R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 6 - Factorization Of Polynomials

Answer:

(a) $\frac{x^2}{2}-\frac{2}{x^2}=\frac{x^2}{2}-2x^{-2}$

Exponent of x cannot be negative.
Therefore, it is not a polynomial.

(b) $\sqrt{2x}-1$

Exponent of x must be a whole number.
Therefore, it is not a polynomial.

(c) $x^2+\frac{3x^{3/2}}{\sqrt{x}}=x^2+3x^{\frac{3}{2}-\frac{1}{2}}=x^2+3x$

It is a polynomial.

(d)  $\frac{x-1}{x+1}$

It is not a polynomial.


Hence, the correct option is (c).



 

Answer:

Given: f(x) = 4x⁴ + 0x³ + 0x⁵ + 5x + 7 = 4x⁴ + 5x + 7

Degree is the highest power of x in the polynomial.

Here, the highest power of x is 4.

Thus, degree of the polynomial is 4.

Hence, the correct option is (a).

Answer:

The zero polynomial is a polynomial in which all the coefficients are zero.
We can't say about powers of x.

Thus, degree of the zero polynomial is not defined.

Hence, the correct option is (d).

Answer:

Given: f(x) = $\sqrt{2}$
It can also be written as $f\left(x\right)=\sqrt{2}{x}^0$.

Degree is the highest power of x in the polynomial.

Here, the highest power of x is 0.

Thus, degree of the polynomial is 0.

Hence, the correct option is (b).

Answer:

Given: zero polynomial i.e., f(x) = 0

Zero of a polynomial is the value of x at which the polynomial becomes zero.

Here, any value of x can be a zero of the zero polynomial.

Hence, the correct option is (c).

Answer:

Given: f(x) = x + 3

f(–x) = –x + 3

Thus, 
f(x) + f(–x) = x + 3 + (–x + 3)
                  = x + 3 – x + 3
                  = 6 

Hence, the correct option is (d).

Answer:

Given: f(x) = 3x + 7

Zero of a polynomial is the value of x at which the polynomial becomes zero.

$$\begin{gathered} f\left(x\right)=0 \\ \Rightarrow 3x+7=0 \\ \Rightarrow 3x=-7 \\ \Rightarrow x=-\frac{7}{3} \end{gathered}$$

Therefore, zero of the polynomial f(x) = 3x + 7 is $-\frac{7}{3}$.

Hence, the correct option is (c).

Answer:

Given: f(x) = 2x² + 7x – 4

Zero of a polynomial is the value of x at which the polynomial becomes zero.

$$\begin{gathered} f\left(x\right)=0 \\ \Rightarrow 2x^2+7x-4=0 \\ \Rightarrow 2x^2+8x-x-4=0 \\ \Rightarrow 2x\left(x+4\right)-1\left(x+4\right)=0 \\ \Rightarrow \left(2x-1\right)\left(x+4\right)=0 \\ \Rightarrow \left(2x-1\right)=0 \mathrm{or} \left(x+4\right)=0 \\ \Rightarrow 2x=1 \mathrm{or} x=-4 \\ \Rightarrow x=\frac{1}{2} \mathrm{or} x=-4 \end{gathered}$$

Therefore, zeroes of the polynomial f(x) = 2x² + 7x – 4 are $\frac{1}{2} \mathrm{and} -4.$
​
Hence, the correct option is (b).

Answer:

Given: f(x) = $x^2-2\sqrt{2}x+1$

$$\begin{gathered} f\left(2\sqrt{2}\right)={\left(2\sqrt{2}\right)}^2-2\sqrt{2}\left(2\sqrt{2}\right)+1 \\ ={\left(2\sqrt{2}\right)}^2-{\left(2\sqrt{2}\right)}^2+1 \\ =1 \end{gathered}$$
​
Hence, the correct option is (b).

Answer:

The polynomial f(x) which becomes zero when x = −1 is the required polynomial whose factor is x + 1.

(a) Let f(x) = x³ + x² – x + 1

f(−1) = (−1)³ + (−1)² − (−1) + 1
         = −1 + 1 + 1 + 1 = 2 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

(b) Let f(x) = x³ + x² + x + 1

f(−1) = (−1)³ + (−1)² + (−1) + 1
         = −1 + 1 − 1 + 1 = 0

Therefore, x + 1 is a factor of the polynomial,

(c) Let f(x) = x⁴ + x³ + x² + 1

f(−1) = (−1)⁴ + (−1)³ + (−1)² + 1
         = 1 − 1 + 1 + 1 =  2 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

(d) Let f(x) = x⁴ + 3x³ + 3x² + x + 1

f(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1
         = 1 − 3 + 3 − 1 + 1 =  1 ≠ 0

Therefore, x + 1 is not a factor of the polynomial,

Hence, the correct option is (b).

