RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL MCQ

 MULTIPLE CHOICE QUESTIONS

PAGE NO-138

Question 1:

If (x + 1) is factor of the polynomial (2x² + kx) then the value of k is
(a) –2
(b) –3
(c) 2
(d) 3

Answer 1:

(c) 2

$$\begin{gathered} \left(x+1\right) \text{is a factor of} 2x^2+kx. \\ \mathrm{So}, -1 \mathrm{is} \mathrm{a} \mathrm{zero} \mathrm{of} 2x^2+kx. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ 2\times {\left(-1\right)}^2+k\times \left(-1\right)=0 \\ \Rightarrow 2-k=0 \\ \Rightarrow k=2 \end{gathered}$$

Question 2:

The value of (249)² – (248)² is
(a) 1²
(b) 477
(c) 487
(d) 497

Answer 2:

(249)² – (248)²
We know
^{$$\begin{gathered} a^2-b^2=\left(a+b\right)\left(a-b\right) \\ So, \\ {\left(249\right)}^2-{\left(248\right)}^2=\left(249-248\right)\left(249+248\right)=497 \end{gathered}$$}
Hence, the correct answer is option (d).

Question 3:

If $\frac{x}{y}+\frac{y}{x}=-1$, where x ≠ 0 and y ≠ 0, then the value of (x³ − y³) is
(a) 1
(b) −1
(c) 0
(d) $\frac{1}{2}$

Answer 3:

 (c) 0

$$\begin{gathered} \frac{x}{y}+\frac{y}{x}=-1 \\ \Rightarrow \frac{x^2+y^2}{xy}=-1 \end{gathered}$$
$\Rightarrow x$² + y² = $-$xy
⇒ x² + y² + xy = 0

Thus, we have:
$\left(x^3-y^3\right)=\left(x-y\right)\left(x^2+y^2+xy\right)$
         $$\begin{gathered} =\left(x-y\right)\times 0 \\ =0 \end{gathered}$$

Question 4:

If a + b + c = 0, then a³ + b³ + c³ = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc

Answer 4:

(d) 3abc

$$\begin{gathered} a+b+c=0 \\ \Rightarrow a+b=-c \end{gathered}$$

$$\begin{gathered} \Rightarrow {\left(a+b\right)}^3={\left(-c\right)}^3 \\ \Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3 \\ \Rightarrow a^3+b^3+3ab\left(-c\right)=-c^3 \\ \Rightarrow a^3+b^3+c^3=3abc \end{gathered}$$

PAGE NO-139

Question 5:

If $\left(3x+\frac{1}{2}\right) \left(3x-\frac{1}{2}\right)=9x^2-p$ then the value of p is
(a) ⁰
(b) $-\frac{1}{4}$
(c) $\frac{1}{4}$
(d) $\frac{1}{2}$

Answer 5:

$\left(3x+\frac{1}{2}\right) \left(3x-\frac{1}{2}\right)=9x^2-p$
$$\begin{gathered} 9x^2-\frac{1}{4}=9x^2-p \left(\because \left(a^2-b^2\right)=\left(a+b\right)\left(a-b\right)\right) \\ \Rightarrow p=\frac{1}{4} \end{gathered}$$
Hence, the correct answer is option (c).

Question 6:

The coefficient of x in the expansion of (x + 3)³ is
(a) 1
(b) 9
(c) 18
(d) 27

Answer 6:

(x + 3)³
$$\begin{gathered} =x^3+3^3+9x\left(x+3\right) \\ =x^3+27+9x^2+27x \end{gathered}$$
So, the coefficient of x in (x + 3)³ is 27.
Hence, the correct answer is option (d). 

Question 7:

Which of the following is a factor of (x + y)³ – (x³ + y³)?
(a) x² + y² + 2xy
(b) x² + y² – xy
(c) xy²
(d) 3xy

Answer 7:

(x + y)³ – (x³ + y³)
$$\begin{gathered} =x^3+y^3+3xy\left(x+y\right)-\left(x^3+y^3\right) \\ =3xy\left(x+y\right) \end{gathered}$$
Thus, the factors of (x + y)³ – (x³ + y³) are 3xy and (x + y).
Hence, the correct answer is option (d). 
 

