Factorise \(x^{3}+6 x^{2}+6 x+5\) given that \(x+5\) is a factor.

We will set up the polynomial long division with the dividend, \(x^3 + 6x^2 + 6x + 5\), and the divisor, \(x + 5\). ``` __________________ x + 5 | x^3 + 6x^2 + 6x + 5 ```

Now, we will perform the long division. 1. Divide the first term of the dividend, \(x^3\), by the first term of the divisor, \(x\). The result is \(x^2\). 2. Multiply the divisor, \(x + 5\), by \(x^2\). The result is \(x^3 + 5x^2\). 3. Subtract the result from the dividend to find the remainder: \((x^3 + 6x^2) - (x^3 + 5x^2) = x^2\). 4. Bring down the next term of the dividend, \(6x\), so our new remainder is \(x^2 + 6x\). 5. Repeat steps 1-4 with the new remainder: divide \(x^2\) by \(x\), which gives \(x\), multiply the divisor by \(x\) to get \(x^2 + 5x\), and subtract it from the remainder to get \(x\). 6. Bring down the last term, \(5\), and repeat steps 1-4 again: divide \(x\) by \(x\), which gives \(1\), multiply the divisor by \(1\) to get \(x + 5\), and subtract it from the remainder to get \(0\). The long division should now look like this: ``` x^2 + x + 1 __________________ x + 5 | x^3 + 6x^2 + 6x + 5 - (x^3 + 5x^2) __________________ x^2 + 6x - (x^2 + 5x) __________________ x + 5 - (x + 5) __________________ 0 ```

Since we have found that the quotient is \(x^2 + x + 1\) and there is no remainder, we can write the polynomial as a product of its factors: \(x^3 + 6x^2 + 6x + 5 = (x + 5)(x^2 + x + 1)\). Thus, the factorisation of \(x^3 + 6x^2 + 6x + 5\) is \((x + 5)(x^2 + x + 1)\).