Find the factor of ${x^3} + {x^2} - x - 1?$
Answer
583.5k+ views
Hint: If a polynomial is of degree $n,$ then it should have upto $n + 1$ terms. In example above, the degree of the polynomial is $3$ and has $4$ terms. Also, it may have terms less than \[n + 1\]. But in case of missing terms, it is difficult to factorise.
Complete step by step solution:
Now, We have to find factors of:
$ \Rightarrow {x^3} + {x^2} - x - 1$
By taking negative sign common from last two terms, we get;_
$ \Rightarrow {x^3} + {x^2} - \left( {x + 1} \right)$
Taking \[\left( {{x^2}} \right)\]common from first two terms,
We get:-
$ \Rightarrow {x^2}\left( {x + 1} \right) - \left( {x + 1} \right)$
Now, as we know, \[\left( {x + 1} \right)\] is common, thus, by taking it common,
We get:-
$ \Rightarrow \left( {x + 1} \right)\left( {{x^2} - 1} \right)$
Here, we use the property
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
We get:-
$ \Rightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)$ $\left\{ {\because \left( {{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right)} \right\}$
Finally we get the result as:-
$ \Rightarrow {\left( {x + 1} \right)^2}\left( {x - 1} \right)$
By factoring the equation we complete it:-
${\left( {x + 1} \right)^2}\left( {x - 1} \right)$
$\therefore $ Factor of given polynomials are $\left( {x + 1} \right),\left( {x + 1} \right)$ and $\left( {x - 1} \right)$
$\because $ Degree of polynomial is $3$ therefore it must have $3$ factors for sure.
And in the above example, two of the factors are repeating i.e. $\left( {x + 1} \right)$ is a repeating factor.
Note: In the above example, by finding value of $f\left( 1 \right)$ we get, $f\left( 1 \right) = {1^3} + {1^2} - 1 - 1 = 0$
$\because f\left( 1 \right) = 0$ $\therefore \left( {x - 1} \right)$ is one of the factors.
Now, by dividing $\left( {{x^3} + {x^2} - x - 1} \right)$ by $\left( {x - 1} \right)$ we get a quadratic equation, which we can solve by formula method.
As there is no formula for finding factors of cubic or higher degree polynomials.
Complete step by step solution:
Now, We have to find factors of:
$ \Rightarrow {x^3} + {x^2} - x - 1$
By taking negative sign common from last two terms, we get;_
$ \Rightarrow {x^3} + {x^2} - \left( {x + 1} \right)$
Taking \[\left( {{x^2}} \right)\]common from first two terms,
We get:-
$ \Rightarrow {x^2}\left( {x + 1} \right) - \left( {x + 1} \right)$
Now, as we know, \[\left( {x + 1} \right)\] is common, thus, by taking it common,
We get:-
$ \Rightarrow \left( {x + 1} \right)\left( {{x^2} - 1} \right)$
Here, we use the property
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
We get:-
$ \Rightarrow \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)$ $\left\{ {\because \left( {{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right)} \right\}$
Finally we get the result as:-
$ \Rightarrow {\left( {x + 1} \right)^2}\left( {x - 1} \right)$
By factoring the equation we complete it:-
${\left( {x + 1} \right)^2}\left( {x - 1} \right)$
$\therefore $ Factor of given polynomials are $\left( {x + 1} \right),\left( {x + 1} \right)$ and $\left( {x - 1} \right)$
$\because $ Degree of polynomial is $3$ therefore it must have $3$ factors for sure.
And in the above example, two of the factors are repeating i.e. $\left( {x + 1} \right)$ is a repeating factor.
Note: In the above example, by finding value of $f\left( 1 \right)$ we get, $f\left( 1 \right) = {1^3} + {1^2} - 1 - 1 = 0$
$\because f\left( 1 \right) = 0$ $\therefore \left( {x - 1} \right)$ is one of the factors.
Now, by dividing $\left( {{x^3} + {x^2} - x - 1} \right)$ by $\left( {x - 1} \right)$ we get a quadratic equation, which we can solve by formula method.
As there is no formula for finding factors of cubic or higher degree polynomials.
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