Answer:

Given: x² + kx + 6 = (x + 2) (x + 3)

$$\begin{gathered} x^2+kx+6=\left(x+2\right)\left(x+3\right) \\ \Rightarrow x^2+kx+6=x^2+2x+3x+6 \\ \Rightarrow x^2+kx+6=x^2+5x+6 \\ \Rightarrow kx=5x \\ \Rightarrow k=5 \end{gathered}$$

Hence, the correct option is (c).

Answer:

As is a factor of

i.e.

Hence, correct option is (d).

Answer:

As is exactly divisible by

i.e., is a factor of f(x).

Therefore,

Hence, the correct option is (c).

Answer:

As is a factor of 

i.e. 

Thus, b = 0

Thus, the correct option is (a).

Answer:

As is divisible by

i.e., is a factor of f(x).

Therefore,

Hence, the correct option is (a).

Answer:

As is a factor of

Therefore,

Hence, the correct option is (c).

Answer:

As is a factor of polynomial

i.e.

Therefore,

Hence, the correct option is (a)

Answer:

As is divided by

The remainder will be

Hence, the correct option is (d).

Answer:

As is a factor of polynomial Therefore,

So,

Hence, the correct option is (d).

Answer:

Asis a factor of polynomial

Therefore, 

Hence, the correct option is (a).

Answer:

As is a factor of polynomial f(x) = 4x³ + 3x² − 4x + k

Therefore,

Hence, the correct option is (d).

Answer:

It is given and are the factors of the polynomial

i.e., and

Now

$$\begin{gathered} -8+40-2m+n=0 \\ \Rightarrow -2m+n=-32 \\ \Rightarrow 2m-n=32 ...\left(i\right) \end{gathered}$$

And

Solving equation (i) and (ii) we get

m = 7 and n = − 18

Hence, the correct option is (c)

Answer:

Let f(x) be a polynomial such that

i.e., is a factor.

On rearranging can be written as

Thus, is a factor of f(x).

Hence, the correct option is (b).

Answer:

If the reminder (x −6) is subtracted from the given polynomial then rest of part of this polynomial is exactly divisible by x² − 2x − 3.

Therefore,

Now,

Therefore, are factors of polynomial p(x).

Now,

And

and

Solving (i) and (ii) we get

Hence, the correct option is (c).

Answer:

It is given that is a factor of the polynomial

Here, reminder is zero. Therefore, is a factor of polynomial.

Thus, the correct option is (a).

Answer:

Asis a factor of polynomial f(x) but not of g(x)

Therefore

Now,

Let

Now

Therefore (x − 1) is also a factor of f(x).g(x).

Hence, the correct option is (a).

Answer:

The linear polynomial is a factor of only if

If n is odd integer, then

Hence, the correct option is (a).

Answer:

Let be the given polynomial,

Since is the factor of f(x). Therefore, re`maider will be zero.

Now,

Now,

Hence, the correct option is (b).

Answer:

Given that,

Putting

We get

Hence, the correct option is (c).

Answer:

As and are the factors of the polynomial

i.e., and

Now,

And 

From equation (i) and (ii), we get

Hence, the correct option is (a).

Answer:

Asis a factor of polynomial

Therefore, 

And 

f(1) = 0

$$\begin{gathered} a{\left(1\right)}^4+b{\left(1\right)}^3+c{\left(1\right)}^2+d\left(1\right)+e=0 \\ \Rightarrow a+b+c+d+e=0 \end{gathered}$$

And

Hence,

The correct option is (a).

Answer:

Given: f(x + 3) = x² − 7x + 2 and divisor is (x + 1).

f(x + 3) = x² − 7x + 2
Replace x be x − 3,
⇒f(x) = (x − 3)² − 7(x − 3) + 2     .....(1)
 

Answer:

Given: $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$
$$\begin{gathered} f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)-2\left(x\right)\left(\frac{1}{x}\right) \\ \Rightarrow f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)-2 \\ \Rightarrow f\left(x+\frac{1}{x}\right)={\left(x+\frac{1}{x}\right)}^2-2 \\ \mathrm{Let} t =x+\frac{1}{x} \\ \Rightarrow f\left(t\right)=t^2-2 \\ \mathrm{Now}, \mathrm{replace} t \mathrm{with} x \\ \Rightarrow f\left(x\right)=x^2-2 .....\left(1\right) \end{gathered}$$
 

