how do yu factor 2x^4-x^3+7x^2-4x-4

Question

Have to factor it to find zeroes

Answer 1

if you add coefficient of all the terms, you will get 0.so x-1 is a factor.

Answer 2

There are positive and negative terms, and the sum of the positive coefficients exactly matches the sum of the negative coefficients: 2+7 and -(1+4+4); this means x=1 is a zero, and x-1 is a factor. We can divide by this root quickly using synthetic division:

1 | 2 -1 7 -4..-4

.....2..2 1..8...4

.....2..1 8..4 | 0

This leaves the cubic: 2x^3+x^2+8x+4. Here we note that there is a common factor between the x^3 and x terms, and the x^2 and constant, because 8/2 is the same as 4/1 and we can see this by rearranging the terms: (2x^3+8x)+(x^2+4)=2x(x^2+4)+(x^2+4)=(2x+1)(x^2+4). This doesn't factorise further into real factors, so we have (x-1)(2x+1)(x^2+4) as the complete factorisation.

Or is it? x^2+4 factorises if we allow complex zeroes involving the imaginary square root of -1: x^2=-4, so x=+2i, where i=sqrt(-1). So the complete answer showing all the zeroes is:

x=1, -1/2, 2i, -2i are the four zeroes. Factorisation is (x-1)(2x+1)(x+2i)(x-2i).