How do you factor #- 6t - 16+ t ^ { 2}#?

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Answer 1 · 2018-05-17T20:26:30

#(t-8)(t+2)#

We can rewrite this as #t^2-6t-16#

We are looking for two numbers that sum to make #-6# and multiply to make #-16#

Naturally, these are #-8# and #2#

So we factor this as:

#t^2-8t+2t-16#

#t(t-8)+2(t-8)#

#(t-8)(t+2)#

Answer 2 · 2018-05-17T20:27:06

take #t^2-6t-16#

find factors of the c constant, -16, which add up to -6.

this gives us -8 and 2, since -8+2 is -6.

replace #-6t# with the 2 values, such that:

#t^2-8t+2t-16#

thus factorizing just the first 2 terms, and the next 2 terms give you identical expressions inside the parenthesizes.

thus giving

#t(t-8)+2(t-8)#

the things not in the brackets can be collected up to give
#(t+2)(t-8)#