
If two zeroes of the polynomial ${{x}^{3}}-4{{x}^{2}}-3x+12$ are $\sqrt{3}$ and $-\sqrt{3}$ then find its third zero.
Answer
555k+ views
Hint: Use the given root as a factor and divide by combining it
we need to use the quotient to calculate the third zero.
If $f(x)= x+a=0$ then $x=-a$ will be a zero of $f(x)$.
Complete step by step answer:
We know that if $x=a$ is the root or zero of any polynomial $f(x)$ then $x+a$ will completely divide the given polynomial and $x+a$ will be termed a factor of that polynomial \[f(x)\].
Here in question, it is given that $\sqrt{3}$ and \[-\sqrt{3}\]are the zeroes of the polynomial $f(x)={{x}^{3}}-4{{x}^{2}}-3x+12$
So according to the factor theorem \[x-\sqrt{3}\]and$x+\sqrt{3}$ will be the factor of the polynomial and product of these two will completely divide the given polynomial\[f(x)\].
$\left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)={{\left( x \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}$.
Using,$\left( a+b \right)(a-b)={{a}^{2}}-{{b}^{2}}$
$={{x}^{2}}-3$
We now divide the polynomial ${{x}^{3}}-4{{x}^{2}}-3x+12$ by ${{x}^{2}}-3$
${{x}^{2}}-3\overset{x}{\overline{\left){\begin{align}
& {{x}^{3}}-4{{x}^{2}}-3x+12 \\
& {{x}^{3}}-3x
\end{align}}\right.}}$
Here first quotient is x since first term of polynomial is${{x}^{3}}$ ,
Now subtracting the term ,we get further division step
as$-4{{x}^{2}}+12$since ${{x}^{3}}$ get cancelled with ${{x}^{3}}$ and $-3x$ with $-3x$
So next quotient should be $-4$ to make same as $-4{{x}^{2}}+12$
${{x}^{2}}-3\overset{x-4}{\overline{\left){\begin{align}
& {{x}^{3}}-4{{x}^{2}}-3x+12 \\
& {{x}^{3}}-3x \\
&\hline\\
& -4{{x}^{2}}+12 \\
& -4{{x}^{2}}+12
\end{align}}\right.}}$
Further subtracting the last term we can clearly see that $-4{{x}^{2}}$ get cancelled with $-4{{x}^{2}}$and $12$ with $12$ so we get remainder as zero hence \[x-4\] will be the quotient as well as the factor of the given polynomial.
Therefore to calculate the third zero of the polynomial we need to put this quotient equals to zero.
$\begin{align}
& x-4=0 \\
& x=4
\end{align}$
Hence, third zero of the polynomial ${{x}^{3}}-4{{x}^{2}}-3x+12$ is 4.
Note:
- We can also use the sum of zeroes and product of zeroes method to find the third zero of the given polynomial.
- We also know it as the relation between zeroes and coefficients of polynomials.
we need to use the quotient to calculate the third zero.
If $f(x)= x+a=0$ then $x=-a$ will be a zero of $f(x)$.
Complete step by step answer:
We know that if $x=a$ is the root or zero of any polynomial $f(x)$ then $x+a$ will completely divide the given polynomial and $x+a$ will be termed a factor of that polynomial \[f(x)\].
Here in question, it is given that $\sqrt{3}$ and \[-\sqrt{3}\]are the zeroes of the polynomial $f(x)={{x}^{3}}-4{{x}^{2}}-3x+12$
So according to the factor theorem \[x-\sqrt{3}\]and$x+\sqrt{3}$ will be the factor of the polynomial and product of these two will completely divide the given polynomial\[f(x)\].
$\left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)={{\left( x \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}$.
Using,$\left( a+b \right)(a-b)={{a}^{2}}-{{b}^{2}}$
$={{x}^{2}}-3$
We now divide the polynomial ${{x}^{3}}-4{{x}^{2}}-3x+12$ by ${{x}^{2}}-3$
${{x}^{2}}-3\overset{x}{\overline{\left){\begin{align}
& {{x}^{3}}-4{{x}^{2}}-3x+12 \\
& {{x}^{3}}-3x
\end{align}}\right.}}$
Here first quotient is x since first term of polynomial is${{x}^{3}}$ ,
Now subtracting the term ,we get further division step
as$-4{{x}^{2}}+12$since ${{x}^{3}}$ get cancelled with ${{x}^{3}}$ and $-3x$ with $-3x$
So next quotient should be $-4$ to make same as $-4{{x}^{2}}+12$
${{x}^{2}}-3\overset{x-4}{\overline{\left){\begin{align}
& {{x}^{3}}-4{{x}^{2}}-3x+12 \\
& {{x}^{3}}-3x \\
&\hline\\
& -4{{x}^{2}}+12 \\
& -4{{x}^{2}}+12
\end{align}}\right.}}$
Further subtracting the last term we can clearly see that $-4{{x}^{2}}$ get cancelled with $-4{{x}^{2}}$and $12$ with $12$ so we get remainder as zero hence \[x-4\] will be the quotient as well as the factor of the given polynomial.
Therefore to calculate the third zero of the polynomial we need to put this quotient equals to zero.
$\begin{align}
& x-4=0 \\
& x=4
\end{align}$
Hence, third zero of the polynomial ${{x}^{3}}-4{{x}^{2}}-3x+12$ is 4.
Note:
- We can also use the sum of zeroes and product of zeroes method to find the third zero of the given polynomial.
- We also know it as the relation between zeroes and coefficients of polynomials.
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