# Ex.2.3 Q2 Polynomials Solution - NCERT Maths Class 10

## Question

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i)\(\begin{align}\quad{{t}^{2}}-3,\;2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12\end{align}\)

(ii)\(\begin{align}\quad{{x}^{2}}+3x+\text{ }1, \;3{{x}^{4}}+5{{x}^{3}}-7{{x}^{2}}+2x+\text{ }2 \end{align}\)

(iii)\(\begin{align}\quad{{x}^{3}}-3x+1,\;{{x}^{5}}-4{{x}^{3}}+{{x}^{2}}+3x+\text{ }1 \\ \end{align}\)

## Text Solution

**What is unknown?**

Whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

**Reasoning:**

To solve this question, follow the same procedure given in question no \(1,\) you have to observe only one thing - if the remainder is \(0\) then the first polynomial is a factor of the second polynomial.

**Steps:**

(i)\(\begin{align}\quad{{t}^{2}}-3,\;2{{t}^{4}}+3{{t}^{3}}-2{{t}^{2}}-9t-12\end{align}\)

\(\begin{align}{ {t^2} - 3 = {t^2} + 0.t - 3}\end{align}\)

Since, the remainder is zero, \({t^2}-3\) is a factor of \(2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\)

(ii)\(\begin{align}\quad{{x}^{2}}+3x+\text{ }1, \;3{{x}^{4}}+5{{x}^{3}}-7{{x}^{2}}+2x+\text{ }2 \end{align}\)

Quotient \(=3{{x}^{2}}-4x+2,\) Remainder \(=0\)

Since, the remainder is zero, \({x^2} +3x +1\) is a factor of \(3{x^4} + 5{x^3}-7{x^2} + 2x +2\)

(iii)\(\begin{align}\quad{{x}^{3}}-3x+1,\;{{x}^{5}}-4{{x}^{3}}+{{x}^{2}}+3x+\text{ }1 \\ \end{align}\)

Quotient \(={{x}^{2}}-1,\) Remainder \(=2\)

Since, the remainder is not zero, \({x^3}-3x + 1\) is not a factor of \({x^5}-4{x^3} + {x^2} + 3x + 1\)