NCERT Solutions For Class 9 Maths Chapter 2

NCERT Solutions Class 9 Maths Polynomials

Ncert Solutions For Class 9 Maths Chapter 2 PDF Download

NCERT maths solutions for Class 9 Chapter 2 Polynomials is one of the most important topic for the students to practice. For solving questions, it is crucial to have good knowledge of the topic. Students can study various concepts of maths by visiting our site BYJU’S. To ease the fear of maths for the students, we at BYJU’S provide NCERT Solution for Class 9 Maths Chapter 2 Polynomials. Student can download the NCERT Solution for class 9 Maths Chapter 2 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 9 maths chapter 2. NCERT solutions for chapter 2 class 9 maths pdf are prepared by our subject experts under the guidelines of NCERT to assists students in their examination. In maths, a polynomial is an equation comprising coefficients, variables and the operations of subtraction, addition, multiplication, and non-negative integer exponents. A polynomial consists of variables, coefficients, and addition, subtraction, multiplication, integer exponents, etc. Polynomials are a necessary language of algebra. They are employed in almost every field of maths to express mathematical operations. These are the building blocks of expressions, like rational expressions.

Various mathematical methods that are performed in daily life can be represented as polynomials. Adding the value of objects on a grocery bill can be described as a polynomial. Estimating the distance covered by a carrier can be represented as a polynomial. The value of an area, perimeter, and volume of geometric figures can be translated as polynomials. A numerical equation is a number expressed by constants, variables, and the mathematical processes conducted on them.

Some examples of Equations:

The polynomials can be recognised by seeing which equations contain the operations of subtraction, addition, non-negative integer exponents, and multiplication. All the non-polynomial equations will be the expressions which include extra operations. We have provided a comprehensive study material of NCERT Solutions For Class 9 Maths Polynomials. The students can also download the pdfs from the link provided above.

 

Exercise – 1

Q.1.Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:

(i)\( 4x^{2}-3x+7\)

Answer: It is a polynomial in one variable.

 

(ii) \( y^{2}+\sqrt{2}\)

Answer: It is a polynomial in one variable.

 

(iii) \( 3\sqrt{t}+t\sqrt{2}\)

Answer: It is not a polynomial since the power of the variable is not a whole number.

 

(iv) \( y+\frac{2}{y}\)

Answer: It is not a polynomial since the power of the variable is not a whole number.

 

(v) \( x^{10}+y^{3}+t^{50}\)

Answer: It is a polynomial in three variables.

 

Q.2. Write the coefficients of a2 in each of the following:

(i) \( 2 + a^{2} + a\)

Answer: Coefficient of \( a^{2}\) is 1.

 

 (ii) \( 2 – a^{2} + a^{3}\)

Answer: Coefficient of \( a^{2}\) is -1.

 

 (iii) \( \frac{\pi }{2}x^{2}+x\)

Answer: Coefficient of \( a^{2}\) is \( \frac{\pi }{2}\).


(iv)
\( \sqrt{2}x-1\)

Answer: Coefficient of \( a^{2}\) is 0.

 

Q.3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer: \( 3x^{35}+5\; and\; 4x^{100}\)

 

Q.4. Write the degree of each of the following polynomials:

(i) \( 5a^{3} + 4a^{2} + 7a\)

Answer : Degree is 3.
(ii) \( 3-b^{2}\)

Answer : Degree is 2.
(iii) \( 5t – \sqrt{8}\)

Answer : Degree is 1.

 

(iv)8

Answer : No Degree.

 

Q.5. Classify the following as linear, quadratic and cubic polynomial.

(i) \( a^{2} + a\)

Answer: Quadratic Polynomial

 

(ii)\( a-a^{3} \)

Answer: Cubic Polynomial

 

(iii) \( y + y^{2} +4\)

Answer: Quadratic Polynomial

 

(iv)1 + x

Answer: Linear Polynomial

 

(v) 3a

Answer: Linear Polynomial

 

(vi) \( a^{2}\)

Answer: Quadratic Polynomial
(vii) \( 6a^{3}\)

Answer: Cubic Polynomial

 

Exercise – 2

Q.1. Find the value of the polynomial at\( f(x)= 5a -4a^{2}+ 3\) at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Answer:

Let \( f(x)= 5a -4a^{2}+ 3\)

 

(i) When a=0

\( f(0) = 5(0) + 4(0)^{2} + 3= 3\)

 

(ii) When a=-1

\( f(a) = 5a + 4a^{2}+ 3\\ f(-1) = 5(-1) + 4(-1)^{2} + 3 = -5 – 4 + 3 = -6\)

