Question

Using factor theorem, factorize the polynomial $x^4+x^3-7x^2-x+6$.
Write down the factors.

• A. (x-1) (x-1) (x-2) (x-3)
• B. (x-1) (x+1) (x-2) (x+3)
• C. (x+1) (x+1) (x+2) (x+3)
• D. (x-1) (x+1) (x+2) (x+3)

Answer 1

The correct option is B (x-1) (x+1) (x-2) (x+3)

Let $f\left(x\right)=x^4+x^3-7x^2-x+6$
the factors of constant term in $f\left(x\right)$ are $\pm 1,\pm 2,\pm 3\text{and}\pm 6$
Now,
$f\left(1\right)=1+1-7-1+6=8-8=0$
$\Rightarrow (x-1)\text{is a factor of f(x)}$
$f(-1)=1-1-7+1+6=8-8=0$
$\Rightarrow x+1\text{is a factor of f(x)}$
$f\left(2\right)=2^4+2^3-7\times 2^2-2+6$
$=16+8-28-2+6=0$
$\Rightarrow x+2\text{is a factor of f(x)}$
$f(-2)=(-2{)}^4+(-2{)}^3-7(-2{)}^2-(-2)+6$
$=16-8-28+2+6=-12\ne 0$
$\Rightarrow x+2\text{is not a factor of f(x)}$
$f\left(3\right)=(-3{)}^4+(-3{)}^3-7(-3{)}^2-(-3)+6$
$81-27-63+3+6=90-90=0$
$\Rightarrow x+3\text{is a factor of}f\left(x\right)$
Since $f\left(x\right)$ is a polynomial of degree $4$. So it cannot have more than $4$ linear factors
Thus, the factors of $f\left(x\right)$ are $(x-1),(x+1),$
$(x-2)$ and $(x+3)$.
Let $f\left(x\right)=k(x-1)(x+1)(x-2)(x+3)$
$\Rightarrow x^4+x^3-7x^2-x+6$
$=k(x-1)(x+1)(x-2)(x+3)$
Putting $x=0$ on both sides, we get
$6=k(-1)\left(1\right)(-2)\left(3\right)\Rightarrow 6=6k\Rightarrow k=1$
Substituting $k=1$ in (i), We get
$x^4+x^3-7x^2-x+6=(x-1)(x+1)(x-2)(x+3)$