RD Sharma 2020 solution class 9 chapter 6 Factorization of polynomial Expressions FBQS

FBQS

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Question 1:

7 – 9x + 2x² is called a _________ polynomial.

Answer 1:

Polynomial with degree 2 is known as a quadratic polynomial.

Hence, 7 – 9x + 2x² is called a quadratic polynomial.

Question 2:

12x – 7x² + 4 – 2x³ is called a ___________ polynomial.

Answer 2:

Polynomial with degree 3 is known as a cubic polynomial.

​Hence, 12x – 7x² + 4 – 2x³ is called a cubic polynomial.

Question 3:

13x² – 88x³ + 9x⁴ is called a _________ polynomial.

Answer 3:

Polynomial with degree 4 is known as a biquadratic polynomial.

​Hence, 12x – 7x² + 4 – 2x³ is called a biquadratic polynomial.

Question 4:

If x + 1 is a factor of the polynomial ax³ + x² – 2x + 4a – 9, then a = __________.

Answer 4:

Let f(x) = ax³ + x² – 2x + 4a – 9

It is given that one factor of f(x) is (x + 1).

Therefore, $f\left(x\right)=0 \mathrm{when} x=-1$.

On putting x = –1 in f(x) = 0, we get

$$\begin{gathered} a{\left(-1\right)}^3+{\left(-1\right)}^2-2\left(-1\right)+4a-9=0 \\ \Rightarrow -a+1+2+4a-9=0 \\ \Rightarrow 3a-6=0 \\ \Rightarrow 3a=6 \\ \Rightarrow a=2 \end{gathered}$$

Hence, if x + 1 is a factor of the polynomial ax³ + x² – 2x + 4a – 9, then a = 2.

Question 5:

If 3x – 1 is a factor of the polynomial 81x³ – 45x² +3a – 6, then the value of a is ___________.

Answer 5:

Let f(x) = 81x³ – 45x² +3a – 6

It is given that one factor of f(x) is (3x – 1).

Therefore, $f\left(x\right)=0 \mathrm{when} x=\frac{1}{3}$.

On putting x = $\frac{1}{3}$ in f(x) = 0, we get

$$\begin{gathered} 81{\left(\frac{1}{3}\right)}^3-45{\left(\frac{1}{3}\right)}^2+3a-6=0 \\ \Rightarrow 81\left(\frac{1}{27}\right)-45\left(\frac{1}{9}\right)+3a-6=0 \\ \Rightarrow 3-5+3a-6=0 \\ \Rightarrow 3a-8=0 \\ \Rightarrow 3a=8 \\ \Rightarrow a=\frac{8}{3} \end{gathered}$$

Hence, if 3x – 1 is a factor of the polynomial 81x³ – 45x² +3a – 6, then the value of a is $\underline{\frac{8}{3}}.$

Question 6:

The remainders obtained when x³ + x² – 9x – 9 is divided by x, x + 1 and x + 2 respectively are _________.

Answer 6:

Let f(x) = x³ + x² – 9x – 9

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x,

we use remainder theorem, put x = 0.

f(0) is the remainder.

Now,

f(0) = (0)³ + (0)² – 9(0) – 9
       = –9

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x is –9.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1,

put x +1 = 0.

f(–1) is the remainder.

Now,

f(–1) = (–1)³ + (–1)² – 9(–1) – 9
         = –1 + 1 + 9 – 9
         = 0

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1 is 0.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2,

put x +2 = 0.

f(–2) is the remainder.

Now,

f(–2) = (–2)³ + (–2)² – 9(–2) – 9
         = –8 + 4 + 18 – 9
         = 5

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2 is 5.


Hence, the remainders obtained when x³ + x² – 9x – 9 is divided by x, x + 1 and x + 2 respectively are –9, 0 and 5.

 

Question 7:

The remainder when f(x) = 4x³ – 3x² + 2x – 1 is divided by 2x + 1 is __________.

Answer 7:

Let f(x) = 4x³ – 3x² + 2x – 1

To find the remainder obtained when 4x³ – 3x² + 2x – 1 is divided by 2x + 1,

we use remainder theorem, put 2x + 1 = 0.

f($-\frac{1}{2}$) is the remainder.

Now,

$$\begin{gathered} f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^3-3{\left(-\frac{1}{2}\right)}^2+2\left(-\frac{1}{2}\right)-1 \\ =4\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)-1-1 \\ =-\frac{1}{2}-\frac{3}{4}-2 \\ =\frac{-2-3-8}{4} \\ =-\frac{13}{4} \end{gathered}$$

Hence, the remainder when f(x) = 4x³ – 3x² + 2x – 1 is divided by 2x + 1 is $\underline{-\frac{13}{4}}.$

Question 8:

The degree of a polynomial f(x) is 7 and that of polynomial f(x) g(x) is 56, then degree of g(x) is ________.

