x4+5x-3 doesn't factor rationally. It should have 4 zeroes (roots) but some may be complex.
If there are complex zeroes, they have to occur in pairs. By observation, when x=0 the polynomial has a value of -3. When x=1, the value is 1+5-3=3, so the value has gone from negative to positive, implying that there must be a zero between x=0 and x=1.
So how do we find these two zeroes? Let x4+5x-3=0 then the solutions for x are the roots of the equation. We can write this x4+5x=3, x(x3+5)=3, x=3/(x3+5). When x is between 0 and 1, x3<1. So we can use this equation iteratively (hopefully), starting with plugging in x=0 into the right-hand side:
x=3/(5)=⅗. Now we plug in x=⅗: x=3/(27/125+5)=375/652=0.5751... Keep repeating the process: x=0.57772 approx becomes a stable value, so it's a valid real zero.
We may need a different process to find another real zero. Let f(x)=x4+5x-3, then f(-1)=1-5-3=-7, and we know f(0)=-3. Both values are negative (no change of sign). f(-2)=16-10-3=3, which is positive so there's a real zero between -1 and -2. Take the average of these two: -1.5 and find f(-1.5)=-5.4375. This places the zero between f(-1.5)=-5.4375 and f(-2)=3. The average of -1.5 and -2=-1.75. f(-1.75)=-2.371... So the zero lies between -2 and -1.75. The average of these is -1.875. f(-1.875)=-0.01538... The zero lies between -2 and -1.875, the average being -1.9375. f(-1.9375)=1.404... This time the zero is between -1.9375 and -1.875, the average is -1.90625, and so on. This process continues until f(c), where c is some value, is as close to zero as necessary (a certain number of decimal places). In this case, x=-1.87572 approximately.
So we have two real zeroes: x=0.57772 and -1.87572. To find the other zeroes we need to reduce the polynomial to a quadratic by dividing it by the quadratic formed by the two discovered zeroes. This can be laborious and is likely to produce complex zeroes, which I think may be beyond the scope of this question.
The question was to factorise the polynomial, and we know that it can't be factorised easily, so there may be an error in the question.