प्रश्नावली 2(I)
मध्य पद को दो भागों में बाँटकर निम्नलिखित बहुपदों के गुणनखण्ड ज्ञात कीजिए ( प्रश्न 1 से 12 तक ) :
प्रश्न 1
$12 x^{2}-7 x+1$
हल :
$\begin{aligned}&=12 x^{2}-7 x+1 \\&=12 x^{2}-3 x-4 x+1 \quad[p+q=-7, p q=12] \\&=3 x(4 x-1)-1(4 x-1)\end{aligned}$
=(4 x-1)(3 x-1)
प्रश्न 2
$3 x^{3}-x^{2}-10 x$
हल :
$\begin{aligned}&=3 x^{3}-x^{2}-10 x \\&=x\left(3 x^{2}-x-10\right) \\&=x\left[3 x^{2}-(6-5) x-10\right] \\3 \times 10 &=30=6 \times 5,6-5=1 \\&=x\left[3 x^{2}-6 x+5 x-10\right] \\&=x[3 x(x-2)+5(x-2)] \\&=x(x-2)(3 x+5)\end{aligned}$
प्रश्न 3
$3 x^{2}-x-4$
हल :
$\begin{aligned}&=3 x^{2}-x-4 \\&=3 x^{2}-4 x+3 x-4 \\&=x(3 x-4)+1(3 x-4) \\&=(3 x-4)(x+1)\end{aligned}$
प्रश्न 4
$\sqrt{3} y^{2}+11 y+6 \sqrt{3}$
हल :
$\begin{aligned}&=\sqrt{3} \times 6 \sqrt{3} \\&=18=2 \times 9,2+9=11 \\&=y(\sqrt{3 y}+2)+3 \sqrt{3}(\sqrt{3 y}+2) \\&=(\sqrt{3} y+2)(y+3 \sqrt{3})\end{aligned}$
प्रश्न 5
$3(a+b)^{2}-5(a+b)+2$
हल :
$3(a+b)^{2}-5(a+b)+2$
$=3(a+b)^{2}-3(a+b)-2(a+b)+2, \quad[\because 3 \times 2=6,3+2=5]$
=3(a+b)[a+b-1]-2[a+b-1]
=(a+b-1)[3(a+b)-2]
प्रश्न 6
$6-35 x-6 x^{2}$
हल :
$\begin{aligned} &=6-35 x-6 x^{2} \\&=6-(36-1) x-6 x^{2}, \quad[\because 6 \times 6=36=36 \times 1,36-1=35] \\ &=6-36 x+x-6 x^{2}\end{aligned}$
=6(1-6x)+x(1-6x)
=(1-6x)(6+x)
प्रश्न 7
$12 x^{2}-7 x+1$.
हल :
$\begin{aligned}&=12 x^{2}-7 x+1 \\&=12 x^{2}-4 x-3 x+1 \\&=4 x(3 x-1)-1(3 x-1) \\&=(4 x-1)(3 x-1)\end{aligned}$
प्रश्न 8
$3 u^{2}-10 u+8$
हल :
$\begin{aligned}&=3 u^{2}-10 u+8 \\&=3 u^{2}-6 u-4 u+8 \\&=3 u(u-2)-4(u-2) \\&=(u-2)(3 u-4)\end{aligned}$
[∵ p+q=-10, p q=24]
प्रश्न 9
$4 p^{2}-17 p-21$.
हल :
$\begin{aligned}&=4 p^{2}-17 p-21 \\&=4 p^{2}+4 p-21 p-21 \\&=4 p(p+1)-21(p+1) \\&=(p+1)(4 p-21)\end{aligned}$
$\left[\begin{array}{rl}p+q & =-17 \\p q & =4(-21) \\& =-84\end{array}\right]$
प्रश्न 10
$2 x^{2}+7 x+3$
हल :
$\begin{aligned}&=2 x^{2}+7 x+3 \\&=2 x^{2}+6 x+x+3 \\&=2 x(x+3)+1(x+3) \\&=(x+3)(2 x+1)\end{aligned}$
प्रश्न 11
$24 x^{2}-65 x+21 .$
हल :
$\begin{aligned}&=24 x^{2}-65 x+21 \\&=24 x^{2}-(56+9) x+21 \\24 \times 21 &=56 \times 9,56+9=65 \\&=24 x^{2}-56 x-9 x+21 \\&=8 x(3 x-7)-3(3 x-7) \\&=(3 x-7)(8 x-3)\end{aligned}$
प्रश्न 12
$6 x^{2}+5 x-6$
हल :
$\begin{aligned}&=6 x^{2}+5 x-6 \\&=6 x^{2}+9 x-4 x-6 \\&=3 x(2 x+3)-2(2 x+3) \\&=(2 x+3)(3 x-2)\end{aligned}$
गुणनखण्ड कीजिए ( प्रश्न 13 से 18 तक ) :
प्रश्न 13
$\sqrt{3 x^{2}}+11 x+6 \sqrt{3}$.
