Question
$x^4 - 2x^3 - 7x^2 + 8x + 12.$
✓
Answer
Let $f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$
The factors of constant term in $f(x)$ are $\pm1,\pm2,\pm3,\pm4,\pm6$ and $\pm12.$
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 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3) ...(1)$
Putting $x = 0$ on both sides, we get,
$12 = k(1)(2)(-2)(-3)$
$12 = 12k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$x^4 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3)$
The factors of constant term in $f(x)$ are $\pm1,\pm2,\pm3,\pm4,\pm6$ and $\pm12.$
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 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3) ...(1)$
Putting $x = 0$ on both sides, we get,
$12 = k(1)(2)(-2)(-3)$
$12 = 12k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$x^4 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3)$
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