Since the polynomial is divisible by `x^2-x+1` we can say;

`x^4+3x^3+ax+3 = (x^2-x+1)(px^2+qx+r)`

`x^4+3x^3+ax+3 = px^4+qx^3+rx^2-px^3-qx^2-rx+px^2+qx+r`

`x^4+3x^3+ax+3 = px^4+(q-p)x^3+(r-q+p)x^2+(q-r)+r`

by comparing coefficients;

`x^4 rarr 1 = p`

`x^3 rarr 3 = q-p` hence q = 4

`x^2 rarr 0 = r-q+p` hence r = 3

So we can say;

`px^2+qx+r...

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Since the polynomial is divisible by `x^2-x+1` we can say;

`x^4+3x^3+ax+3 = (x^2-x+1)(px^2+qx+r)`

`x^4+3x^3+ax+3 = px^4+qx^3+rx^2-px^3-qx^2-rx+px^2+qx+r`

`x^4+3x^3+ax+3 = px^4+(q-p)x^3+(r-q+p)x^2+(q-r)+r`

by comparing coefficients;

`x^4 rarr 1 = p`

`x^3 rarr 3 = q-p` hence q = 4

`x^2 rarr 0 = r-q+p` hence r = 3

So we can say;

`px^2+qx+r = x^2+4x+3 = (x+3)(x+1)`

So we can say the factorized answer is;

`p(x) = (x+3)(x+1)(x^2-x+1)`

For roots of p(x) = 0;

`(x+3)(x+1)(x^2-x+1) = 0`

`x = -3`

`x = -1`

`x^2-x+1 = 0`

Let `q(x) = x^2-x+1`

`Delta = 1-4xx1xx1 = -3`

So there is no real solutions for `x^2-x+1 = 0` .

*So real roots of p(x) is x = -3 and x = -1.*