Answer 8:

Given:
Degree of a polynomial f(x) = 7
Degree of polynomial f(x) g(x) = 56

Degree of polynomial f(x) g(x) = Degree of ​polynomial f(x) × Degree of ​polynomial g(x)
⇒ 56 = 7 × Degree of ​polynomial g(x)
⇒ Degree of ​polynomial g(x) = $\frac{56}{7}$
⇒ Degree of ​polynomial g(x) = 8

Hence, degree of g(x) is 8.

Question 9:

The remainder when x¹⁵ is divided by x + 1 is __________.

Answer 9:

Let f(x) = x¹⁵

To find the remainder obtained when x¹⁵ is divided by x + 1,

we use remainder theorem, put x + 1 = 0.

f(−1) is the remainder.

Now,

$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^{15} \\ =-1 \end{gathered}$$

Hence, the remainder when x¹⁵ is divided by x + 1 is −1.

Question 10:

If $p\left(x\right)=x^2-4x+3, \mathrm{then} p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 10:

Let p(x) =  x² – 4x + 3

$$\begin{gathered} p\left(2\right)={\left(2\right)}^2-4\left(2\right)+3 \\ =4-8+3 \\ =-1 \\ p\left(-1\right)={\left(-1\right)}^2-4\left(-1\right)+3 \\ =1+4+3 \\ =8 \\ p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+3 \\ =\frac{1}{4}-2+3 \\ =\frac{1}{4}+1 \\ =\frac{1+4}{4} \\ =\frac{5}{4} \\ \mathrm{Now}, \\ p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4} \\ =-9+\frac{5}{4} \\ =\frac{-36+5}{4} \\ =-\frac{31}{4} \end{gathered}$$

Hence, if $p\left(x\right)=x^2-4x+3, \mathrm{then} p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\underline{-\frac{31}{4}}.$

Question 11:

If the polynomial f(x) = 5x⁵ – 3x³ + 2x² – k gives remainder 1 when divided by x + 1, then k = __________.

Answer 11:

Let f(x) = 5x⁵ – 3x³ + 2x² – k

To find the remainder obtained when 5x⁵ – 3x³ + 2x² – k is divided by x + 1,

we use remainder theorem, put x + 1 = 0.

f(−1) is the remainder.

Now,

$$\begin{gathered} f\left(-1\right)=5{\left(-1\right)}^5-3{\left(-1\right)}^3+2{\left(-1\right)}^2-k \\ \Rightarrow 1=5\left(-1\right)-3\left(-1\right)+2\left(1\right)-k \\ \Rightarrow 1=-5+3+2-k \\ \Rightarrow 1=-k \\ \Rightarrow k=-1 \end{gathered}$$

Hence, k = –1.

Question 12:

The remainder when f(x) = x⁴⁵ is divided by x² – 1 is  ____________.

Answer 12:

Let f(x) =  x⁴⁵

To find the remainder obtained when  x⁴⁵ is divided by  x² – 1,

Let the remainder (r) be ax + b.

Then,
f(x) =  (x² – 1) q + r
$$\begin{gathered} \Rightarrow x^{45}=\left(x^2-1\right) q+\left(ax+b\right) ...\left(1\right) \\ \mathrm{Putting} x=1 \mathrm{in} \mathrm{equation} \left(1\right), \mathrm{we} \mathrm{get} \\ {\left(1\right)}^{45}=\left({\left(1\right)}^2-1\right) q+\left(a\left(1\right)+b\right) \\ \Rightarrow 1=0+a+b \\ \Rightarrow a+b=1 ...\left(2\right) \\ \mathrm{Putting} x=-1 \mathrm{in} \mathrm{equation} \left(1\right), \mathrm{we} \mathrm{get} \\ {\left(-1\right)}^{45}=\left({\left(-1\right)}^2-1\right) q+\left(a\left(-1\right)+b\right) \\ \Rightarrow -1=0-a+b \\ \Rightarrow -a+b=-1 ...\left(3\right) \\ \mathrm{Solving} \left(2\right) \mathrm{and} \left(3\right), \mathrm{we} \mathrm{get} \\ b=0 \mathrm{and} a=1 \end{gathered}$$

Therefore, the remainder (r) = 1(x) + 0 = x.

Hence, the remainder when f(x) = x⁴⁵ is divided by x² – 1 is x.