Answer:

Given: $f\left(x-\frac{1}{x}\right)=x^2+\frac{1}{x^2}$
$$\begin{gathered} f\left(x-\frac{1}{x}\right)=x^2+\frac{1}{x^2}+2\left(x\right)\left(\frac{1}{x}\right)-2\left(x\right)\left(\frac{1}{x}\right) \\ f\left(x-\frac{1}{x}\right)=x^2+\frac{1}{x^2}-2\left(x\right)\left(\frac{1}{x}\right)+2 \\ f\left(x-\frac{1}{x}\right)={\left(x-\frac{1}{x}\right)}^2+2 \\ \mathrm{Let} t =x-\frac{1}{x} \\ \Rightarrow f\left(t\right)=t^2+2 \\ \mathrm{Now} \mathrm{replace} t \mathrm{with} x \\ \Rightarrow f\left(x\right)=x^2+2 .....\left(1\right) \end{gathered}$$
 

Answer:

Given: f(x – 2) = 2x² – 3x + 4, and the divisor is (x – 1).

f(x – 2) = 2x² – 3x + 4
Let t = x – 2 ⇒ t + 2 = x
f(t + 2– 2) = 2(t + 2)² – 3(t + 2) + 4
⇒ f(t) = 2(t + 2)² – 3(t + 2) + 4

Now, replacing t by x, we get
f(x) = 2(x + 2)² – 3(x + 2) + 4                 .....(1)
 

Answer:

Given: polynomial p(x) = ax² + bx + c is divided by x, x – 2 and x + 3, the remainders obtained are 7, 9 and 49 respectively.

Answer:

Given: (x – a) and (x – b) are factors of x² + ax + b
Let p(x) = x² + ax + b

According to the Factor theorem, if (x – a) and (x – b) are factors of x² + ax + b then a and b are roots of the equation x² + ax + b.
i.e., p(a) = 0 and p(b) = 0

$$\begin{gathered} \Rightarrow a^2+a\left(a\right)+b=0 \\ \Rightarrow 2a^2+b=0 \\ \Rightarrow b=-2a^2 .....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \Rightarrow b^2+a\left(b\right)+b=0 \\ \Rightarrow b\left(b+a+1\right)=0 \\ \Rightarrow \left(-2a^2\right)\left[-2a^2+a+1\right]=0 \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \left(-2a^2\right)\left(2a+1\right)\left(a-1\right)=0 \\ \Rightarrow \left(-2a^2\right)=0 \mathrm{or} \left(2a+1\right)=0 \mathrm{or} \left(a-1\right)=0 \\ \Rightarrow a=0 \mathrm{or} a=\frac{-1}{2} \mathrm{or} a=1 .....\left(2\right) \end{gathered}$$
Substituting (2) in (1),

$$\begin{gathered} \mathrm{At} a=0, b=0 \\ \mathrm{At} a=\frac{-1}{2}, b=-2{\left(\frac{-1}{2}\right)}^2 =\frac{-1}{2} \\ \mathrm{At} a=1, b=-2{\left(1\right)}^2=-2 \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Given: (x – a) and (x – b) are factors of  x² + ax – b
Let p(x) =  x² + ax – b

According to the Factor theorem, if (x – a) and (x – b) are factors of  x² + ax – b then a and b are roots of the equation  x² + ax – b.
i.e., p(a) = 0 and p(b) = 0

$$\begin{gathered} \Rightarrow a^2+a\left(a\right)-b=0 \\ \Rightarrow 2a^2-b=0 \\ \Rightarrow b=2a^2 .....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \Rightarrow b^2+a\left(b\right)-b=0 \\ \Rightarrow b\left(b+a-1\right)=0 \\ \Rightarrow \left(2a^2\right)\left[2a^2+a-1\right]=0 \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \left(2a^2\right)\left(2a-1\right)\left(a+1\right)=0 \\ \Rightarrow \left(2a^2\right)=0 \mathrm{or} \left(2a-1\right)=0 \mathrm{or} \left(a+1\right)=0 \\ \Rightarrow a=0 \mathrm{or} a=\frac{1}{2} \mathrm{or} a=-1 .....\left(2\right) \end{gathered}$$

Substituting (2) in (1),

$$\begin{gathered} \mathrm{At} a=0, b=0 \\ \mathrm{At} a=\frac{1}{2}, b=2{\left(\frac{1}{2}\right)}^2 =\frac{1}{2} \\ \mathrm{At} a=-1, b=2{\left(-1\right)}^2=2 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Given: The ratio of remainders when f(x) = x² + ax + b is divided by (x – 2) and (x – 3) respectively is 5 : 4.

