Chapter 2: Polynomials

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Polynomials CBSE NCERT Solutions

Question:

If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.

Answer:

Solution: The given polynomial is x3 – 3x2 + x + 1.  ………. (I)


As (a−b), a and (a+b) are the zeros of the given polynomial.

Let α = (a−b), β = a, and \gamma  = (a+b). 

We know,

The cubic polynomial with roots α, β, and \gamma  is:

x3 – (α + β + \gamma  )x2 + (αβ+ β\gamma  + α\gamma)x – (αβ\gamma ) ………. (II)

On substituting the values of α = (a−b), β = a, and  \gamma = (a+b) in (II). We get,

x3 – (a - b + a + a + b)x2 + [(a−b)a + a(a+b) + (a+b) (a−b)]x – [a(a−b)(a+b)]

=  x3 – 3ax2 + [a2 – ab + a2 + ab + a2 – b2]x –  a3 – ab2

=  x3 – 3ax2 + (3a2 – b2)x  –  (a3 + ab2) …… (III)

Now we can Compare the polynomial x3 – 3x2 + x + 1 with x3 – 3ax2 + (3a2 – b2)x  –  (a3 + ab2), we obtain,

So, 3a = 3  \Rightarrow a = 1,  (3a2 – b2) = 1 and  (a3 + ab2) = -1

Now on substituting the value of a = 1 in (3a2 – b2) = 1 then we get

(3x12 – b2) = 1

\Rightarrow 3 - b2 = 1
\Rightarrow 3 - 1 =  b2

\Rightarrow b2 = 2  \Rightarrow b =  \pm \sqrt{2}

Hence, a = 1 and b = \pm \sqrt{2}

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