# Ex.2.4 Q3 Polynomials Solution - NCERT Maths Class 10

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## Question

If the zeroes of the polynomial

$$x^{3}-3 x^{2}+x+1$$ are $$a-b, \;a,\;a+b,$$

find $$a$$ and $$b$$.

Video Solution
Polynomials
Ex 2.4 | Question 3

## Text Solution

What is known?

Zeroes of the polynomial $$x^{3}-3 x^{2}+x+1$$ are $$a- b, \; a,\; a+b$$

What is unknown?

$$a$$ and $$b$$.

Reasoning:

First compare the given polynomial with the general equation of the cubic polynomial $$p{x^3} + q{x^2} + rx + t$$ and you will get the values of $$p, \;q,\; r$$ and $$t.$$

Now put this value in the equation of sum of zeroes and product of zeroes, you will get the value of $$a$$ and $$b$$.

Steps:

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

Zeroes are $$a - b,{{ }}a,\;a + b.$$

On comparing the given polynomial with $$p{x^3} + q{x^2} + rx + t$$

we get,

$$p = 1, \;q = - 3,\;r = 1$$ and $$t = 1$$

Sum of zeroes

\begin{align}&= a - b + a + a + b\\& = \frac{{ - {\text{coeficient of }}{x^2}}}{{{{ }}{\text{coeficient of }}{x^3}}} = \frac{{ - q}}{p}\\{\frac{{ - q}}{p}}&={ 3{{a}}}\\{\frac{{ - ( - 3)}}{1}} &= {3{{a}}}\\a{{ }}& = {{ }}1\end{align}

Since the value of $$a$$ is found to be $$1,$$ the zeroes are $$1 - b,\;1,\;1 + b$$

Multiplication of zeroes

\begin{align}&= 1\left( {1 - b} \right)\left( {1 + b} \right)\\& = \frac{{ - {\text{constant term}}}}{\text{coeficient of }{x^3}}\\&= \frac{{ - t}}{p}\\{\frac{{ - t}}{p}} &= {1-{{ }}{b^2}}\\\frac{{ - 1}}{1} &= {1-{{ }}{b^2}}\\1 - {b^2}& = 1\\1 + 1 &= {b^2}\\{b^2} &= 2\end{align}

$$b = \pm \sqrt 2 .$$ Hence, $$a = 1, \;b = \sqrt 2$$ or $$- \sqrt 2$$

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