# Check whether 7+3x is a factor of \[3{{x}^{3}}+7x\].

Last updated date: 26th Mar 2023

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Answer

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Hint: Relate factor of any polynomial with its zeroes(If x = a is zero of polynomial then x-a will be a factor of that polynomial).

Let us first find out the relationship between the zero and factor of any polynomial.

Let us assume we have a polynomial

\[F\left( x \right)={{a}_{o}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+.....{{a}_{n}}\]

Let us assume \[\left( x+\alpha \right)\] is a factor of the above polynomial i.e. if we divide the given polynomial by \[\left( x+\alpha \right)\] then remainder will be zero.

As we know the division algorithm that Dividend \[=\]Divisor \[\times \]Quotient \[+\]Remainder

\[F\left( x \right)=\left( x+\alpha \right)Q\left( x \right)+0\] [\[Q\left( x \right)\]\[=\] Quotient]

Hence, \[F\left( x \right)=\left( x+\alpha \right)\left( Q\left( x \right) \right)....\left( i \right)\]

Let us find out the zeros / roots of a given polynomial \[F\left( x \right)\]and \[\left( x+\alpha \right)\].

Since, zero is the value of the variable term at which polynomial / expression will become zero.

Zero of \[\left( x+\alpha \right)\]\[\Rightarrow x+\alpha =0\]

\[x=-\alpha ....\left( ii \right)\]

Let us find out zero of\[F\left( x \right)\].

\[F\left( x \right)=0\]

\[{{a}_{o}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+....{{a}_{n}}=0\]

As we can write \[F\left( x \right)=\left( x+\alpha \right)Q\left( x \right)\]

Hence, \[F\left( x \right)=0\]

Or, \[\left( x+\alpha \right)Q\left( x \right)=0\]

Since, \[F\left( x \right)\] is a degree of \[n\] then \[F\left( x \right)\]have \[n\] roots and \[Q\left( x \right)\] will become degree of \[\left( n-1 \right)\]as divided by \[\left( x+\alpha \right)\].

Hence, \[\left( x+\alpha \right)\left( Q\left( x \right) \right)=0\]

\[x+\alpha =0\] or \[Q\left( x \right)=0\]

\[x=-\alpha \]and \[Q\left( x \right)=0\] which have \[\left( n-1 \right)\] roots.

Therefore, \[F\left( x \right)\] has one root \[\left( -\alpha \right)\] and \[\left( n-1 \right)\] roots which can be obtained from \[Q\left( x \right)=0\].

Now, it is proved that if a polynomial is a factor of another, then the roots or zeros of them is also common.

Now, coming to the question part. We need to check whether \[\left( 7+3x \right)\] is a factor of \[3{{x}^{2}}+7x\] or not.

Let us find out the root / zero of \[\left( 7+3x \right)\].

\[7+3x=0\]

\[3x=-7\]

\[x=\dfrac{-7}{3}\]

Root / Zero of \[7+3x\] is \[\left( \dfrac{-7}{3} \right)\].

Now, if \[\left( 7+3x \right)\] is a factor of given polynomial, then root of \[7+3x\] i.e. \[\left( \dfrac{-7}{3} \right)\] is also zero / root of the polynomial.

Hence, we have root \[=\dfrac{-7}{3}\]

Polynomial \[=F\left( x \right)=3{{x}^{3}}+7x\]

\[=-3\times \dfrac{49\times 7}{27}-\dfrac{49}{3}\]

\[F\left( \dfrac{-7}{3} \right)=\dfrac{-49}{9}(7+3)\ne 0\]

As \[\dfrac{-7}{3}\] is not a root of a given polynomial as it is not satisfying the given polynomial.

Hence, \[7+3x\] is not a factor of \[3{{x}^{3}}+7x\].

Note: One can check by dividing the given polynomial \[3{{x}^{3}}+7x\] by \[7+3x\] and relate it with the remainder. If the remainder will be zero, then \[7+3x\] is a factor otherwise not. (Remainder will not be zero). We can prove it by factoring \[3{{x}^{3}}+7x\] as well in the following way:

\[3{{x}^{3}}+7x\]

\[x\left( 3{{x}^{2}}+7 \right)\]

Now, further we cannot factorize the expression. And hence \[\left( 7+3x \right)\] is not a factor of \[3{{x}^{3}}+7x\] by observing the factors of \[3{{x}^{3}}+7x\].

