Question

Using factor theorem, factorize : x cube + 2 x sq. - x - 2

Answer 1

Step-by-step explanation:

f(x)=x³+2x²-x-2

f(1)=(1)³+2(1)²-(1)-2=1+2-1-2=0

f(1)=0 then x-1 is a factor

f(-2)=(-2)³+2(-2)²-(-2)+2=-8+8+2-2=0

f(-2)=0 then x+2 is a factor

f(-1)=(-1)³+2(-1)²-(-1)-2=-1+2+1-2=0

f(-1)=0 then x+1 is a factor

So (x-1),(x+1),(x+2) are factors

Factorisation--> (x-1)(x+1)(x+2)

Answer 2

Step-by-step explanation:

Given : A polynomial $x^{3} +2x^{2} -x-2$.

To find : Factors of given polynomial by using Factor theorem.

• Solving equation for first root of equation,

Let us assume,

$f(x)=x^{3} +2x^{2} -x-2$

By using factor theorem we have to find value of x where $f(x)=0$.

$f(x)=x^{3} +2x^{2} -x-2$

Finding $f(1)$ where the $x=1$

$f(1)=(1)^{3} +2*(1)^{2} -1-2$

      $=1+2-1-2$

      $=0$

We get a root of polynomial $x=1$ where it is zero and the factor of equation is $(x-1)$ .

• Calculating other two roots

Now dividing $x^{3} +2x^{2} -x-2$ by $(x-1)$.

By this division we get quotient is $x^{2} +3x+2$ and remainder is zero. (we can see in attached picture).

quotient is a quadratic equation so we can equate it with 0,

$x^{2} +3x+2=0$

we can split 3 as  $2+1$ whose multiplication is 2.

$x^{2} +2x+x+2=0$

$x(x+2)+1(x+2)$

$(x+1)(x+2)$

equating both factors with 0 separately,

$x+1=0$

$x=-1$

$x+2=0$

$x=-2$

so the roots of polynomial are $x=1,-1,-2$.

Hence factors of given polynomial are $(x-1)$ , $(x+1)$, $(x+2)$.