# Is it possible to factor a anything that has root x in it?

I'm doing some calculus homework and I got stuck on a question, but eventually figured it out on my own. My textbook doesn't have all the answers included (it only gives answers to even numbered questions for some reason). Anyways I got stuck when I needed to solve for x for this function.

$${\ -3x^3+8x-4{\sqrt{x}}-1=0}$$

I tried to factor it but I couldn't see a way to remove the radical. However, intuitively I could see it the answer to this question was just one, after a long time of confusion. Is there a possible way to factor this? Is there any way to solve this other than just looking at it and seeing the correct answer?

If you are curious here is the question in my textbook:

"Find the equation of the tangent line to the curve at the point (1,5)" $${y=(2-{\sqrt{x})}}{(1+{\sqrt{x}}+3x)}$$

Thank you for your time.

• There is likely no general way to factor/solve an equation like your first, as it is equivalent to solving a 6th degree polynomial, which is not possible in general, though it might be in this special case. – Mathily Mar 30 '17 at 21:41

## 3 Answers

To remove the radical: use a change of variable $x=t^2$. The equation becomes polynomial

$$-3t^6+8t^2-4t-1=0$$ and only positive roots can be accepted.

By inspection, $x=t=1$ is a root. You can deflate the polynomial, but for the other roots, you are stuck.

Your first equation is not the derivative of the second. $$y'=\frac{1}{2\sqrt{x}}+5-\frac32\sqrt{x}$$

If you chose to set this equal to 0, (which is not what is necessary for the question from the book as noted by A---B) $$0=\frac{1}{2\sqrt{x}}+5-\frac32\sqrt{x}$$ $$0=\frac12+5\sqrt{x}-\frac32 x$$ substitute $x=u^2$ for comfort, $$0=\frac12 +5u-\frac32 u^2$$ $$0=1+10u-3u^2$$ We can now use the quadratic formula to proceed.

• Sorry for being unclear, but the first equation was not my derivative. It was just some algebra I was doing on the side in my work. I just couldn't factor it but eventually saw the answer. – Peter Wang Mar 30 '17 at 22:06

Use $$l(x) = f^{'}(x_0)(x- x_0) + f(x_0)$$

Where $x_0$ is the abscissa of the point.

And $f$ is the function given.

$$f^{'}(x) = -{1\over 2\sqrt{x}}(1+ \sqrt{x} + 3x) + \left({1\over 2\sqrt{x}} + 3\right)(2- \sqrt{x})$$

$$f^{'}(1) = 1$$

$$f(1) = 5$$

So, $$l(x)= (x - 1) + 5 = x- 4$$

• The graph is weird. – A---B Mar 30 '17 at 21:45