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:

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

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

\[\begin{align} 103 \times 107 &=(100+3)(100+7) \\ &=(100)^{2}+(3+7)(100)+(3)(7) \\(\text { taking } x&=100, a=3, b=7 ) \\ &=10000+1000+21 \\ &=11021 \end{align}\]

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

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

\[\begin{align}95 \times 96&=(100-5)(100-4) \\ &=(100)^{2}+(-5-4)(100)+(-5)(-4) \\ \text { (Taking } x=100, a&=-5, b=-4 ) \\&= 10000-900+20 \\&= 9120 \end{align}\]

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

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

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