According to the Factor theorem, if (x – a) is a factor of  f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (x – 1) is a factor of f(x) then f(1) = 0
⇒ f(1) = (1)² + a(1) + b
⇒ ​0 = 1 + a + b
⇒ −1 − a = b                  .....(1)

According to remainder theorem, if f(x) is divided by x − a, the remainder will be f(a).
Thus, remainders of  f(x) = x² + ax + b when divided by (x – 2) is f(2).
∴ f(2) = (2)² + a(2) + b
⇒ f(2) = 4 + 2a + b                  .....(2)

Thus, remainders of  f(x) = x² + ax + b when divided by (x – 3) is f(3).
∴ f(3) = (3)² + a(3) + b
⇒ f(3) = 9 + 3a + b                  .....(2)

Given,
$$\begin{gathered} \Rightarrow f\left(2\right):f\left(3\right)=5:4 \\ \Rightarrow \frac{f\left(2\right)}{f\left(3\right)}=\frac{5}{4} \\ \Rightarrow \frac{4+2a+b}{9+3a+b}=\frac{5}{4} \\ \Rightarrow 4\left(4+2a+b\right)=5\left(9+3a+b\right) \\ \Rightarrow 16+8a+4b=45+15a+5b \\ \Rightarrow 16+8a+4b=45+15a+5b \\ \Rightarrow 0=7a+b+29 \\ \Rightarrow 0=7a-1-a+29 \left[From \left(1\right)\right] \\ \Rightarrow 0=6a+28 \\ \Rightarrow 6a=-28 \\ \Rightarrow 3a=-14 \\ \Rightarrow a=\frac{-14}{3} .....\left(2\right) \\ \mathrm{Substituting} \left(2\right) \mathrm{in} \left(1\right), \\ b=-1-\left(\frac{-14}{3}\right) \\ \Rightarrow b=-1+\frac{14}{3} \\ \Rightarrow b=\frac{11}{3} \end{gathered}$$

Thus, $a=-\frac{14}{3}, b=\frac{11}{3}$
Hence, the correct answer is option (b).

Answer:

Given:  f(x) = x⁴⁵ + x²⁵ + x¹⁴ + x⁹ + x and divisor g(x) = x² – 1
 

Let the remainder is in the form of ax+b.
The function is divided by x² – 1 so it is a factor of the function.
Thus,
x² – 1 = 0
⇒ x² – 1² = 0
​⇒ (x – 1) (x + 1) = 0
​​⇒ (x – 1) = 0 or (x + 1) = 0
​​​⇒ x = 1 or x = –1

Now, by division algorithm,

Dividend = Divisor × Quotient + Remainder

Since f(x) = (x – 1) (x + 1) × q(x) + ax+b
At x = –1
∴  f(–1) = (–1 – 1) (–1 + 1) × q(–1) – a+b
​​​⇒ f(–1) =  –a+b                                          .....(1)

Putting,x= –1 in f(x)
 f(–1) = (–1)⁴⁵ + (–1)²⁵ + (–1)¹⁴ + (–1)⁹ + (–1) 
⇒ f(–1) = –1 – 1 + 1 – 1 – 1
⇒ f(–1) = –3                                                 .....(2)

From (1) and (2)
⇒ –a+b = –3                                               .....(3)

f(x) = (x – 1) (x + 1) × q(x) + ax+b
∴  f(1) = (1 – 1) (1 + 1) × q(1) + a+b
​​​⇒ f(1) =  a+b                                          .....(4)

Putting,x= 1 in f(x)
 f(1) = (1)⁴⁵ + (1)²⁵ + (1)¹⁴ + (1)⁹ + (1) 
⇒ f(1) = 1 + 1 + 1 + 1 + 1
⇒ f(1) = 5                                                 .....(5)

From (4) and (4)
⇒ a+b = 5                                             .....(6)

Adding (3) and (4),
b = 1
⇒ a = 4 
∴ The remainder = ax+b
                            = 4x + 1

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): If f(x) = ax² + b + c is exactly divisible by 2x – 3 then 4a + 6b + 9c = 0.

