Question

(i) Write the surd $\sqrt{98}$ in the simplest form.
(ii) Write the ${{x}^{4}}-5{{x}^{2}}+7$ polynomial in coefficient form.

Answer 1

Hint: (i) In the first question, the surd has square root, so firstly we have to calculate the prime factors of the number which is in square root. After factorization, the factors that are coming two times, they will come out of the square root because the surd is ${{\left( \dfrac{1}{n} \right)}^{th}}$ power of any factor where$n=2$. So if the factor multiplies two times, it will come out. Here we use this formula also, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$.
(ii) In the second question, we have to calculate the coefficient form of the polynomial, for which we have to get coefficients of all the degree terms.
Let $p\left( x \right)$ is a polynomial of degree $n$.
So, $p\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+{{a}_{n-2}}{{x}^{n-2}}+...+{{a}_{1}}x+{{a}_{0}}x$
So, coefficient from of $p\left( x \right)$will be
$\left( {{a}_{n}},{{a}_{n-1}},{{a}_{n-2}},{{a}_{n-3}},...,{{a}_{2}},{{a}_{1}},{{a}_{0}} \right)$.

Complete step-by-step answer:
(i) So let us learn about the surd. If a number \[N\] is $nth$ root of any positive integer where $n\ge 2$, and that cannot be simplified in pure rational numbers, then it is called surd. So, \[{{\left( N \right)}^{\dfrac{1}{n}}}\] is a surd.
Now the given surd is$\sqrt{98}$. We have to do prime factorization of$98$.
\[\begin{align}
  & 2\left| \!{\underline {\,
  98 \,}} \right. \\
 & 2\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
\end{align}\]
   $1$

So, $98=2\times 7\times 7$.
So, prime factorization of $98$ is $2\times 7\times 7$. Now we have to simplify the square root of $98$. For that, we have to see which factor has a square in factorization.

Here, $7$ has a square in factorization.
$98=2\times {{\left( 7 \right)}^{2}}$.
So, square root of $98$ is –
$\sqrt{98}=\sqrt{2\times {{\left( 7 \right)}^{2}}}$
\[\Rightarrow \sqrt{98}={{\left\{ 2\times {{\left( 7 \right)}^{2}} \right\}}^{\dfrac{1}{2}}}\]

$\because {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
So, we have \[\sqrt{98}={{2}^{\dfrac{1}{2}}}\times {{\left( 7 \right)}^{2}}^{\times \dfrac{1}{2}}\]
\[\Rightarrow \sqrt{98}=7\sqrt{2}\]


(ii) Let us assume that the given polynomial is $q\left( x \right)$ .
$q\left( x \right)={{x}^{4}}-5{{x}^{2}}+7$.
This has to be represented in coefficient form.
We know that the degree of the polynomial is 4. Let us take a general polynomial of degree 4 as$p\left( x \right)$.

$p\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e$ …(1)
And its coefficient form will be
$\left( a,b,c,d,e \right)$
Now we can write the given equation in this form as well.
${{x}^{4}}-5{{x}^{2}}+7=p\left( x \right)={{x}^{4}}+0{{x}^{3}}-5{{x}^{2}}+0x+7$
$q\left( x \right)={{x}^{4}}+0{{x}^{3}}-5{{x}^{2}}+0x+7$ …(2)

Now we can compare both polynomials $p\left( x \right)$ and$q\left( x \right)$. We get,
$\left( a=1,b=0,c=-5,d=0,e=7 \right)$ .

Hence, the coefficient form of $p\left( x \right)$is$\left( a,b,c,d,e \right)$ and $q\left( x \right)$ is $\left( 1,0,-5,0,7 \right)$.

Note: (i) In the first question, students should remember, surd is not only square root, but it can be cube root, fourth root, etc. also. So, we have to take factors out of root according to surd’s root.
(ii) In the second question, students should take care of missing degree terms, and take coefficient of missing degree term as zero.