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Answer:

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Step-by-step explanation:

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Answer:  " [tex]3x^{2} -4x +7[/tex] " .

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Step-by-step explanation:

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We are to find the following sum:

  Add:  [tex]f(x) + g(x)[/tex]  .

Given:  [tex]f(x)= 8-4x +4x^{2}[/tex]  ;

Rewrite as follows:

             [tex]f(x) =x^{2}-4x+8[/tex] ;

{Note:  Since:  [tex]f(x) = +8 + (-4x) + x^{2}[/tex] ;  we can rearrange the terms;

   and rewrite by starting in the order of the highest degree monomial

(i.e., the term among the 3 (three) monomials containing the variable with the highest exponential value—which is:  "[tex]x^{2}[/tex] " ; and continue in descending order; with   " [tex](-4x^{1})[/tex]" ;  which is:  " [tex]-4x[/tex] " ; [with the 'implied exponent of "1" ; since any value, raised to the exponent, "1" ; ["first power"]; resulting in that same value.   Then, we finish with the last monomial, " [tex](+8x^0)[/tex] ";  which is:  "(+8x) " —with the implied exponent of "0" ; since any non-zero value; raised to the "0th" exponent ['raised to the power of zero']; results in the value of "1" .  

       →  As such:  "  +8x  =  +8x⁰ " ;

                      ↔ "  +8x⁰ = (+8) * (x⁰) = (+8) * (1) =  " (+8) ;

→  {since any value, multiplied by "1" ; results in that same value;  this refers to the "identity property" of multiplication.}.

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So, plug in:  " [tex]x^{2}-4x+8[/tex] " ;  for:  " [tex]f(x)[/tex] " ;

And plug in the given:  " [tex]2x^{2} + 5x -1[/tex] " ;  for " [tex]g(x)[/tex] " ;

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And add the sum:

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→        [tex]f(x) +g(x) = (x^{2}-4x+8)+(2x^2+5x-1)[/tex]  ;

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                            [tex]= x^{2} -4x+8+2x^{2} -1[/tex] ;

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         →  Now, combine the "like terms" ; and simplify:

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         →   [tex]+x^{2} +2x^{2} =+3x^{2}[/tex] ;

         →   [tex]-4x[/tex]  ;  stands alone;

         →   [tex]+8 -1 = + 7[/tex] ;

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→  Now; we have accounted for All of the terms in our expression;

and we can write out our answer:

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   →  which is:  " [tex]3x^{2} -4x +7[/tex] " .

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Hopefully—this answer and explanation is helpful to you.

  Wishing you the best in your academic pursuits!

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