Question
Without actual division, prove that $2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2.$
✓
Answer
Let $f(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$
$g(x) = x^2 - 3x + 2$
$= x^2 - 2x - x + 2$
$= x(x - 2) - 1(x - 2)$
$= (x - 2)(x - 1)$
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 $g(x).$
$g(x) = x^2 - 3x + 2$
$= x^2 - 2x - x + 2$
$= x(x - 2) - 1(x - 2)$
$= (x - 2)(x - 1)$
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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Class interval (km/ h)
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Frequency
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30-40
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50-60
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