Ex.2.5 Q1 Polynomials Solution - NCERT Maths Class 9

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Question

Use suitable identities to find the following products:

(i) \(\begin{align}(x+4)(x+10)\end{align}\) 

(ii) \(\begin{align}(x+8)(x-10)\end{align}\)

(iii) \(\begin{align}(3 x+4)(3 x-5)\end{align}\)

(iv) \(\begin{align}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\end{align}\) 

(v) \(\begin{align}(3-2 x)(3+2 x)\end{align}\)

 

 Video Solution
Polynomials
Ex 2.5 | Question 1

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:

\(\begin{align}\text{(i)}\;\;(x+4)(x+10)\end{align}\)

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

Here \(\begin{align}\text{a} = {4}, \text{b} = 10\end{align}\)

\[\begin{align}(x+4)(x+10) &=x^{2}+(4+10) x+4 \times 10 \\ &=x^{2}+14 x+40 \end{align}\]

\(\begin{align}\text{(ii)}\;\;(x+8)(x-10)\end{align}\)

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

Here \(a =8, b=-10\)

\[\begin{align}(x+8)(x-10) &=x^{2}+(8-10) x+(8)(-10) \\ &=x^{2}-2 x-80 \end{align}\]

\(\begin{align}\text{(iii)}\;\;(3 x+4)(3 x-5)\end{align}\)

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

Here \(\begin{align}x \rightarrow 3 x, a=4, b=-5\end{align}\)

\[\begin{align}(3 x+4)(3 x-5) &=(3 x)^{2}+(4-5)(3 x)+(4)(-5) \\ &=9 x^{2}-3 x-20 \end{align}\]

(iv) \(\begin{align}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)\end{align}\)

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

Here \(\begin{align}a=y^{2}, b=\frac{3}{2}\end{align}\)

\[\begin{align}\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) &=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2} \\ &=y^{4}-\frac{9}{4} \end{align}\]

(v) \(\begin{align}(3-2 x)(3+2 x) \end{align}\)

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

Here \(\begin{align}a=3, b=2 x\end{align}\)

\[\begin{align}(3-2 x)(3+2 x) &=(3)^{2}-(2 x)^{2} \\ &=9-4 x^{2} \end{align}\]