Question

Let ​ f(x) = 4(x^2) - 3x

Answer 1

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)

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

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

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

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

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

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