Question
Without actual division, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2.$
✓
Answer
Let $f(x)=2 x^4-5 x^3+2 x^2-x+2$
$g(x)=x^2-3 x+2=x^2-2 x-x+2=x(x-2)-1(x-2)=(x-2)(x-1)$
Clearly, $(x-2)$ and $(x-1)$ are factors of $g(x)$.
In order to prove that $f(x)$ is exactly divisible by $g(x)$,
it is sufficient to prove that $f(x)$ is exactly divisible by $(x-2)$ and $(x-1)$.
Thus, we will show that $(x-2)$ and $(x-1)$ are factors of $f(x)$.
Now, $f(2)=2(2)^4-5(2)^3+2(2)^2-2+2=32-40+8=0$ and $f(1)=2(1)^4-5(1)^3+2(1)^2-1+2=2-5+2-1+2=0$
Therefore, $(x-2)$ and $(x-1)$ are factors of $f(x)$.
$\Rightarrow g(x)=(x-2)(x-1)$ is a factor of $f(x)$.
Hence, $f(x)$ is exactly divisible by $g(x)$.
$g(x)=x^2-3 x+2=x^2-2 x-x+2=x(x-2)-1(x-2)=(x-2)(x-1)$
Clearly, $(x-2)$ and $(x-1)$ are factors of $g(x)$.
In order to prove that $f(x)$ is exactly divisible by $g(x)$,
it is sufficient to prove that $f(x)$ is exactly divisible by $(x-2)$ and $(x-1)$.
Thus, we will show that $(x-2)$ and $(x-1)$ are factors of $f(x)$.
Now, $f(2)=2(2)^4-5(2)^3+2(2)^2-2+2=32-40+8=0$ and $f(1)=2(1)^4-5(1)^3+2(1)^2-1+2=2-5+2-1+2=0$
Therefore, $(x-2)$ and $(x-1)$ are factors of $f(x)$.
$\Rightarrow g(x)=(x-2)(x-1)$ is a factor of $f(x)$.
Hence, $f(x)$ is exactly divisible by $g(x)$.
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