According to the Factor theorem, if (x – a) is a factor of  f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (2x – 3) is a factor of f(x) then f($\frac{3}{2}$) = 0
$$\begin{gathered} \Rightarrow f\left(\frac{3}{2}\right)=a{\left(\frac{3}{2}\right)}^2+b\left(\frac{3}{2}\right)+c \\ \Rightarrow 0=\frac{9}{4}a+\frac{3}{2}b+c \\ \Rightarrow 0=\frac{9a+6b+4c}{4} \\ \Rightarrow 0=9a+6b+4c \end{gathered}$$

Thus, Statement-2 is false.

Statement-1 (Assertion): If the polynomial p(x) = x³ + ax² – 2x + a + 4 has (x + a) as one of its factors, then $a=-\frac{4}{3}$.

Given that, p(x) = x³ + ax² – 2x + a + 4 has (x + a) as one of its factors.
According to the Factor theorem, if (x – a) is a factor of  f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (x + a) is a factor of p(x) then p(−a) = 0
⇒ p(−a) = (−a)³ + a(−a)² – 2(−a) + a + 4
⇒ 0 = −a³ + a³ + 2a + a + 4
⇒ 0 = 3a + 4
$\Rightarrow a=\frac{-4}{3}$

Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): If a polynomial is divided by (x – a), the remainder is f(a).

Answer:

Statement-2 (Reason):

Answer:

Statement-2 (Reason): If x – 2 and 2x – 1 are factors of f(x) = px² + 5x + r, then p = r.

Given that, x – 2 and 2x – 1 are factors of f(x) = px² + 5x + r
According to the Factor theorem, if (x – a) is a factor of  f(x) then a is the root of the equation.
i.e., f(a) = 0.
$\Rightarrow f\left(2\right)=0 \mathrm{and} f\left(\frac{1}{2}\right)=0$
So, f(2) = p(2)² + 5(2) + r
⇒ 0 = 4p + 10 + r                            .....(1)

$$\begin{gathered} f\left(\frac{1}{2}\right)=p{\left(\frac{1}{2}\right)}^2+5\left(\frac{1}{2}\right)+r \\ \Rightarrow 0=\frac{1}{4}p+\frac{5}{2}+r \\ \Rightarrow 0=p+10+4r .....\left(2\right) \end{gathered}$$

Multiplying (1) by 4, we get

 0 = 16p + 40 + 4r                            .....(3)

Subtracting (2) from (3),
0 − 0 = 16p + 40 + 4r − (p + 10 + 4r )
⇒ 0 = 16p + 40 + 4r − p − 10 − 4r
⇒ 0 = 15p + 30
⇒ p = −2                          .....(4)

From (4) and (1), we get
⇒ 0 = 4(−2 ) + 10 + r     
⇒ 0 = −8 + 10 + r    
⇒ 0 = 2 + r   
⇒ r = −2 
⇒ p = r = −2

Thus, Statement-2 is true.

Statement-1 (Assertion): If x +1 is a factor of f(x) = px² + 5x + r, then p + r + 5 = 0.

Given that, x +1 is a factor of f(x) = px² + 5x + r
According to the Factor theorem, if (x – a) is a factor of  f(x) then a is the root of the equation.
i.e., f(a) = 0.
$\Rightarrow f\left(-1\right)=0$
So, f(−1) = p(−1)² + 5(−1) + r
⇒ 0 = p − 5 + r

Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).












 

Answer:

Statement-2 (Reason): If f(x) is divisible by (ax + b), then $f\left(-\frac{b}{a}\right)=0.$
According to the Factor theorem, if (x – a) is a factor of  f(x) then a is a root of the equation.
i.e., f(a) = 0.

So, if (ax + b) is a factor of  f(x) then $\left(-\frac{b}{a}\right)$ is a root of the equation.
Thus, Statement-2 is true.

Statement-1 (Assertion): If x + 2a is a factor of f(x) = x⁵ – 4a²x³ + 2x + 2a + 3, then 2a – 3 = 0.

Given that,  x + 2a is a factor of f(x) = x⁵ – 4a²x³ + 2x + 2a + 3.
According to the Factor theorem, if (x – a) is a factor of  f(x) then a is a root of the equation.
i.e., f(a) = 0.
Since x + 2a is a factor of f(x) = x⁵ – 4a²x³ + 2x + 2a + 3 then –2a is the root of f(x), i.e., f(–2a) = 0.

⇒ f(–2​a) = (–2a)⁵ – 4a²(–2a)³ + 2(–2a) + 2a + 3
⇒ 0 = –32a⁵ + 32a⁵ – 4a + 2a + 3
⇒ 0 =  –2a + 3
⇒ 2a – 3 = 0

Thus, Statement-2 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).