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Mathematics 20 Online
OpenStudy (anonymous):

factoring for n^2 - 15n + 56 i was thinking n - 8 and n -7

OpenStudy (anonymous):

(n-8)(n-7) You are correct.

OpenStudy (anonymous):

what about for q^2 - 8q + 12 i was thing ( q-4 ) (q-3) but also maybe (q-2) (q-6)

OpenStudy (anonymous):

thinking*

OpenStudy (anonymous):

It is (q-2)(q-6). Not sure if I can explain factoring to you in my own terms but this is because -4 - 3 does not equal 8. -2 - 6 = -8 Which satisfies the original equation.

OpenStudy (anonymous):

does not equal negative 8 ***

OpenStudy (anonymous):

thank you i understand :)

OpenStudy (anonymous):

You can always check your answers by multiplying across. (q-2)(q-6) q*q = q^2 q * -6 = -6q -2 * q = -2q -2*-6 = 12 = q^2 - 6q - 2q + 12 = q^2 - 8q + 12

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