# Ex.2.4 Q1 Polynomials Solution - NCERT Maths Class 10

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## Question

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) $$\quad 2{x^3} + {x^2} - 5x + 2;\,\,\,\frac{1}{2},\,\,1,\,\, - 2$$

(ii)$$\quad {x^3} - 4{x^2} + 5x - 2;\;,2,\;1,\;1$$

Video Solution
Polynomials
Ex 2.4 | Question 1

## Text Solution

What is known?

Cubic polynomials and their zeroes are given

What is unknown?

Verify that the numbers given alongside of the cubic polynomials are their zeroes and verify the relationship between the zeroes and the coefficients in each case.

Reasoning:

This question is very simple, cubic polynomials are already given. Put the values of zeroes in the polynomial, you will get zero because it will satisfy the equation of the given polynomial. Now compare the polynomial with $$a{x^3} + b{x^2} + cx + d,$$ you will get the values of the coefficients of $$x$$ i.e. $$a, \,b,\, c$$ and $$d.$$

Steps:

$$p\left( x \right) = 2{x^3} + {x^2}-5x + 2$$

Given numbers are $$= \frac{1}{2},\;1,\; - 2$$

Put $$x= \frac{1}{2}$$ in $$p\left( x \right) = 2{x^3} + {x^2}-5x + 2$$

\begin{align}p\left( {\frac{1}{2}} \right) & \!\!=\!\! 2 \! {\left( {\frac{1}{2}} \right)^3} \!\!+\!\! {\left( {\frac{1}{2}} \right)^2}\! \!- 5 \! \left( {\frac{1}{2}} \right) \!\!+\! \! 2\\p\left( {\frac{1}{2}} \right) &\!\!= \!\!2\left( {\frac{1}{8}} \right) \!\!+ \!\! \left( {\frac{1}{4}} \right) - 5\left( {\frac{1}{2}} \right) \!+\!\! 2\\p\left( {\frac{1}{2}} \right) &= \frac{1}{4} + \left( {\frac{1}{4}} \right) - \frac{5}{2} + 2\\p\left( {\frac{1}{2}} \right)& = \frac{{1 + 1 - 10 + 8}}{4}\\p\left( {\frac{1}{2}} \right) &= 0\end{align}

Put $$x= 1$$ in $$p\left( x \right) = 2{x^3} + {x^2}-5x + 2$$

\begin{align}p\left( 1 \right)&= 2{\left( 1 \right)^3} + {\left( 1 \right)^2} - 5\left( 1 \right) + 2\\p\left( {\frac{1}{2}} \right) &= 2 + 1 - 5 + 2\\p\left( 1 \right) &= 3 - 5 + 2\\p\left( 1 \right) &= 0\end{align}

Put $$x= - 2$$ in $$p\left( x \right) = 2{x^3} + {x^2}-5x + 2$$

\begin{align}{p}\left( { - 2} \right)&= \!\!2\left( - 2 \right)^3 + \left( - 2 \right)^2 - 5( - 2) \!\!+ \!\!2\\{{p}\left(- 2 \right)} &= {16 - 8 + 4 + 10 + 2}\\{{p}\left( { - 2} \right)}& ={ - 16 + 4 + 10 + 2}\\{{p}\left( { - 2} \right)}&={ - 16 + 16}\\{p\left( { - 2} \right)} &= 0\end{align}

Therefore, \begin{align}\frac{1}{2},\;1,\; - 2 \end{align} are the zeroes of the polynomial.

Now let  \begin{align}\alpha = \frac{1}{2},\,\,\beta = 1 \end{align} and \begin{align}\gamma = 2.\end{align}

\begin{align}\alpha + \beta + \gamma &= \frac{1}{2} + 1 + \left( { - 2} \right)\\ &= \frac{{ - 1}}{2}\\&= \frac{{ -{ \text{coeficient of }}{x^2}}}{{{\text{coeficient of }}{x^3}}}\end{align}

\begin{align} & \alpha \beta + \beta \gamma + \gamma \alpha \\ \\ &=\! \frac{1}{2}\! \!\times\!\! 1 \!+ \!1\!\! \times\! \left( {\!\! - \!2} \right) \!+\! \left( {\!\! -\! 2} \right) \!\!\times\! \!\frac{1}{2}\\ &= \frac{{ - 5}}{2}\\& = \frac{{ - {\text{coeficient of }}x}}{{{\text{coeficient of }}{x^3}}}\end{align}

\begin{align}\alpha .\beta .\gamma &= \frac{1}{2} \times 1 \times \left( { - 2} \right)\\&= \frac{{ - 2}}{2}\\ &= \frac{{ - {\text{ constant term }}}}{{{\text{ coeficient of }}{x^3}}}\end{align}

Hence, the relation between zeroes and coefficient is verified.

(ii)$$\quad {x^3} - 4{x^2} + 5x - 2;\;,2,\;1,\;1$$

Given numbers are \begin{align} = 2,1,1\end{align}

Put $$x = 2$$ in $${x^3}-4{x^2} + 5x-2$$

\begin{align}{p\left( 2 \right)} &= {{{\left( 2 \right)}^3} - 4{{\left( 2 \right)}^2} + 5\left( 2 \right) - 2}\\{p\left( 2 \right)}& = {8 - 16 + 10 - 2}\\{p\left( 2 \right)} & ={ 2 - 2}\\{p\left( 2 \right)} &= 0\end{align}

Put $$x = 1$$ in $${x^3}-4{x^2} + 5x-2$$

\begin{align}p\left( 1 \right) &= {{\left( 1 \right)}^3} - 4{{\left( 1 \right)}^2} + 5\left( 1 \right) - 2\\p\left( 1 \right)& = 1 - 4 + 5 - 2\\p \left( 1 \right)&= - 3 + 3\\p \left( 1 \right) &= 0\end{align}

Therefore, $$2,\,1$$ and $$1$$ are the zeroes of the polynomial.

Now let  $$\alpha = 2,\;\beta = 1$$ and $$\gamma = 1.$$

\begin{align}\alpha + \beta + \gamma &= 2 + 1 + 1\\& = \frac{{ - \left( { - 1} \right)}}{1}\\&= \frac{{ - \,{\text{coeficient of }}{x^2}}}{{{\text{coeficient of }}{x^3}}}\end{align}

\begin{align}\alpha \beta + \beta \gamma + \gamma \alpha&= 2\! \times \!1\! + \!1\! \times \!1 \!+\! 1\! \times\! 2 \! \\& = \!5\\&= \frac{5}{1}\\&= \frac{{ - \,{\text{coeficient of }}x}}{{{\text{coeficient of }}{x^3}}}\end{align}

\begin{align}\alpha \,.\,\,\beta \,.\,\,\gamma &= \;2\; \times \;1\; \times \;1\;\; = \;2\\ &= \frac{{ - \left( { - 2} \right)}}{1}\\& = \frac{ -\,\text{constant term}}{\text{coeficient of }{x^3}}\end{align}

Hence, the relation between zeroes and coefficient is verified.

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