How do you simplify $(3 x^{2} + 2 x + 1) + (x - 8)$ $(3x^2 + 2x + 1) + (x-8)$ ?

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How do you simplify $(3 x^{2} + 2 x + 1) + (x - 8)$ $(3x^2 + 2x + 1) + (x-8)$ ?

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Answer 1 · 2017-07-08T13:52:56

$f (x) = 3 x^{2} + 2 x + 1 + (x - 8)$ $f(x) = 3x^2 + 2x + 1 + (x - 8)$
$f (x) = 3 x^{2} + 3 x - 7$ $f(x) = 3x^2 + 3x - 7$
$D = b^{2} - 4 a c = 9 + 84 = 93$ $D = b^2 - 4ac = 9 + 84 = 93$
Because D isn't a perfect square, so, this trinomial can't be factored, and, F(x) can't be simplified further.

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How do you simplify $(3 x^{2} + 2 x + 1) + (x - 8)$ $(3x^2 + 2x + 1) + (x-8)$ ?

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