हल :
$\begin{aligned}&=\sqrt{3 x^{2}}+11 x+6 \sqrt{3} \\p+q &=11 \\p q &=\sqrt{3 \times 6} \sqrt{3}=18\end{aligned}$
$\begin{aligned} 11 &=9+2 \\ &=\sqrt{3 x^{2}}+11 x+6 \sqrt{3} \\ &=\sqrt{3 x^{2}}+9 x+2 x+6 \sqrt{3} \\ &=\sqrt{3 x^{2}}+3 \times \sqrt{3} \times \sqrt{3 x}+2 x+6 \sqrt{3} \\ &=\sqrt{3 x}(x+3 \sqrt{3})+2(x+3 \sqrt{3)}\\ &=(x+3 \sqrt{3})(\sqrt{3 x}+2) \end{aligned}$
प्रश्न 14
$x^{2}-\left(a-\frac{1}{a}\right) x-1$.
हल :
$\begin{aligned}&{x}^{2}-\left(a-\frac{1}{a}\right) x-1 \\&=x^{2}-a x+\frac{1}{a} x-1 \\&=x^{2}-a x+\frac{1}{a} x-\frac{1}{a} \times a \\&=x(x-a)+\frac{1}{a}(x-a) \\&=(x-a)\left(x+\frac{1}{a}\right) .\end{aligned}$
प्रश्न 15
$x^{2}+\frac{1}{x^{2}}-7\left(x-\frac{1}{x}\right)+8$.
हल :
$x^{2}+\frac{1}{x^{2}}-7\left(x-\frac{1}{x}\right)+8$
$\begin{aligned}=\left(x-\frac{1}{x}\right)^{2}+2-7\left(x-\frac{1}{x}\right)+8 \\=\left(x-\frac{1}{x}\right)^{2}-7\left(x-\frac{1}{x}\right)+10 \\ &=\left(x-\frac{1}{x}\right)^{2}-5\left(x-\frac{1}{x}\right)-2\left(x-\frac{1}{x}\right)+10 \\ &=\left(x-\frac{1}{x}\right)\left[x-\frac{1}{x}-5\right]-2\left[x-\frac{1}{x}-5\right] \\ &=\left(x-\frac{1}{x}-5\right)\left(x-\frac{1}{x}-2\right) \end{aligned}$
प्रश्न 16
$7 x^{2}+2 \sqrt{14} x+2$
हल :
$7 x^{2}+2 \sqrt{14 x}+2=7 x^{2}+\sqrt{14 x}+\sqrt{14 x}+2$
$\begin{aligned}&=7 x^{2}+\sqrt{7} \times \sqrt{2 x}+\sqrt{7} \times \sqrt{2 x}+2 \\&=\sqrt{7} \times \sqrt{7 x^{2}}+\sqrt{7} \times \sqrt{2 x}+\sqrt{7} \times \sqrt{2 x}+\sqrt{2 \times}\sqrt{2}\end{aligned}$
$=\sqrt{\overline{7}}(\sqrt{7 x}+\sqrt{2})+\sqrt{2}-\sqrt{7 x}+\sqrt{2})$
$=(\sqrt{7 x}+\sqrt{2})(\sqrt{7 x}+\sqrt{2})$
प्रश्न 17
$a x^{2}+\left(4 a^{2}-3 b\right) x-12 a b$.
हल :
$\begin{aligned}&=a x^{2}+\left(4 a^{2}-3 b\right) x-12 a b \\&=a x^{2}+4 a^{2} x-3 b x-12 a b \\&=a x(x+4 a)-3 b(x+4 a) \\&=(x+4 a)(a x-3 b)\end{aligned}$
प्रश्न 18
$x^{2}+\frac{12}{35} x+\frac{1}{35}$
हल :
$x^{2}+\frac{12}{35} x+\frac{1}{35}$
$\begin{aligned} p+q &=\frac{12}{35} \\ pq &=\frac{1}{35}=\frac{1}{5} \times \frac{1}{7} \\ \frac{12}{35} &=\frac{1}{5}+\frac{1}{7} \\ &=x^{2}+\frac{12}{35} x+\frac{1}{35} \\ &=x^{2}+\left(\frac{1}{5}+\frac{1}{7}\right) x+\frac{1}{35} \\ &=x^{2}+\frac{1}{5} x+\frac{1}{7} x+\frac{1}{35} \\ &=x\left(x+\frac{1}{5}\right)+\frac{1}{7}\left(x+\frac{1}{5}\right) \\ &=\left(x+\frac{1}{5}\right)\left(x+\frac{1}{7}\right)\end{aligned}$
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