Answer
397.8k+ views
Hint: Here we will factorize the given polynomial by using the factor theorem. Firstly we will find a value of \[x\] which satisfies the polynomial by hit and trial method. Then we will take that factor and divide the polynomial by it. Finally, we will factorize the quotient, which we obtained after dividing the polynomial and multiply it with the first factor to get the required answer.
Complete step-by-step answer:
We have to factorize polynomial \[{x^3} - 49x\]
First, we need to know at least one zero of the polynomial, so we will use the hit and trial method.
Let us check whether \[x = 0\] satisfies the polynomial or not
Putting \[x = 0\] in \[{x^3} - 49x\] we get,
\[{0^3} - 49 \times 0 = 0\]
So \[x = 0\] is a factor as it satisfies the polynomial.
So, one factor is \[\left( {x - 0} \right)\] or we can say \[x\] is a factor.
Now we will divide the polynomial \[{x^3} - 49x\] by \[x\] and take the common factor out from the polynomial. Therefore, we get
\[\dfrac{{{x^3} - 49x}}{x} = \dfrac{{x\left( {{x^2} - 49} \right)}}{x}\]
Cancelling out common terms, we get
\[ \Rightarrow \dfrac{{{x^3} - 49x}}{x} = \left( {{x^2} - 49} \right)\]
So we get our quotient as \[{x^2} - 49\].
Now we will factorize the obtained quotient \[{x^2} - 49\].
\[{x^2} - 49 = \left( {x - 7} \right)\left( {x + 7} \right)\]
To find the zeros we will put the above equation equal to zero.
\[\left( {x - 7} \right)\left( {x + 7} \right) = 0\]
\[\begin{array}{l} \Rightarrow \left( {x - 7} \right) = 0\\ \Rightarrow x = 7\end{array}\]
or
\[\begin{array}{l} \Rightarrow \left( {x + 7} \right) = 0\\ \Rightarrow x = - 7\end{array}\]
So, we get the zeroes of the given polynomial as \[x = 0\], \[x = 7\] and \[x = - 7\]
Hence, the factors of the polynomial are \[\left( x \right)\left( {x - 7} \right)\left( {x + 7} \right)\].
Note: Factor theorem states that if a polynomial \[f\left( x \right)\] of degree\[n \ge 1\] and ‘\[a\]’ is any real number then, if \[f\left( a \right) = 0\] then only \[\left( {x - a} \right)\] is a factor of the polynomial. The two problems where Factor Theorem is usually applied are when we have to factorize a polynomial and also when we have to find the roots of the polynomial. It is also used to remove known zeros from a polynomial while leaving all unknown zeros intact. It is a special case of a polynomial remainder theorem.
Complete step-by-step answer:
We have to factorize polynomial \[{x^3} - 49x\]
First, we need to know at least one zero of the polynomial, so we will use the hit and trial method.
Let us check whether \[x = 0\] satisfies the polynomial or not
Putting \[x = 0\] in \[{x^3} - 49x\] we get,
\[{0^3} - 49 \times 0 = 0\]
So \[x = 0\] is a factor as it satisfies the polynomial.
So, one factor is \[\left( {x - 0} \right)\] or we can say \[x\] is a factor.
Now we will divide the polynomial \[{x^3} - 49x\] by \[x\] and take the common factor out from the polynomial. Therefore, we get
\[\dfrac{{{x^3} - 49x}}{x} = \dfrac{{x\left( {{x^2} - 49} \right)}}{x}\]
Cancelling out common terms, we get
\[ \Rightarrow \dfrac{{{x^3} - 49x}}{x} = \left( {{x^2} - 49} \right)\]
So we get our quotient as \[{x^2} - 49\].
Now we will factorize the obtained quotient \[{x^2} - 49\].
\[{x^2} - 49 = \left( {x - 7} \right)\left( {x + 7} \right)\]
To find the zeros we will put the above equation equal to zero.
\[\left( {x - 7} \right)\left( {x + 7} \right) = 0\]
\[\begin{array}{l} \Rightarrow \left( {x - 7} \right) = 0\\ \Rightarrow x = 7\end{array}\]
or
\[\begin{array}{l} \Rightarrow \left( {x + 7} \right) = 0\\ \Rightarrow x = - 7\end{array}\]
So, we get the zeroes of the given polynomial as \[x = 0\], \[x = 7\] and \[x = - 7\]
Hence, the factors of the polynomial are \[\left( x \right)\left( {x - 7} \right)\left( {x + 7} \right)\].
Note: Factor theorem states that if a polynomial \[f\left( x \right)\] of degree\[n \ge 1\] and ‘\[a\]’ is any real number then, if \[f\left( a \right) = 0\] then only \[\left( {x - a} \right)\] is a factor of the polynomial. The two problems where Factor Theorem is usually applied are when we have to factorize a polynomial and also when we have to find the roots of the polynomial. It is also used to remove known zeros from a polynomial while leaving all unknown zeros intact. It is a special case of a polynomial remainder theorem.
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