 

(iii) When a=2

\( f(a) = 5a + 4a^{2} + 3\\ f(2) = 5(2) + 4(2)^{2} + 3 = 10 – 16 + 3 = -3\)

 

Q.2. Find p(0), p(1) and p(2) for each of the following polynomials:

 (i) \( p(y) = y^{2}- y + 1\)

Answer: \( p(y) = y2 – y + 1\\ ∴ p(0) = (0)^{2}- (0) + 1 = 1\\ p(1) = (1)^{2} – (1) + 1 = 1\\ p(2) = (2)^{2} – (2) + 1 = 3\)

 

 (ii) \( p(a) = 2 + a + 2a^{2}- a^{3}\)

Answer: \( p(a) = 2 +a + 2a^{2}-a^{3}\\ ∴ p(0) = 2 + 0 + 2 (0)^{2} – (0)^{3} = 2\\ p(1) = 2 + 1+ 2(1)^{2} – (1)^{3} = 2 + 1 + 2 – 1 = 4\\ p(2) = 2 + 2 + 2(2)^{2} – (2)^{3} = 2 + 2 + 8 – 8 = 4\)

 

(iii) \( p(x) = x^{3}\)

Answer: \( p(x) = x3\\ ∴ p(0) = (0)^{3}= 0\\ p(1) = (1)^{3} = 1\\ p(2) = (2)^{3} = 8\)

 

(iv) \( p(a) = (a- 1) (a+ 1)\)

Answer: \(p(a) = (a – 1) (a + 1)\\ ∴ p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1\\ p(1) = (1 – 1) (1 + 1) = 0 (2) = 0\\ p(2) = (2 – 1 ) (2 + 1) = 1(3) = 3\)

 

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) \(p(x) = 3x + 1, x = -\frac{1}{3}\)

Answer: For, \(x = -\frac{1}{3}\)

\(p(x) = 3x + 1\\ ∴ p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1=-1+1=0\)

\(∴ -\frac{1}{3}\) is a zero of p(x).

 

 (ii) \(p(x) = 5x – \pi , x = \frac{4}{5}\)

Answer: For, \(x = \frac{4}{5}\)

\(p(x) = 5x – \pi \\ ∴ p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi=4-\pi\)

\(∴ \frac{4}{5}\) is not a zero of p(x).

 

 (iii) \(p(x) = x^{2}- 1, x = 1, -1\)

Answer: For, \(x = 1,-1\)

\(p(x) = x^{2}- 1\\ ∴ p(1) = x^{1}- 1=1-1=0\\ p(-1) = x^{-1}- 1=1-1=0\)

\(∴ 1,-1\) are zeros of p(x).

 

(iv) \(p(x) = (x + 1) (x – 2), x = -1, 2\)

Answer: For, \(x = -1,2\)

\(p(x) = (x + 1) (x – 2)\\ ∴ p(-1) = (-1 + 1) (-1 – 2)\\ =((0)(-3))\\ =0\\ \\ p(2) = (2 + 1) (2 – 2)\\ =(3)(0)=0\)

\(∴ -1,2\) are zeros of p(x).

 

(v) \(p(x) = x^{2}, x = 0\)

Answer: For, \(x = 0\)

\(p(x) = 0^{2}=0\)

\(∴ 0\) is a  zero of p(x).

 

(vi) \(p(x)=lx+m,x=-\frac{m}{l}\)

Answer: For, \( x=-\frac{m}{l} \)

\(p(x)=lx+m\\ ∴ p(-\frac{m}{l} )=l(-\frac{m}{l} )+m=-m+m=0\)

\(∴-\frac{m}{l}\) is a zero of p(x).

 

(vii) \(p(x)=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}\)

Answer: For, \( x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}} \)
\(p(x)=3x^{2}-1\\ ∴ p(-\frac{1}{\sqrt{3}})=3(-\frac{1}{\sqrt{3}})^{2}-1\\ =3(\frac{1}{3})-1=1-1=0\\ \\ ∴ p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^{2}-1\\ =3(\frac{4}{3})-1=4-1=3\neq 0\)

\(∴ -\frac{1}{\sqrt{3}}\) is a zero of p(x) but \(\frac{2}{\sqrt{3}}\) is not a zero of p(x).

