Question

Without actual division, prove that 2x⁴ − 6x³ + 3x² + 3x − 2 is exactly divisible by x² − 3x + 2.

Answer 1

Let $f\left(x\right)=2x^4-6x^3+3x^2+3x-2\mathrm{and}g\left(x\right)=x^2-3x+2$.
$$\begin{gathered} g\left(x\right)=x^2-3x+2 \\ =x^2-2x-x+2 \\ =x\left(x-2\right)-1\left(x-2\right) \\ =\left(x-1\right)\left(x-2\right) \end{gathered}$$
Thus, f(x) is divisible by g(x) if f(1) = 0 and f(2) = 0. Now
$$\begin{gathered} f\left(1\right)=2{\left(1\right)}^4-6{\left(1\right)}^3+3{\left(1\right)}^2+3\left(1\right)-2=2-6+3+3-2=0 \\ f\left(2\right)=2{\left(2\right)}^4-6{\left(2\right)}^3+3{\left(2\right)}^2+3\left(2\right)-2=32-48+12+6-2=0 \end{gathered}$$
Hence, 2x⁴ − 6x³ + 3x² + 3x − 2 is exactly divisible by x² − 3x + 2.