What is the standard form of #y= (x-14)(x-2) #?

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Answer 1 · 2017-05-02T14:30:21

#y=x^2-16x+28#

To find the standard form from this form (factored form), we simply multiply the sets of brackets. If you're unsure how to do that, see this link

#y=(x-14)(x-2)#
#y=x^2-14x-2x+28#
Then collect the x terms, to get:
#y=x^2-16x+28#

Answer 2 · 2017-05-02T14:31:28

See the entire solution process below:

To get to the standard form of the equation you must multiply the two terms on the right side of the equation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#y = (color(red)(x) - color(red)(14))(color(blue)(x) - color(blue)(2))# becomes:

#y = (color(red)(x) xx color(blue)(x)) - (color(red)(x) xx color(blue)(2)) - (color(red)(14) xx color(blue)(x)) + (color(red)(14) xx color(blue)(2))#

#y = x^2 - 2x - 14x + 28#

We can now combine like terms:

#y = x^2 + (-2 - 14)x + 28#

#y = x^2 + (-16)x + 28#

#y = x^2 - 16x + 28#