How do you factor the trinomial #x² - 8x + 7#?

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Answer 1 · 2015-11-28T18:41:07

#(x-7)(x-1) #
See explanation for the logic.

The only factors of 7 are {1,7} as 7 is prime.

If you add 1 and 7 you get 8

The sign of the constant (7) is positive so both the factors are either positive or negative.

The sum of these factors must be 8 and 8 is negative.

Thus both factors are negative

There is no coefficient of #x^2#, o,r you could argue that there is one but its value is 1

So we have:

#(x-1)(x-7)#

You could if you so wish write them the other way round. It would make no difference to the outcome.

#(x-7)(x-1) #

Expanding these gives: #color(blue)(x^2-x-7x+7)color(green)( = x^2-8x+7)#