Ex.2.3 Q4 Polynomials Solution - NCERT Maths Class 10

Go back to  'Ex.2.3'

Question

On dividing \(x^{3} -3 x^{2}+x +2\) by a

polynomial \(g(x),\) the quotient and remainder were \(x – 2\) and \(–2x + 4,\) respectively. Find \(g(x).\)

 Video Solution
Polynomials
Ex 2.3 | Question 4

Text Solution

What is unknown?

Divisor \(g(x)\) of a polynomial \(p(x).\)

Reasoning:

This question is straight forward, you can solve it by using division algorithm

Dividend \(=\) Divisor \(\times\) Quotient \(+\) Remainder

Put the given values in the above equation and simplify it, get the value of \(g(x).\)

Steps:

Dividend \(=\) Divisor \(\times\) Quotient \(+\) Remainder

\(\begin{align}&x^{3}- 3 x^{2}\!+\!x\!+\!2 = \begin{bmatrix} g(x) \!\times\! x\! \\ -\!2\!+\!(-2 x\!+\!4) \end{bmatrix} \\& \begin{bmatrix} \left(x^{3}\!-\!3 x^{2}\!+\!x\!+\!2\right) \\ - (\!\!-\!2 x\!+\!4) 
 \end{bmatrix} =\!g(x)\! \times\! x\!-\!2\\&  \begin{bmatrix} x^{3}\!-\!3 x^{2}\!+\!x\!+\!2 x \\ +\!2\!-\!4\end{bmatrix} =\!g(x)\! \times\! x\!-\!2\\&\left(x^{3}\!-\!3 x^{2}\!+\!3 x\!-\!2\right)\!=\!g(x) \times x\!-\!2 \end{align}\)

Therefore, \(g\left( x \right) = {x^2} - x + 1\)

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school