Can someone please help?

Let's see. The roots of a polynomial means that when we plug in x, we essentially get p(x) = 0. 

Thus, setting the polynomial to 0, we get the equation $4x^3-11x^2+2x+3 = 0$

Now, we have to find factors. We could do this in two ways. We could first have

$-4x^{3}+3x^{2}+8x^{2}-2x-3=0\\ -4x^{3}+3x^{2}+8x^{2}-6x+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+8x^{2}-6x+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+2x\cdot \left(4x-3\right)+4x-3=0\\-x^{2}\cdot \left(4x-3\right)+2x\cdot \left(4x-3\right)+1\left(4x-3\right)=0\\-\left(4x-3\right)\cdot \left(x^{2}-2x-1\right)=0\\\left(4x-3\right)\cdot \left(x^{2}-2x-1\right)=0$

We could also test for factors. If we plug in 1, we get that $P(1) = -2$ and plugging in 0 gets $p(0) = 3$

Thus, the number must be in between 0 and 1. Testing 1/2, we get $P(1/2)=1.75$ so it must be in between 1/2 and 1. That's how we get 3/4. 

We can split these into two equations to find x. We have

$4x-3=0\\ x^{2}-2x-1=0$

Now, the second quadratic is now factorable over whole numbers, so we must use the quadratic equation. 

Using it, we get that 

$x=1+\sqrt{2}\\ x=1-\sqrt{2}$

From the first equation, we know x = 3/4 as well. 

Thus, our final 3 answers are

$x=\frac{3}{4}\\ x=1+\sqrt{2}\\ x=1-\sqrt{2}$

Thanks! :)