Dr Manohar Re Solution CLASS 9 CHAPTER 2 बहुपद (POLYNOMIALS AND THEIR FACTORS) प्रश्नावली 2(G)

 प्रश्नावली 2(G)

बहुधिकल्पीय प्रश्न :

प्रत्येक प्रश्न के चार उत्तर दिये हुए हैं, सही उत्तर छाँटिए :

प्रश्न 1

$81 x^{3}-x$ के गुणनखण्ड हैं :

(i) (9x+1)(9x-1)

(ii) x(9x+1)(9x-1)

(iii) $(9x-1)^{2}$

(iv) $(9 x+1)^{2}$

उत्तर : विकल्प (ii) x(9x+1)(9x-1)

हल :

$\begin{aligned}81 x^{3}-x &=x\left(81 x^{2}-1\right) \\&=x\left[(9 x)^{2}-(1)^{2}\right] \\&=x(9 x-1)(9 x+1)\end{aligned}$

हल :

$\begin{aligned} 16 p^{2}-9 q^{2} &=(4 p)^{2}-(3 q)^{2} \\ &=(4 p-3 q)(4 p+3 q) \end{aligned}$

अति लघु उत्तरीय प्रश्न :

प्रश्न 3

गुणनखण्ड ज्ञात कीजिए :

(i) $4 x^{2}-\frac{1}{9}$

(ii) $9 x^{2}-(y+z)^{2}$

(iii) $(x+y)^{2}-9 z^{2}$

(iv) $x^{2}-\frac{y^{2}}{100}$

(v) $16 x^{2}-25 y^{2}$

(vi) $18 x^{2} a^{2}-32$

हल :

(i)

$\begin{aligned}4 x^{2}-\frac{1}{9} &=(2 x)^{2}-\left(\frac{1}{3}\right)^{2} \\&=\left(2 x+\frac{1}{3}\right)\left(2 x-\frac{1}{3}\right)\end{aligned}$

(ii)

$\begin{aligned}9 x^{2}-(y+z)^{2} &=(3 x)^{2}-(y+z)^{2} \\&=(3 x+y+z)(3 x-y-z)\end{aligned}$

(iii)

$\begin{aligned}(x+y)^{2}-9 z^{2} &=(x+y)^{2}-(3 z)^{2} \\&=(x+y+3 z)(x+y-3 z)\end{aligned}$

(iv)

$\begin{aligned}x^{2}-\frac{y^{2}}{100} &=(x)^{2}-\left(\frac{y}{10}\right)^{2} \\&=\left(x-\frac{y}{10}\right)\left(x+\frac{y}{10}\right)\end{aligned}$

(v)

$\begin{aligned}16 x^{2}-25 y^{2} &=(4 x)^{2}-(5 y)^{2} \\&=(4 x-5 y)(4 x+5 y)\end{aligned}$

(vi)

$\begin{aligned}18 x^{2} a^{2}-32 &=2\left[9 x^{2} a^{2}-16\right] \\&=2\left[(3 x a)^{2}-(4)^{2}\right] \\&=2(3 x a+4)(3 x a-4)\end{aligned}$

लघु उत्तरीय प्रश्न

प्रश्न 4

गुणनखण्ड ज्ञात कीजिए :

(i) $64 x^{2}-(7 y+4 z)^{2}$

(ii) $3 x^{3} y-243 x y^{3}$

(ii) $\frac{49}{81} x^{2}-\frac{1}{25}$

(iv) $a^{4}-b^{4}$

(v) $x^{4}-25 y^{4}$

हल : 

(i)

$\begin{aligned} 64 x^{2}-(7 y+4 z)^{2} &=(8 x)^{2}-(7 y+4 z)^{2} \\ &=(8 x-7 y-4 z)(8 x+7 y+4 z) \end{aligned}$

 

(ii) 

$\begin{aligned} 3 x^{3} y-243 x y^{3} &=3 x y\left(x^{2}-81 y^{2}\right) \\ &=3 x y\left[x^{2}-(9 y)^{2}\right] \\ &=3 x y(x+9 y)(x-9 y) . \end{aligned}$

 

(iii) 

$\begin{aligned} \frac{49}{81} x^{2}-\frac{1}{25} &=\left(\frac{7}{9} x\right)^{2}-\left(\frac{1}{5}\right)^{2} \\ &=\left(\frac{7}{9} x+\frac{1}{5}\right)\left(\frac{7}{9} x-\frac{1}{5}\right) \end{aligned}$

