Mathematics: F(x) = x^3 + 6x^2 + x1/2 and g(x)= x^1/2. find f(x) ÷ g(x) a) x^5/2 + 6x^3/2-1 b) (x^5+6x^3 +1)1/2 c) x5/2 +6x^3/2 + 1 d) x^3/2 + 6x+ x1/4 e) x3/2 - 6x + x1/4

F(x) = x^3 + 6x^2 + x1/2 and g(x)= x^1/2. find f(x)

glogaming16

31.12.2021

$\frac{f(x)}{g(x)}=x^{\frac{5}{2}} +6 x^{\frac{3}{2}} +1}$

Step-by-step explanation:

$f(x)=x^3+6x^2+\sqrt{x}$

$g(x)= \sqrt{x}$

we are asked to determine

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

Let us do it step by step.

$f(x)=x^3+6x^2+\sqrt{x}$

$f(x)=x^2 \times x+6x \times x +\sqrt{x}$

$f(x)=x^2 \times \sqrt{x} \times \sqrt{x} +6x \times \sqrt{x} \times \sqrt{x} +\sqrt{x}$

Taking $\sqrt{x}$ as GCF

$f(x)= \sqrt{x}(x^2 \times \sqrt{x} +6x \times \sqrt{x} +1)$

Hence

$\frac{f(x)}{g(x)}=\frac{\sqrt{x}(x^2 \times \sqrt{x} +6x \times \sqrt{x} +1)}{ \sqrt{x}}$

$\frac{f(x)}{g(x)}=x^2 \times \sqrt{x} +6x \times \sqrt{x} +1$

$\sqrt{x}=x^\frac{1}{2}$

$\frac{f(x)}{g(x)}=x^2 \times x^{\frac{1}{2}} +6 \times x \times x^{\frac{1}{2}} +1$

Using law of exponents

$a^m \times a^n = a^{m+n}$

$\frac{f(x)}{g(x)}=x^{2+\frac{1}{2}} +6 \times x^{1+\frac{1}{2}} +1}$

$\frac{f(x)}{g(x)}=x^{\frac{5}{2}} +6 x^{\frac{3}{2}} +1}$

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Anne4912

25.03.2021

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Meeting code

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Step-by-step explanation:

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