Courses
Courses for Kids
Free study material
Free LIVE classes
More

# Check whether 7+3x is a factor of $3{{x}^{3}}+7x$.

Last updated date: 26th Mar 2023
Total views: 308.7k
Views today: 5.86k
Verified
308.7k+ 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 $3{{x}^{3}}+7x$.