What is the standard form of $y = (x - 3) (4 - x) + 3 x - 2$ $y= (x-3)(4-x)+3x-2$ ?

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Answer 1 · 2017-05-24T04:22:19

$- x^{2} + 10 x - 14$ $-x^2+10x-14$

Let's first do the multiplication of the brackets using FOIL, then add in the remaining terms:

FOIL

$F$ $color(red)(F)$ - First terms - $(a + b) (c + d)$ $(color(red)(a)+b)(color(red)(c)+d)$
$O$ $color(brown)(O)$ - Outside terms - $(a + b) (c + d)$ $(color(brown)(a)+b)(c+color(brown)d)$
$I$ $color(blue)(I)$ - Inside terms - $(a + b) (c + d)$ $(a+color(blue)b)(color(blue)(c)+d)$
$L$ $color(green)(L)$ - Last terms - $(a + b) (c + d)$ $(a+color(green)b)(c+color(green)d)$

and so $(x - 3) (4 - x)$ $(x-3)(4-x)$ becomes:

which adds to:

$4 x - x^{2} - 12 + 3 x = - x^{2} + 7 x - 12$ $4x-x^2-12+3x=-x^2+7x-12$

Now let's add in the remaining terms:

$(- x^{2} + 7 x - 12) + 3 x - 2 = - x^{2} + 10 x - 14$ $(-x^2+7x-12)+3x-2=-x^2+10x-14$

What is the standard form of y=(x−3)(4−x)+3x−2y= (x-3)(4-x)+3x-2?