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Ex.2.4 Q3 Polynomials Solution - NCERT Maths Class 9

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Question

Find the value of \(k,\) if \( x – 1\) is a factor of \(p(x)\) in each of the following cases:

 

 Video Solution
Polynomials
Ex 2.4 | Question 3

Text Solution

(i) \(\begin{align} p(x) = {x^2} + x + k\end{align}\)

(ii) \(\begin{align} p(x) = 2{x^2} + kx + \sqrt 2 \end{align}\)

(iii)  \(\begin{align} p(x) = k{x^2} - \sqrt {2x} + 1  \end{align}\) 

(iv)  \(\begin{align} p(x) = k{x^2} - 3x + k\end{align}\)

Reasoning:

By factor theorem, if \(x-1\) is a factor of \(p(x),\) then \( p\left( 1 \right) = 0\).

Steps:

(i) \(\begin{align} p(x) = {x^2} + x + k\end{align}\)

\[\begin{align}\,\,\,\, p(x) &= {x^2} + x + k\\ p(1) &= {(1)^2} + (1) + k\\ \,\,\,\,\,\,\,0\,\, &= 2 + k\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= - 2\end{align}\]

(ii) \(\begin{align} p(x) = 2{x^2} + kx + \sqrt 2 \end{align}\)

\[\begin{align}\,\,\,\, p(x) &= 2{x^2} + kx + \sqrt 2 \\ p(1) &= 2{(1)^2} + k(1) + \sqrt 2 \\ \,\,\,\,\,\,\,0\,\, &= 2 + k + \sqrt 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= - (2 + \sqrt 2 )\end{align}\]

(iii)  \(\begin{align} p(x) = k{x^2} - \sqrt {2x} + 1  \end{align}\)

\[\begin{align}\,\,\,\,\,\,\,\, p(x) &= k{x^2} - \sqrt {2x} + 1\\ p(1) &= k{(1)^2} - \sqrt {2(1)} + 1\\ \,\,\,\,\,\,\,\,0 &= k - \sqrt 2 + 1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= \sqrt 2 - 1\end{align}\]

(iv)  \(\begin{align} p(x) = k{x^2} - 3x + k\end{align}\)

\[\begin{align}\,\,\, p(x) &= k{x^2} - 3x - k\\ p(1) &= k({1^2}) - 3(1) - k\\ \,\,\,\,\,\,\,\,0 &= 2k - 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow k &= \frac{3}{2}\end{align}\]

 Video Solution
Polynomials
Ex 2.4 | Question 3
  
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