Question
Using factor theorem, factorize the following polynomials: $x^4 - 7x^3_+ 9x^2 + 7x - 10$
✓
Answer
Let $f(x)=x^4-7 x^3+9 x^2+7 x-10$ The factors of constant term in $f ( x )$ are $\pm 1, \pm 2, \pm 5$ and $\pm 10$.
We have, $f(1)=1-7+9+7-10=0$
$\Rightarrow( x -1)$ is a factor of $f ( x ) f (-1)=1+7+9-7-10=0$
$\Rightarrow(x+1)$ is a factor of $f(x) f(2)=16-56+36+14-10=0$
$\Rightarrow( x -2)$ is a factor of $f ( x ) f (-2)=16+56-36-14-10=10$
$\Rightarrow(x+2)$ is not a factor of $f(x) f(5)=625-875+225+35-10=0$
$\Rightarrow(x-5)$ 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 $(x-1),(x+1),(x-2)$ and $(x-5)$.
Therefore, $f(x)=k(x-1)(x+1)(x-2)(x-5) x^4-7 x^3+9 x^2+7 x-10$ $=k(x-1)(x+1)(x-2)(x-5) \ldots(1)$
Putting $x=0$ on both sides,
we get, $-10=k(-1)(1)(-2)(-5)-10=-10 k k=1$
Substituting $k=1$ in $(1),$
we get, $x^4-7 x_{+}^3 9 x^2+7 x-10$
$=(x-1)(x+1)(x-2)(x-5)$
We have, $f(1)=1-7+9+7-10=0$
$\Rightarrow( x -1)$ is a factor of $f ( x ) f (-1)=1+7+9-7-10=0$
$\Rightarrow(x+1)$ is a factor of $f(x) f(2)=16-56+36+14-10=0$
$\Rightarrow( x -2)$ is a factor of $f ( x ) f (-2)=16+56-36-14-10=10$
$\Rightarrow(x+2)$ is not a factor of $f(x) f(5)=625-875+225+35-10=0$
$\Rightarrow(x-5)$ 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 $(x-1),(x+1),(x-2)$ and $(x-5)$.
Therefore, $f(x)=k(x-1)(x+1)(x-2)(x-5) x^4-7 x^3+9 x^2+7 x-10$ $=k(x-1)(x+1)(x-2)(x-5) \ldots(1)$
Putting $x=0$ on both sides,
we get, $-10=k(-1)(1)(-2)(-5)-10=-10 k k=1$
Substituting $k=1$ in $(1),$
we get, $x^4-7 x_{+}^3 9 x^2+7 x-10$
$=(x-1)(x+1)(x-2)(x-5)$
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