Question 8:

One of the factors of $\left(25x^2-1\right)+{\left(1+5x\right)}^2$ is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x
 

Answer 8:

$$\begin{gathered} \left(25x^2-1\right)+{\left(1+5x\right)}^2 \\ =\left(5x-1\right)\left(5x+1\right)+{\left(1+5x\right)}^2 \\ =\left(5x+1\right)\left[\left(5x-1\right)+\left(1+5x\right)\right] \\ =\left(5x+1\right)\left(10x\right) \end{gathered}$$
So, the factors of $\left(25x^2-1\right)+{\left(1+5x\right)}^2$ are (5x + 1) and 10x
Hence, the correct answer is option (d). 

Question 9:

If (x + 5) is a factor of p(x) = x³ − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

Answer 9:

(b) 5

$$\begin{gathered} \left(x+5\right) \text{is a factor of }p\left(x\right)=x^3-20x+5k. \\ \therefore p\left(-5\right)=0 \\ \Rightarrow {\left(-5\right)}^3-20\times \left(-5\right)+5k=0 \\ \Rightarrow -125+100+5k=0 \\ \Rightarrow 5k=25 \\ \Rightarrow k=5 \end{gathered}$$

Question 10:

If (x + 2) and (x − 1) are factors of (x³ + 10x² + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

Answer 10:

(b) m = 7, n = −18

Let:
$p\left(x\right)=x^3+10x^2+mx+n$
Now,
$x+2=0\Rightarrow x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0
$$\begin{gathered} \Rightarrow {\left(-2\right)}^3+10\times {\left(-2\right)}^2+m\times \left(-2\right)+n=0 \\ \Rightarrow -8+40-2m+n=0 \\ \Rightarrow 32-2m+n=0 \\ \Rightarrow 2m-n=32 .....\left(\mathrm{i}\right) \end{gathered}$$
Now,
$x-1=0\Rightarrow x=1$
Also, 
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0
$$\begin{gathered} \Rightarrow 1^3+10\times 1^2+m\times 1+n=0 \\ \Rightarrow 1+10+m+n=0 \\ \Rightarrow 11+m+n=0 \\ \Rightarrow m+n=-11 .....\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ 3m=21\Rightarrow m=7 \end{gathered}$$
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

Question 11:

104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884

Answer 11:

(b) 9984

$104\times 96=\left(100+4\right)\left(100-4\right)$
         $$\begin{gathered} ={100}^2-4^2 \\ =\left(10000-16\right) \\ =9984 \end{gathered}$$

Question 12:

305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840

Answer 12:

(c) 93940

$$\begin{gathered} 305\times 308=\left(300+5\right)\times \left(300+8\right) \\ ={\left(300\right)}^2+300\times \left(5+8\right)+5\times 8 \\ =90000+3900+40 \\ =93940 \end{gathered}$$

Question 13:

207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951

Answer 13:

(b) 39951

$$\begin{gathered} 207\times 193 \\ =\left(200+7\right)\left(200-7\right) \\ ={\left(200\right)}^2-{\left(7\right)}^2 \\ =40000-49 \\ =39951 \end{gathered}$$

Question 14:

4a² + b² + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)²
(b) (2a − b + 2)²
(c) (a + 2b + 2)²
(d) none of these

Answer 14:

(a) (2a + b + 2)²

$$\begin{gathered} 4a^2+b^2+4ab+8a+4b+4 \\ =4a^2+b^2+4+4ab+4b+8a \\ ={\left(2a\right)}^2+b^2+2^2+2\times 2a\times b+2\times b\times 2+2\times 2a\times 2 \\ ={\left(2a+b+2\right)}^2 \end{gathered}$$

Question 15:

(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these

Answer 15:

(c) (x − 7)(x + 3)

