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JulsSmile [24]
3 years ago
8

Let ​ f(x) = 4(x^2) - 3x

Mathematics
1 answer:
posledela3 years ago
7 0

Answer:

(f+g)(x)=5x²-4x+3

(f-g)(x)=3x²-2x+3

(fg)(x)=4x^4-7x^3+15x^2-9x

\frac{f}{g}(x) =\frac{4x^2-3x}{x^2-x+3}

Step-by-step explanation:

Given that,

f(x)=4x²-3x

g(x)=x²-x+3

(f+g)(x)

=f(x)+g(x)

=4x²-3x+x²-x+3

=(4x²+x²)+(-3x-x)+3   [ combined the like terms]

=5x²-4x+3

(f-g)(x)

=f(x)-g(x)

=4x²-3x-(x²-x+3)

=4x²-3x-x²+x-3

=(4x²-x²)+(-3x+x)-3   [ combined the like terms]

=3x²-2x+3

(fg)(x)

=f(x).g(x)

=(4x²-3x).(x²-x+3)

=4x²(x²-x+3)-3x(x²-x+3)

=4x^4-4x^3+12x^2-3x^3+3x^2-9x

=4x^4+(-4x^3-3x^3)+(12x^2+3x^2)-9x

=4x^4-7x^3+15x^2-9x

\frac{f}{g}(x)

=\frac{f(x)}{g(x)}

=\frac{4x^2-3x}{x^2-x+3}

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