Question
$x^4 - 2x^3 - 7x^2 + 8x + 12.$
✓
Answer
Let $f(x)=x^4-2 x^3-7 x^2+8 x+12$
The factors of constant term in $f(x)$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ and $\pm 12$.
We have,
$f(1)=1-2-7+8+12=12$
$\Rightarrow( x -1)$ is not a factor of $f ( x )$
$f(-1)=1+2-7-8+12=0$
$\Rightarrow(x+1)$ is a factor of $f(x)$
$f(2)=16-16-28-16+12=0$
$\Rightarrow( x -2)$ is a factor of $f ( x )$
$f(-2)=16+16-28-16+12=0$
$\Rightarrow(x+2)$ is a factor of $f(x)$
$f(3)=81-54-63+24+12=0$
$\Rightarrow(x-3)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree 4 . So, it cannot have more than 4 linear factors.
Thus, factors of $f(x)$ are $(x+1),(x-2),(x+2)$ and $(x-3)$.
Therefore,
$f(x)=k(x+1)(x+2)(x-2)(x-3)$
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
Putting $x=0$ on both sides, we get,
$12=k(1)(2)(-2)(-3)$
$12=12 k$
$k=1$
Substituting $k =1$ in (1), we get,
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
The factors of constant term in $f(x)$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$ and $\pm 12$.
We have,
$f(1)=1-2-7+8+12=12$
$\Rightarrow( x -1)$ is not a factor of $f ( x )$
$f(-1)=1+2-7-8+12=0$
$\Rightarrow(x+1)$ is a factor of $f(x)$
$f(2)=16-16-28-16+12=0$
$\Rightarrow( x -2)$ is a factor of $f ( x )$
$f(-2)=16+16-28-16+12=0$
$\Rightarrow(x+2)$ is a factor of $f(x)$
$f(3)=81-54-63+24+12=0$
$\Rightarrow(x-3)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree 4 . So, it cannot have more than 4 linear factors.
Thus, factors of $f(x)$ are $(x+1),(x-2),(x+2)$ and $(x-3)$.
Therefore,
$f(x)=k(x+1)(x+2)(x-2)(x-3)$
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
Putting $x=0$ on both sides, we get,
$12=k(1)(2)(-2)(-3)$
$12=12 k$
$k=1$
Substituting $k =1$ in (1), we get,
$x^4-2 x^3-7 x^2+8 x+12=k(x+1)(x+2)(x-2)(x-3)$
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