Question

Factorise the polynomial $x^4+x^3-7x^2-x+6$ using factor theorem.

Answer 1

Here we have $p\left(x\right)=x^4+x^3-7x^2-x+6$ (say)

Here we have constant term $6$ and coefficient of $x^4$ is $1$ ,

So, the factors of constant term i.e. $6$ are $\pm 1,\pm 2,\pm 3,\pm 6$

Case 1: Put $x=1$ and check if that satisfied our equation $x^4+x^3-7x^2-x+6$

$p\left(1\right)=(1{)}^4+(1{)}^3-7(1{)}^2-(1)+6$

$=1+1-7-1+6$

$=0$

So $(x-1)$ is a factor of the given equation p(x).

[Factor theorem, if $f\left(a\right)=0,$ then $(x-a)$ is a factor of polynomial $f\left(x\right)$ ]

Case 2: Put $x=-1$ and check if that satisfied our equation , $x^4+x^3-7x^2-x+6$

$p(-1)=(-1{)}^4+(-1{)}^3-7(-1{)}^2-(-1)+6$

$=1-1-7+1+6$

$=0$

So $(x+1)$ is a factor of the given equation p(x).

Case 3: Put $x=2$ and check if that satisfied our equation , $x^4+x^3-7x^2-x+6$

$p\left(2\right)=(2{)}^4+(2{)}^3-7(2{)}^2-(2)+6$

$=16+8-28-2+6$

$=0$

So $(x-2)$ is a factor of the given equation p(x).

Case 4: Put $x=-2$ and check if that satisfied our equation , $x^4+x^3-7x^2-x+6$

$p(-2)=(-2{)}^4+(-2{)}^3-7(-2{)}^2-(-2)+6$

$=16-8-28+2+6$

$=-12$

$\ne 0$

So $(x+2)$ is not a factor of the given equation p(x).

Case 5: Put $x=-3$ and check if that satisfied our equation , $x^4+x^3-7x^2-x+6$

$p(-2)=(-3{)}^4+(-3{)}^3-7(-3{)}^2-(-3)+6$

$=81-27-63+3+6$

$=0$

So $(x+3)$ is a factor of the given equation p(x).

So we have, the factors of $x^4+x^3-7x^2-x+6$ are $(x-1)(x+1)(x-2)(x+3)$