Question
Factorise: $2 x^3-3 x^2-17 x+30$
✓
Answer
Let $f(x)=2 x^3-3 x^2-17 x+30$ be the given polynomial.
The factors of the constant term $+30$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$.
The factor of coefficient of $x^3$ is $2 .$
Hence, possible rational roots of $f(x)$ are: $\pm 1, \pm 3, \pm 5, \pm 15, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}$
We have $f(2)=2(2)^3-3(2)^2-17(2)+30$
$=2(8)-3(4)-17(2)+30$
$=16-12-34+30=0$
And $f(-3)=2(-3)^3-3(-3)^2-17(-3)+30$
$=2(-27)-3(9)-17(-3)+30$
$=-54-27+51+30=0$
So, $(x-2)$ and $(x+3)$ are factors of $f(x)$.
$\Rightarrow x ^2+ x -6$ is a factor of $f ( x )$.
Let us noe divide $f(x)=2 x^3-3 x^2-17 x+30$ by $x^2+x-6$ to get the other factors of $f(x)$.
Factors of $f ( x )$.
By long division, we have
$\therefore 2 x^3-3 x^2-17 x+30=\left(x^2+x-6\right)(2 x-5)$
$\Rightarrow 2 x^3-3 x^2-17 x+30=(x-2)(x+3)(2 x-5)$
Hence, $2 x^3-3 x^2-17 x+30=(x-2)(x+3)(2 x-5)$
The factors of the constant term $+30$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$.
The factor of coefficient of $x^3$ is $2 .$
Hence, possible rational roots of $f(x)$ are: $\pm 1, \pm 3, \pm 5, \pm 15, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}$
We have $f(2)=2(2)^3-3(2)^2-17(2)+30$
$=2(8)-3(4)-17(2)+30$
$=16-12-34+30=0$
And $f(-3)=2(-3)^3-3(-3)^2-17(-3)+30$
$=2(-27)-3(9)-17(-3)+30$
$=-54-27+51+30=0$
So, $(x-2)$ and $(x+3)$ are factors of $f(x)$.
$\Rightarrow x ^2+ x -6$ is a factor of $f ( x )$.
Let us noe divide $f(x)=2 x^3-3 x^2-17 x+30$ by $x^2+x-6$ to get the other factors of $f(x)$.
Factors of $f ( x )$.
By long division, we have
$\therefore 2 x^3-3 x^2-17 x+30=\left(x^2+x-6\right)(2 x-5)$
$\Rightarrow 2 x^3-3 x^2-17 x+30=(x-2)(x+3)(2 x-5)$
Hence, $2 x^3-3 x^2-17 x+30=(x-2)(x+3)(2 x-5)$
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