find \\( \\frac{f(x)}{g(x)} \\) when \\( f(x)=2 x^{3}+7 x^{2}+2 x+9 \\) and \\( g(x)=2 x+3 \\).\n\\(…

find \\( \\frac{f(x)}{g(x)} \\) when \\( f(x)=2 x^{3}+7 x^{2}+2 x+9 \\) and \\( g(x)=2 x+3 \\).\n\\( \\frac{f(x)}{g(x)}= \\)

find \\( \\frac{f(x)}{g(x)} \\) when \\( f(x)=2 x^{3}+7 x^{2}+2 x+9 \\) and \\( g(x)=2 x+3 \\).\n\\( \\frac{f(x)}{g(x)}= \\)

Answer

Explanation:

Step1: Dividir (2x^{3}) por (2x)

$$\frac{2x^{3}}{2x}=x^{2}$$

Step2: Multiplicar (x^{2}) por (2x + 3)

$$x^{2}(2x + 3)=2x^{3}+3x^{2}$$

Step3: Restar (2x^{3}+3x^{2}) de (2x^{3}+7x^{2}+2x + 9)

$$(2x^{3}+7x^{2}+2x + 9)-(2x^{3}+3x^{2})=4x^{2}+2x + 9$$

Step4: Dividir (4x^{2}) por (2x)

$$\frac{4x^{2}}{2x}=2x$$

Step5: Multiplicar (2x) por (2x + 3)

$$2x(2x + 3)=4x^{2}+6x$$

Step6: Restar (4x^{2}+6x) de (4x^{2}+2x + 9)

$$(4x^{2}+2x + 9)-(4x^{2}+6x)=-4x + 9$$

Step7: Dividir (-4x) por (2x)

$$\frac{-4x}{2x}=-2$$

Step8: Multiplicar (-2) por (2x + 3)

$$-2(2x + 3)=-4x - 6$$

Step9: Restar (-4x - 6) de (-4x + 9)

$$(-4x + 9)-(-4x - 6)=15$$

Answer:

$$x^{2}+2x - 2+\frac{15}{2x + 3}$$