$x^2-4x-21=x^2-7x+3x-21$
             $$\begin{gathered} =x\left(x-7\right)+3\left(x-7\right) \\ =\left(x-7\right)\left(x+3\right) \end{gathered}$$

Question 16:

(4x² + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these

Answer 16:

(c) (2x + 3) (2x − 1)

$4x^2+4x-3=4x^2+6x-2x-3$
             $$\begin{gathered} =2x\left(2x+3\right)-1\left(2x+3\right) \\ =\left(2x+3\right)\left(2x-1\right) \end{gathered}$$

Question 17:

6x² + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these

Answer 17:

(b) (2x + 5)(3x + 1)

$6x^2+17x+5=6x^2+15x+2x+5$
              $$\begin{gathered} =3x\left(2x+5\right)+1\left(2x+5\right) \\ =\left(2x+5\right)\left(3x+1\right) \end{gathered}$$

Question 18:

(x + 1) is a factor of the polynomial
(a) x³ − 2x² + x + 2
(b) x³ + 2x² + x − 2
(c) x³ − 2x² − x − 2
(d) x³ − 2x² − x + 2

Answer 18:

(c) x³ − 2x² − x − 2

Let:
$f\left(x\right)=x^3-2x^2+x+2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3-2\times {\left(-1\right)}^2+\left(-1\right)+2 \\ =-1-2-1+2 \\ =-2\ne 0 \end{gathered}$$
Hence, (x + 1) is not a factor of $f\left(x\right)=x^3-2x^2+x+2$.

Now,
Let:
$f\left(x\right)=x^3+2x^2+x-2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3+2\times {\left(-1\right)}^2+\left(-1\right)-2 \\ =-1+2-1-2 \\ =-2\ne 0 \end{gathered}$$
Hence, (x + 1) is not a factor of $f\left(x\right)=x^3+2x^2+x-2$.

Now,
Let:
$f\left(x\right)=x^3+2x^2-x-2$
By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3+2\times {\left(-1\right)}^2-\left(-1\right)-2 \\ =-1+2+1-2 \\ =0 \end{gathered}$$
Hence, (x + 1) is a factor of $f\left(x\right)=x^3+2x^2-x-2$.

PAGE NO-140

Question 19:

3x³ + 2x² + 3x + 2 = ?
(a) (3x − 2)(x² − 1)
(b) (3x − 2)(x² + 1)
(c) (3x + 2)(x² − 1)
(d) (3x + 2)(x² + 1)

Answer 19:

(d) (3x + 2)(x² + 1)

$3x^3+2x^2+3x+2=x^2\left(3x+2\right)+1\left(3x+2\right)$
                   $=\left(3x+2\right)\left(x^2+1\right)$

Question 20:

If a + b + c = 0, then $\left(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\right)$=?

Answer 20:

(d) 3

$a+b+c=0\Rightarrow a^3+b^3+c^3=3abc$

Thus, we have:
$\left(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\right)=\frac{a^3+b^3+c^3}{abc}$
                     $$\begin{gathered} =\frac{3abc}{abc} \\ =3 \end{gathered}$$

Question 21:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x³ + y³ + z³ − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729

Answer 21:

(a) 108

$x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$
                   $$\begin{gathered} =\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right] \\ =9\times \left(81-3\times 23\right) \\ =9\times 12 \\ =108 \end{gathered}$$

Question 22:

If $\frac{a}{b}+\frac{b}{a}=-1$ then (a³ − b³) = ?
(a) −3
(b) −2
(c) −1
(d) 0

Answer 22:

 $$\begin{gathered} \frac{a}{b}+\frac{b}{a}=-1 \\ \Rightarrow \frac{a^2+b^2}{ab}=-1 \end{gathered}$$
$\Rightarrow a$² + b² = $-$ab
$\Rightarrow a$² + b² + ab = 0

Thus, we have:
$\left(a^3-b^3\right)=\left(a-b\right)\left(a^2+b^2+ab\right)$
         $$\begin{gathered} =\left(a-b\right)\times 0 \\ =0 \end{gathered}$$