Answer
Verified
424.5k+ views
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 \[3{{x}^{3}}+7x\].
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE