I have to complete a factoring packet for AP Calculus, and I'm having trouble with three of the questions...
Find the missing factor:
1. $2\sqrt{x} + 6x^\frac 32 = 2\sqrt{x}$(_____________)
I was thinking that the missing factor would be 1+3x? I'm not sure.
2. $\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2 +1}}=\frac{1}{\sqrt{x^2+1}}$ (__________)
3. $(2x+1)^\frac 32\left(x^\frac 12\right)+(2x +1)^\frac 52(x^\frac {-1}{2})=(2x+1)^\frac 32(x^\frac {-1}{2})$(______________)
On
HINTS: I assume the question is to solve the equations.
You have $2\sqrt{x}+6x\sqrt{x}=2\sqrt{x}$. Subtract $2\sqrt{x}$ and then factor out $\sqrt{x}$
$\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2 +1}}=\frac{1}{\sqrt{x^2+1}}$ Multiply by $\sqrt{x^2+1}$
$(2x+1)^{\frac{3}{2}} \left( x^\frac{1}{2}\right)+(2x +1)^\frac{5}{2}(x^\frac{-1}{2})=(2x+1)^\frac{3}{2}(x^\frac{-1}{2})$ Subtact $(2x +1)^\frac{5}{2}x^\frac{-1}{2}$ and then factor out $(2x +1)^\frac{5}{2}x^\frac{-1}{2}$.
1. Let's see if the missing factor is $1+3x$:
$$2\sqrt{x}+6x^{\frac{3}{2}}=2\sqrt{x}(1+3x)=2 \sqrt{x}+6x \sqrt{x}=2 \sqrt{x}+6x^{1+\frac{1}{2}}=2 \sqrt{x}+6x^{\frac{3}{2}}$$
So, it is correct!
At the first term, you could multiply and divide by $\sqrt{x^2+1}$:
$$\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2 +1}}=\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2 +1}}=\frac{1}{\sqrt{x^2+1}} \left ( \left (\sqrt{x^2+1} \right )^2-x^2\right )=\frac{1}{\sqrt{x^2+1}} \left ( x^2+1-x^2\right )=\frac{1}{\sqrt{x^2+1}} $$
3. $$(2x+1)^\frac 32\left(x^\frac 12\right)+(2x +1)^\frac{5}{2}(x^\frac{-1}{2})$$
Use that $(2x +1)^\frac{5}{2}=(2x +1)^\frac{3}{2}(2x +1)^\frac{2}{2}=(2x +1)^\frac{3}{2}(2x +1)$ and $x^{\frac{1}{2}}=x^{-\frac{1}{2}}x$
Can you continue?