(iv) 

$\begin{aligned} a^{4}-b^{4} &=\left(a^{2}\right)^{2}-\left(b^{2}\right)^{2} \\ &=\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right) \\ &=(a-b)(a+b)\left(a^{2}+b^{2}\right) \end{aligned}$

(v)

$\begin{aligned} x^{4}-25 y^{4} &=\left(x^{2}\right)^{2}-\left(5 y^{2}\right)^{2} \\ &=\left(x^{2}+5 y^{2}\right)\left(x^{2}-5 y^{2}\right) \end{aligned}$

प्रश्न 5

गुणनखण्ड ज्ञात कीजिए :

(i) $(x+4)^{2}-(r+8)^{2}$

(ii) $2 x^{4}-32 y^{4}$

(iii) $1-2 a b-\left(a^{2}+b^{2}\right)$

(iv) $2 x y-\left(x^{2}+y^{2}-z^{2}\right)$

हल : 

(i) $\begin{aligned}(x+4)^{2}-(x+8)^{2} &=(x+4+x+8)(x+4-x-8) \\&=(2 x+12)(-4) \\&=-8(x+6)\end{aligned}$

(ii)

$\begin{aligned}2 x^{4}-32 y^{4} &=2\left[x^{4}-16 y^{4}\right]=2\left[\left(x^{2}\right)^{2}-\left(4y^{2}\right)^{2}\right] \\&=2\left[\left(x^{2}-4 y^{2}\right)\left(x^{2}+4 y^{2}\right)\right] \\&=2\left[(x)^{2}-(2 y)^{2}\right]\left(x^{2}+4 y^{2}\right) \\&=2(x-2 y)(x+2 y)\left(x^{2}+4 y^{2}\right)\end{aligned}$

(iii)

$\begin{aligned}1-2 a b-\left(a^{2}+b^{2}\right) &=1-\left(a^{2}+b^{2}+2 a b\right) \\&=(1)^{2}-(a+b)^{2} \\&=(1+a+b)(1-a-b)\end{aligned}$

(iv)

$\begin{aligned} 2 x y-\left(x^{2}+y^{2}-z^{2}\right) &=2 x y-x^{2}-y^{2}+z^{2} \\ &=z^{2}-\left(x^{2}+y^{2}-2 x y\right) \\ &=z^{2}-(x-y)^{2} \\ &=(z+x-y)(z-x+y) \end{aligned}$

प्रश्न 6

गुणनखण्ड ज्ञात कीजिए :

(i) $16 x^{4}-y^{4}$

(ii) $x^{8}-\frac{1}{x^{8}}$

हल : 

(i) 

$\begin{aligned} 6 x^{4}-y^{4} &=\left(4 x^{2}\right)^{2}-\left(y^{2}\right)^{2} \\ &=\left(4 x^{2}-y^{2}\right)\left(4 x^{2}+y^{2}\right) \\ &=(2 x-y)(2 x+y)\left(4 x^{2}+y^{2}\right) \end{aligned}$

हल : 

(ii) 

$\begin{aligned} x^{8}-\frac{1}{x^{8}} &=\left(x^{4}\right)^{2}-\left(\frac{1}{x^{4}}\right)^{2} \\ &=\left(x^{4}-\frac{1}{x 4}\right)\left(x^{4}+\frac{1}{x^{4}}\right) \\ &=\left(x^{2}-\frac{1}{x^{2}}\right)\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{4}+\frac{1}{x^{4}}\right) \\ &=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{4}+\frac{1}{x^{4}}\right) . \end{aligned}$

प्रश्न 7

गुणनखण्ड कीजिए :

(i) $x^{3}-2 x^{2}-x+2$

(ii) $2 y^{3}+y^{2}-2 y-1$

हल : 

(i)

$\begin{aligned}x^{3}-2 x^{2}-x+2 &=\left(x^{3}-2 x^{2}\right)-(x-2) \\&=x^{2}(x-2)-1(x-2) \\&=\left(x^{2}-1\right)(x-2) \\&=(x-1)(x+1)(x-2)\end{aligned}$

(ii)

 $\begin{aligned} 2 y^{3}+y^{2}-2 y-1 &=\left(2 y^{3}+y^{2}\right)-(2 y+1) \\ &=y^{2}(2 y+1)-1(2 y+1) \\ &=\left(y^{2}-1\right)(2 y+1) \\ &=(y+1)(y-1)(2 y+1) \end{aligned}$