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6a2 + 5ab - 6b2 = (3a - 2b)(2a + 3b)

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2015-07-29 19:48:37
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How do you factor 6a2 plus 5ab - 6b2?
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Which is a factor of 6a2 5ab - 6b2?

(3a - 2b)(2a + 3b)


Which is a factor of 6a2 plus 5ab - 6b2?

Considering the minus sign between 5ab and 6b2 then we have the polynomial as 6a2 + 5ab - 6b2. The polynomial is a quadratic polynomial.Steps to factorize a quadratic polynomial:1 - Multiply first term by third term. 6a2 x (-6b2) = -36a2b22 - If possible break the second term into two terms such that they multiple to -36a2b2. If not then it is factorized by Sridharacharya's formula.5ab can be broken as 9ab + (-4ab).These two terms multiply to give -36a2b2.So we can write 6a2 + 5ab - 6b2 = 6a2 + 9ab + (-4ab) - 6b2.6a2 + 9ab - 4ab - 6b2 = 3a(2a + 3b) - 2b(2a + 3b) = (2a + 3b)(3a - 2b).So the factors are (2a + 3b) and (3a - 2b).


What is the factor of b3 plus 6b2?

b2(b + 6)


How do you factor 6ab - 6ac plus 6b2 - 6bc?

6(a + b)(b - c)


What is the factor polynomial 6b2-15b3?

-3b2(5b - 2)


How do you factor out 6b2-15b-6?

3(2b2 - 5b - 2)


How do you factor 6b2-7b-20?

(-3b - 4)(-2b + 5)


How do you factor the trinomial 6b2-13bs-63s2?

6b^2-13bs-63s^2 is factorised to (2b-9s)(3b+7s)


Solve for a a2-8a plus 12 a-2 -6b2 6 c -3 -4 d 3 4?

a2-8a plus 12, if done correctly according to calculus, results in the number 2.


Prove that square root of 6 is irrational number?

If a/b=sqrt(6), then a2=6b2 On the other hand, given integers ''a'' and ''b'', because the valuation (i.e., highest power of 2 dividing a number) of 6b2 is odd, while the valuation of a2 is even, they must be distinct integers. Contradiction.


What are all the factors of the monomial 6ab to the 2nd power?

They are 1, 2, 3, 4, 6, 9, 12, 18, 36,1a, 2a, 3a, 4a, 6a, 9a, 12a, 18a, 36a,1b, 2b, 3b, 4b, 6b, 9b, 12b, 18b, 36b,1a2, 2a2, 3a2, 4a2, 6a2, 9a2, 12a2, 18a2, 36a2,1ab, 2ab, 3ab, 4ab, 6ab, 9ab, 12ab, 18ab, 36ab,1b2, 2b2, 3b2, 4b2, 6b2, 9b2, 12b2, 18b2, 36b2.


Solve by factoring 6b2-13b 6 equals 0?

The question is ambiguous because there is no sign shown before the final 6. Assume it is +6 since -6 would make factorisation very difficult. Thus the equation is 6b2 - 13b + 6 = 0 Then 6b2 - 4b - 9b + 6 = 0 or 2b(3b - 2) - 3(3b - 2) = 0 ie (3b - 2)(2b - 3) = 0 then 3b -2 = 0 or 2b - 3 = 0 so 3b =2 or 2b = 3 ie b = 2/3 or b = 3/2

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