Question
Using factor theorem, factorize the polynomial: $x^3-6 x^2+3 x+10$
✓
Answer
Let, $f(x)=x^3-6 x^2+3 x+10$
The constant term in $f(x)$ is $10$
The factors of $10$ are $\pm 1, \pm 2, \pm 5, \pm 10$
Let, $x+1=0$
$\Rightarrow x=-1$
Substitute the value of $x$ in $f(x)$
$f(-1)=(-1)^3-6(-1)^2+3(-1)+10$
$=-1-6-3+10$
$=0$
Similarly, $(x – 2)$ and $(x – 5)$ are other factors of $f(x)$
Since, $f(x)$ is a polynomial having a degree $3,$ it cannot have more than three linear factors.
$\therefore f(x)=k(x+1)(x-2)(x-5)$
Substitute $x = 0$ on both sides
$\Rightarrow x ^3-6 x ^2+3 x +10= k ( x +1)( x -2)( x -5)$
$\Rightarrow 0-0+0+10= k (1)(-2)(-5)$
$\Rightarrow 10= k (10)$
$\Rightarrow k =1$
Substitute $k=1$ in $f(x)=k(x+1)(x-2)(x-5)$
$f(x)=(1)(x+1)(x-2)(x-5)$
so, $x^3-6 x^2+3 x+10=(x+1)(x-2)(x-5)$
This is the required factorisation of $f(x)$
The constant term in $f(x)$ is $10$
The factors of $10$ are $\pm 1, \pm 2, \pm 5, \pm 10$
Let, $x+1=0$
$\Rightarrow x=-1$
Substitute the value of $x$ in $f(x)$
$f(-1)=(-1)^3-6(-1)^2+3(-1)+10$
$=-1-6-3+10$
$=0$
Similarly, $(x – 2)$ and $(x – 5)$ are other factors of $f(x)$
Since, $f(x)$ is a polynomial having a degree $3,$ it cannot have more than three linear factors.
$\therefore f(x)=k(x+1)(x-2)(x-5)$
Substitute $x = 0$ on both sides
$\Rightarrow x ^3-6 x ^2+3 x +10= k ( x +1)( x -2)( x -5)$
$\Rightarrow 0-0+0+10= k (1)(-2)(-5)$
$\Rightarrow 10= k (10)$
$\Rightarrow k =1$
Substitute $k=1$ in $f(x)=k(x+1)(x-2)(x-5)$
$f(x)=(1)(x+1)(x-2)(x-5)$
so, $x^3-6 x^2+3 x+10=(x+1)(x-2)(x-5)$
This is the required factorisation of $f(x)$
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