# Ex.2.5 Q2 Polynomials Solution - NCERT Maths Class 9

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## Question

Evaluate the following products without multiplying directly:

(i) \begin{align}103 \times 107\end{align}

(ii)  \begin{align}95 \times 96\end{align}

(iii)  \begin{align}104 \times 96\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 2

## Text Solution

Reasoning:

Identities:

\begin{align}(x+a)(x+b)&=x^{2}+(a+b) x+a b \\ (a+b)(a-b)&=a^{2}-b^{2}\end{align}

Steps:

(i) \begin{align}103 \times 107\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

\begin{align} 103 \times 107 &=(100+3)(100+7) \\ &=(100)^{2}\!\!+\!(3\!+\!7)\!(100)\!+\!\!(3)(7) \\\\text {Taking } x&=100, a=3, b=7 ) \\ &=10000+1000+21 \\ &=11021 \end{align} (ii) \(\begin{align}95 \times 96\end{align}

Identity: \begin{align}(x+a)(x+b)=x^{2}+(a+b) x+a b\end{align}

\begin{align}{95 \times 96}&={ (100\!-\!5)(100\!-\!4)}\\&={\left[\!\begin{array}{l} {(100)^2}\!\!+\!(\!- 5\!- 4)\\(100)\!+\!( - 5)( - 4)\end{array}\!\right]}\\\\{\rm{(Taking }\;x \!=\! 100,a}&{ = - 5,b = - 4)}\\&{ = 10000 - 900 + 20}\\&{ = 9120}\end{align}

(iii)  \begin{align}104 \times 96\end{align}

Identity: \begin{align}(a+b)(a-b)=a^{2}-b^{2}\end{align}

\begin{align} 104 \times 96 &=(100+4)(100-4) \\ &=(100)^{2}-(4)^{2} \\\\(\text { Taking } a&=100, b=4 ) \\ &=10000-16 \\ &=9984 \end{align}

Video Solution
Polynomials
Ex 2.5 | Question 2

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