find all zeros of f(x)=x^4-3x^3+5x^2-27x-36
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The clue is to look at the coefficients: 1-3+5-27-36. The odd powers of x have coefficients -3 and -27. Now we look at 36 and note that 1+3+5+27=36. When we add 36 to -36 we get zero, so.that means we need to make -3 and -27 into +3 and +27. We can do this if we put x=-1 into the cubic, so x=-1 is a root and (x+1) is a factor. If we divide by this root, or by the factor x+1, we will get a cubic. Use synthetic division to divide by the root:

-1 | 1 -3 5 -27 -36

 ......1 -1 4..-9...36

......1 -4 9 -36 | 0

The cubic is x^3-4x^2+9x-36. This time we can't get zero that easily. But there's something we can make use of. Rearrange the terms: x^3+9x-4x^2-36. Note that 4*9 is 36 and that means the constant term is 9 times the x^2 term; but the x term is 9 times the x^3 term: x(x^2+9)-4(x^2+9)=(x-4)(x^2+9)=0. So we have x=4 as a root and x^2=-9 will give us two imaginary roots: 3i and -3i. We now have all roots: real: -1 and 4; imaginary: 3i and -3i.

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