Let us first find out the relationship between the zero and factor of any polynomial.

Let us assume we have a polynomial

\[F\left( x \right)={{a}_{o}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+.....{{a}_{n}}\]

Let us assume \[\left( x+\alpha \right)\] is a factor of the above polynomial i.e. if we divide the given polynomial by \[\left( x+\alpha \right)\] then remainder will be zero.

As we know the division algorithm that Dividend \[=\]Divisor \[\times \]Quotient \[+\]Remainder

\[F\left( x \right)=\left( x+\alpha \right)Q\left( x \right)+0\] [\[Q\left( x \right)\]\[=\] Quotient]

Hence, \[F\left( x \right)=\left( x+\alpha \right)\left( Q\left( x \right) \right)....\left( i \right)\]

Let us find out the zeros / roots of a given polynomial \[F\left( x \right)\]and \[\left( x+\alpha \right)\].

Since, zero is the value of the variable term at which polynomial / expression will become zero.

Zero of \[\left( x+\alpha \right)\]\[\Rightarrow x+\alpha =0\]

\[x=-\alpha ....\left( ii \right)\]

Let us find out zero of\[F\left( x \right)\].

\[F\left( x \right)=0\]

\[{{a}_{o}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+....{{a}_{n}}=0\]

As we can write \[F\left( x \right)=\left( x+\alpha \right)Q\left( x \right)\]

Hence, \[F\left( x \right)=0\]

Or, \[\left( x+\alpha \right)Q\left( x \right)=0\]

Since, \[F\left( x \right)\] is a degree of \[n\] then \[F\left( x \right)\]have \[n\] roots and \[Q\left( x \right)\] will become degree of \[\left( n-1 \right)\]as divided by \[\left( x+\alpha \right)\].

Hence, \[\left( x+\alpha \right)\left( Q\left( x \right) \right)=0\]

\[x+\alpha =0\] or \[Q\left( x \right)=0\]

\[x=-\alpha \]and \[Q\left( x \right)=0\] which have \[\left( n-1 \right)\] roots.

Therefore, \[F\left( x \right)\] has one root \[\left( -\alpha \right)\] and \[\left( n-1 \right)\] roots which can be obtained from \[Q\left( x \right)=0\].

Now, it is proved that if a polynomial is a factor of another, then the roots or zeros of them is also common.

Now, coming to the question part. We need to check whether \[\left( 7+3x \right)\] is a factor of \[3{{x}^{2}}+7x\] or not.

Let us find out the root / zero of \[\left( 7+3x \right)\].

\[7+3x=0\]

\[3x=-7\]

\[x=\dfrac{-7}{3}\]

Root / Zero of \[7+3x\] is \[\left( \dfrac{-7}{3} \right)\].

Now, if \[\left( 7+3x \right)\] is a factor of given polynomial, then root of \[7+3x\] i.e. \[\left( \dfrac{-7}{3} \right)\] is also zero / root of the polynomial.

Hence, we have root \[=\dfrac{-7}{3}\]

Polynomial \[=F\left( x \right)=3{{x}^{3}}+7x\]

\[=-3\times \dfrac{49\times 7}{27}-\dfrac{49}{3}\]

\[F\left( \dfrac{-7}{3} \right)=\dfrac{-49}{9}(7+3)\ne 0\]

As \[\dfrac{-7}{3}\] is not a root of a given polynomial as it is not satisfying the given polynomial.

Hence, \[7+3x\] is not a factor of \[3{{x}^{3}}+7x\].

Note: One can check by dividing the given polynomial \[3{{x}^{3}}+7x\] by \[7+3x\] and relate it with the remainder. If the remainder will be zero, then \[7+3x\] is a factor otherwise not. (Remainder will not be zero). We can prove it by factoring \[3{{x}^{3}}+7x\] as well in the following way:

\[3{{x}^{3}}+7x\]

\[x\left( 3{{x}^{2}}+7 \right)\]

Now, further we cannot factorize the expression. And hence \[\left( 7+3x \right)\] is not a factor of \[3{{x}^{3}}+7x\] by observing the factors of \[3{{x}^{3}}+7x\].

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