus 3ac?
5 + 10ac + 3b
What is the degree of this polynomial 5a2 plus 6ab-7b2?
2 is.
Factor 3ab plus 3ac plus 2b2 plus 2bc?
3a(b+c)+2b(b+c)
Would a x c x 3?
3ac
Factorize 3ab plus 3ac?
Factorizing 3ab + 3ac gives 3a (b + c).Factorizing is to express a number or expression as a product of factors.When factorizing always look for common factors. To factorize (3ab + 3ac) look for the highest factor between the two terms (3a). 3ab + 3ac = 3a (b + c)
Degree of a polynomial 5a2 plus 6ab minus 7b2?
This has a degree of 2.
A-3b plus 4 6ab?
6a^b-18ab^2+24ab
Find the difference (-2ab-2a-5)-(6ab plus 2)?
4ab - 2a - 7
How do you factor 12abx squared plus 6abx minus 30ab?
6ab(2x2+x-5)
How do you simplify 6a2 - 6ab plus 7a2?
To simplify the expression 6a^2 - 6ab + 7a^2, first combine like terms. Combine the terms with the same variable (a) raised to the same power. This results in 13a^2 - 6ab as the simplified expression. Remember to keep the terms in standard form with the variable term first, followed by any constant terms.
