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It is "a squared minus eighteen b".
The expression is faulty because we don't have + or - before 10a. Also have to assume a2 is a2 . Use the standard formula for solving a quadratic = 0, to get two answers p and q. Then the factorization is (a-p)(a-q).
Assuming it's a2-1 you mean. If a2-1 is even, it can be expressed as a2-1 = 2 * t, where t is a natural number. a2-1 = (a - 1)(a + 1) = 2 * t You immediately see that at least one of the terms in parenthesis has to be even, and then, that actually both have to be even, because (a + 1) - (a - 1) = 2. Therefore, you can express both of the terms as: a - 1 = 2 * s a + 1 = 2 *r, which gives us: 2 * s * 2 * r = a2 - 1, simplify: 4 * r * s = a2 - 1. Therefore, if a2-1 is even, a2-1 is also divisible by 4.
2AB
2 a2 is a monomial, not a binomial but 2 + a2 is a binomial, so is 2 - a2 .
To evaluate the expression ( 10ba^2 \cdot b^6 \cdot a^2 ), first combine the like terms. The ( b ) terms can be combined as ( b^{1+6} = b^7 ), and the ( a ) terms as ( a^{2+2} = a^4 ). Thus, the expression simplifies to ( 10b^7a^4 ).
a2-5a-14 = (a-7)(a+2) when factored
It is "a squared minus eighteen b".
The expression ( a^2 + 2a^2 ) can be simplified by combining like terms. Since both terms contain ( a^2 ), you add their coefficients: ( 1 + 2 = 3 ). Therefore, ( a^2 + 2a^2 = 3a^2 ).
(A+C)-B
To simplify the expression (2a^2b + a^2 + 5ab + 3ab^2 + b^2 + 2(a^2b + 2ab)), first distribute the 2 in the last term to get (2a^2b + 4ab). Combining like terms, we group all terms with (a^2b), (ab), (ab^2), and constant terms. The final simplified expression is (3a^2b + 9ab + 3ab^2 + b^2).
(a - b + 2)(a + b + 2)
It is an expression and a term that are of equal value
First you would see is there is a Greatest Common Factor. Then you would factor the difference of squares binomial. After factoring out the GCF, the square root of the first term is 'a' and the square root of the second term is 'b'. Once 'a' and 'b' are found, substitute them into the binomial factors (a + b)(a - b).
The expression is faulty because we don't have + or - before 10a. Also have to assume a2 is a2 . Use the standard formula for solving a quadratic = 0, to get two answers p and q. Then the factorization is (a-p)(a-q).
a3 - b3 = (a - b)(a2 + ab + b2) a3 + b3 = (a + b)(a2 - ab + b2)
Assuming it's a2-1 you mean. If a2-1 is even, it can be expressed as a2-1 = 2 * t, where t is a natural number. a2-1 = (a - 1)(a + 1) = 2 * t You immediately see that at least one of the terms in parenthesis has to be even, and then, that actually both have to be even, because (a + 1) - (a - 1) = 2. Therefore, you can express both of the terms as: a - 1 = 2 * s a + 1 = 2 *r, which gives us: 2 * s * 2 * r = a2 - 1, simplify: 4 * r * s = a2 - 1. Therefore, if a2-1 is even, a2-1 is also divisible by 4.