Question
Using factor theorem, factorize the following polynomials:
$x^4 - 7x^3_+ 9x^2 + 7x - 10$
$x^4 - 7x^3_+ 9x^2 + 7x - 10$
✓
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,
$f(x)=k(x-1)(x+1)(x-2)(x-5)$
$x^4-7 x^3+9 x^2+7 x-10=k(x-1)(x+1)(x-2)(x-5)$
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,
$x^4-7 x^3+9 x^2+7 x-10=(x-1)(x+1)(x-2)(x-5)$
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,
$f(x)=k(x-1)(x+1)(x-2)(x-5)$
$x^4-7 x^3+9 x^2+7 x-10=k(x-1)(x+1)(x-2)(x-5)$
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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