SChand CLASS 9 Chapter 4 Factorisation TEST

 TEST

Question 1

$8 x^{2} y^{3}-x^{5}$

Sol :

$\begin{aligned} =& 8 x^{2} y^{3}-x^{5} \\=& x^{2}\left(8 y^{3}-x^{3}\right) \\=& x^{2}\left[(2 y)^{3}-(x)^{3}\right] \\=& x^{2}(2 y-3)\left(y y^{2}+2 x y+x^{2}\right) \end{aligned}$

Question 2

$x^{2}+\frac{1}{x^{2}}+2-2 x-\frac{2}{x}$

Sol :

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

Question 3

2x² – x – 6

Sol :

$=2 x^{2}-x-6$

$=2 x^{2}-4 x+3 x-6$

=2x(x-2)+3(x-2)

=(x-2)(2x+3)

Question 4

a³ – 0.216

Sol :

$\begin{aligned} =& a^{3}-0.226 \\=&(a)^{3}-(0.6)^{3} \\=&(a-0.6)\left(a^{2}+a \times 0.6+(0.6)^{2}\right] \\=&(a-0.6)\left(a^{2}+0.6 a+0.36\right) \end{aligned}$

Question 5

6x²y – xy – 2y

Sol :

$\begin{aligned} =& 6 x^{2} y-x y-2 y \\=& y\left[6 x^{2}-x-2\right) \\=& y\left[6 x^{2}-4 x+3 x-2\right] \\=& y[2 x(3 x-2)+2(3 x-2)\} \\=& y[2 x(3 x-2)+1(3 x-2)] \\=& y(3 x-2)(2 x+1) \end{aligned}$

Question 6

(x² – 3x)² – 8(x² – 3x) – 20

Sol :

$\left(x^{2}-3 x\right)^{2}-8\left(x^{2}-3 x\right)-20$

Let $x^{2}-3 x=y$, then

$\begin{aligned} =& y^{2}-8 y-20 \\=& y^{2}-10 y+2 y-20 \\=& y(y-10)+2(y-10) \\=&(y-10)(y+2) \end{aligned}$

Now,

$=\left(x^{2}-3 x-10\right)\left(x^{2}-3 x+2\right)$

$=\left\{x^{2}-5 x+2 x-10\right\}\left\{x^{2}-x-2 x+2\right\}$

$=\{x(x-5)+2(x-5)\}\{x(x-1)-2(x-1)\}$

$=(x-5)(x+2)(x-1)(x-2)$

Question 7

One of the factors of (x – 1) – (x² – 1) is

(a) x² – 1

(b) x + 1

(c) x – 1

(d) x + 4

Sol :

$=(x-1)-\left(x^{2}-1\right)$

=(x-1)-(x+1)(x-1)

=(x-1)[1-x+1]

So, x-1 is its factor 

option (C) is correct

Question 8

If $\frac{x}{y}+\frac{y}{x}=-1$ (x, y ≠ 0), then the value of x³ – y³ is

(a) 1

(b) – 1

(c) $\frac{1}{2}$

(d) 0

Sol :

$=\frac{x}{y}+\frac{y}{x}=-1$

$=\frac{x^{2}+y^{2}}{x y}=-1$

$=x^{2}+y^{2}=-x y$

$=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$....(formula)

=(x-y)×0=0

Option (d) is correct

Question 9

The product of $=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

(a) $x^4+\frac{1}{x^4}$

(b) $x^3+\frac{1}{x^3}-2$

(c) $x^{4}-\frac{1}{x^{4}}$

(d) $x^2+\frac{1}{x^2}+2$

Sol :

$=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

$=\left(x^{2}-\frac{1}{x^{2}}\right)\left(x^{2}+\frac{1}{x^{2}}\right)$

$=\left(x^{2}\right)^{2}-\left(\frac{1}{x^{2}}\right)^{2}=x^{4}-\frac{1}{x^{4}}$

Option (C) is correct

Question 10

If x – 2y= 11 and xy = 8, then the value of x³ – 8y³ is

(a) 1860

(b) 1600

(c) 1859

(d) 2000

Sol :

x-2y=11 , xy=8

x-2y=11

⇒$(x-2 y)^{3}=(11)^{3}$ {(i) using both sides}

⇒$x^{3}-8 y^{3}-3 x x \times(2 y)(x-2 y)=1331$

⇒$x^{3}-8 y^{3}-6 x y(x-2 y)=1331$

⇒$x^{3}-8 y^{3}-6 \times 8 \times 11=1331$

⇒$x^{3}-8 y^{3}-528=1331$

⇒$x^{3}-8 y^{3}=1331+528$

⇒$x^{3}-8 y^{3}=1859$

So, option (C) is correct