Decide whether \[\left( {x + 1} \right)\] is a factor of the polynomial \[\left( {{x^3} + {x^2} - x - 1} \right)\] or not.
Answer
534.6k+ views
Hint: To find out the given expression is polynomial or not, we need to apply factor theorem; factor theorem is used when factoring the polynomials completely, it is a theorem that links factors and zeros of the polynomial. Hence applying the theorem and substituting the obtained value of x, to decide whether \[\left( {x + 1} \right)\] is a factor of the polynomial or not.
Complete step by step answer:
Given,
\[\left( {x + 1} \right)\] and \[\left( {{x^3} + {x^2} - x - 1} \right)\]
Using factor theorem,
\[x + 1 = 0\]
\[ \Rightarrow x = 0 - 1\]
\[ \Rightarrow x = - 1\]
Now, let
\[p\left( x \right) = \left( {{x^3} + {x^2} - x - 1} \right)\]
Substitute the obtained value of x as:
\[ \Rightarrow p\left( { - 1} \right) = \left( {{{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^2} - \left( { - 1} \right) - 1} \right)\]
Evaluating the terms, we get:
\[ \Rightarrow p\left( { - 1} \right) = - 1 + 1 + 1 - 1\]
\[ \Rightarrow p\left( { - 1} \right) = 0\]
Therefore,
\[\left( {x + 1} \right)\]is a factor of the polynomial \[\left( {{x^3} + {x^2} - x - 1} \right)\].
Note: We must note that the factor theorem is used when factoring the polynomials completely. It is a theorem that links factors and zeros of the polynomial. The key point to find the given equation is polynomial or not we must know that if \[f\left( x \right)\] is a polynomial of degree \[n \geqslant 1\] and \['a'\] is any real number, then, \[\left( {x - a} \right)\] is a factor of \[f\left( x \right)\] if \[f\left( a \right) = 0\].
Complete step by step answer:
Given,
\[\left( {x + 1} \right)\] and \[\left( {{x^3} + {x^2} - x - 1} \right)\]
Using factor theorem,
\[x + 1 = 0\]
\[ \Rightarrow x = 0 - 1\]
\[ \Rightarrow x = - 1\]
Now, let
\[p\left( x \right) = \left( {{x^3} + {x^2} - x - 1} \right)\]
Substitute the obtained value of x as:
\[ \Rightarrow p\left( { - 1} \right) = \left( {{{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^2} - \left( { - 1} \right) - 1} \right)\]
Evaluating the terms, we get:
\[ \Rightarrow p\left( { - 1} \right) = - 1 + 1 + 1 - 1\]
\[ \Rightarrow p\left( { - 1} \right) = 0\]
Therefore,
\[\left( {x + 1} \right)\]is a factor of the polynomial \[\left( {{x^3} + {x^2} - x - 1} \right)\].
Note: We must note that the factor theorem is used when factoring the polynomials completely. It is a theorem that links factors and zeros of the polynomial. The key point to find the given equation is polynomial or not we must know that if \[f\left( x \right)\] is a polynomial of degree \[n \geqslant 1\] and \['a'\] is any real number, then, \[\left( {x - a} \right)\] is a factor of \[f\left( x \right)\] if \[f\left( a \right) = 0\].
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