Divide 2x^4 - x^3 - 5x^2 + 2x + 2 by x - 1.

Step-by-step solution

1. 1

   Choose synthetic division and find the right test value. The divisor $x-1$ is linear and monic, so synthetic division applies. Set $x-1=0$ to get the value $x=1$ — note the sign: the divisor $x-1$ uses $+1$, not $-1$.

2. 2

   Write the coefficients in order, including any zeros. The dividend has every degree present, so the row is

   $2,\ -1,\ -5,\ 2,\ 2$

   (If a power were missing, a $0$ would have to be inserted in its place — omitting it is the classic synthetic-division error.)

3. 3

   Bring down and repeat: multiply by 1, add to the next coefficient.

   $2;\quad -1+2=1;\quad -5+1=-4;\quad 2+(-4)=-2;\quad 2+(-2)=0$

   Because the test value is $1$, each multiplication step is trivial and the process is just a running sum of the coefficients.

4. 4

   Read off the quotient and remainder. The first four numbers are the quotient coefficients, one degree lower than the dividend, and the last is the remainder:

   $\text{quotient}=2x^{3}+x^{2}-4x-2,\qquad \text{remainder}=0$

5. 5

   Interpret the zero remainder. By the factor theorem, remainder $0$ means $x=1$ is a root of the dividend and $(x-1)$ is a genuine factor:

   $2x^{4}-x^{3}-5x^{2}+2x+2=(x-1)\left(2x^{3}+x^{2}-4x-2\right)$

   Indeed the coefficients of the dividend sum to $2-1-5+2+2=0$, which is exactly the shortcut test for $x=1$.

6. 6

   Check by multiplying back and go one step further. Expanding $(x-1)\left(2x^{3}+x^{2}-4x-2\right)$ reproduces the original quartic ✓. The cubic factors again by grouping: $x^{2}(2x+1)-2(2x+1)=(2x+1)\left(x^{2}-2\right)$, so the full factorisation is $(x-1)(2x+1)\left(x^{2}-2\right)$ and the roots are $1,\ -\tfrac12,\ \pm\sqrt2$.