Respuesta :

mberisso

Answer:

[tex]a^2+3\,a\,b-5\,a\,b-15\,b^2=(a-5\,b)\,(a+3\,b)[/tex]

Step-by-step explanation:

Work via factoring by groups:

!) re arrange the terms as follows:

[tex]a^2-5ab+3ab-15b^2[/tex]

then extract the common factor for the first two terms (a), and separately the common factors for the last two terms (3 b):

[tex]a^2-5ab+3ab-15b^2\\a\,(a-5\,b)+3\,b\,(a-5\,b)[/tex]

Now notice that the binomial factor (a-5 b) is in both expressions, so extract it:

[tex]a\,(a-5\,b)+3\,b\,(a-5\,b)\\(a-5\,b)\,(a+3\,b)[/tex]

which is the final factorization.

Cathenna

Answer:

[tex] \boxed{\sf (a + 3b)(a - 5b)} [/tex]

Step-by-step explanation:

[tex] \sf Factor \: the \: following: \\ \sf \implies {a}^{2} + 3ab - 5ab - 15 {b}^{2} \\ \\ \sf Grouping \: like \: terms, \\ \sf {a}^{2} + 3ab - 5ab - 15 {b}^{2} = {a}^{2} + (3ab - 5ab) - 15 {b}^{2} : \\ \sf \implies {a}^{2} + (3ab - 5ab) - 15 {b}^{2} \\ \\ \sf 3ab - 5ab = - 2ab : \\ \sf \implies {a}^{2} - 2ab - 15 {b}^{2} \\ \\ \sf The \: factors \: of \: - 15 \: that \: sum \: to \: - 2 \: are \: 3 \: and \: - 5. \\ \\ \sf So, \\ \sf \implies {a}^{2} + (3 - 5)ab - 15 {b}^{2} \\ \\ \sf \implies {a}^{2} + 3ab - 5ab - 15 {b}^{2} \\ \\ \sf \implies a(a + 3b) - 5b(a + 3b) \\ \\ \sf \implies (a + 3b)(a - 5b)[/tex]