 

(viii) \(p(x) = 2x + 1, x = \frac{1}{2}\)
Answer: For, \( x=\frac{1}{2} \)

\( p(x) = 2x + 1\\ ∴ p(\frac{1}{2} ) = 2(\frac{1}{2}) + 1=1+1=2\neq 0\)

\(∴\frac{1}{2}  \) is not a  zero of p(x).

 

Q.4. Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5 

Answer: \( p(x) = x + 5 \\ \Rightarrow x + 5=0 \Rightarrow x=-5\)

\(∴\)-5 is a zero polynomial of the polinomila p(x).

 

(ii) p(x) = x – 5

Answer: \( p(x) = x -5 \\ \Rightarrow x -5=0 \Rightarrow x=5\)

\(∴\)5 is a zero polynomial of the polinomila p(x).

 

(iii) p(x) = 2x + 5

Answer: \( p(x) =2 x + 5 \\ \Rightarrow 2x + 5=0 \Rightarrow 2x=-5\\ \Rightarrow x=\frac{-5}{2} \)

\(∴\)\( x=\frac{-5}{2} \) is a zero polynomial of the polynomial p(x).

 

(iv)p(x) = 3x – 2 

Answer: \( p(x) =3 x – 2 \\ \Rightarrow 3x -2=0 \Rightarrow 3x=2\\ \Rightarrow x=\frac{2}{3} \)

\(∴\)\( x=\frac{2}{3} \) is a zero polynomial of the polynomial p(x).

 

 (v)p(x) = 3x 

Answer: \( p(x) =3 x \\ \Rightarrow 3x =0 \Rightarrow x=0 \)

\(∴\)0 is a zero polynomial of the polynomial p(x).

 

Exercise – 3

Q.1.Find the remainder when \(x^{3}+3x^{2}+3x+1\) is divided by

(i) x+1

Answer:

\(x+1=0\\ \Rightarrow x=-1\)

\(∴\;Remainder=p(-1)=(-1)^{3}+3(-1)^{2}+3(-1)+1=-1+3-3+1=0\)

 

(ii) \(x-\frac{1}{2}\)

Answer:

\(x-\frac{1}{2}=0\\ \Rightarrow x=\frac{1}{2}\)

\(∴\;Remainder (\frac{1}{2})^{3}+3(\frac{1}{2})^{2}+3(\frac{1}{2})+1\\ =\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{27}{8}\)

 

(iii) x

Answer:

\(∴\;Remainder= (0)^{3}+3(0)^{2}+3(0)+1\\ =1\)

 

(iv) \(x+\pi\)

Answer:

\(x+\pi=0\\ \Rightarrow x=-\pi\)

\(∴ \;Remainder= (-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1=-\pi ^{3}+3\pi ^{2}-3\pi +1\)

 

(v) 5+2x

Answer:

\(5+2x=0\\ \Rightarrow 2x=-5\Rightarrow x=-\frac{5}{2}\)

\(∴\; Remainder=(-\frac{5}{2})^{3}+3(-\frac{5}{2})^{2}+3(-\frac{5}{2})+1=-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1=-\frac{27}{8}\)

 

Q.2.Find the remainder when \(x^{3}-ax^{2}+6x-a\) is divided by x-a.

Answer:

Let \(p(x)=x^{3}-ax^{2}+6x-a\\ x-a=0\)

\(∴ x=a\)

\(Remainder=(a)^{2}-a(a^{2})+6(a)-a\\ =a^{3}-a^{3}+6a-a\\ =5a\)

 

Q.3.Check whether 7+3x is a factor of \(3x^{3}+7x\).

Answer:

\(7+3x=0\Rightarrow 3x=-7\) only if 7+3x divides \(3x^{3}+7x\) leaving no reaminder.

Let \(p(x)=3x^{3}+7x\)

\(7+3x=0\Rightarrow 3x=-7\Rightarrow x=-\frac{7}{3}\)

\(∴ Remainder=3(-\frac{7}{3})^{3}+7(-\frac{7}{3})=-\frac{343}{9}-\frac{49}{3}=-\frac{490}{9}\neq 0\)

\(∴\)7+3x is not a factor of \(3x^{3}+7x\)

 

Exercise – 4

Q.1. Determine which of the following polynomials has (x + 1) a factor:


(i)
\(x^{3} + x^{2}+ x + 1\)

Answer:

Let p(x)= \(x^{3} + x^{2}+ x + 1\)

The zero of x+1 is -1.

\(p(-1)=(-1)^{3} + (-1)^{2}+ (-1) + 1\\ =-1+1-1+1=0\)

\(∴\)By factor theorem, x+1 is a factor of \(x^{3} + x^{2}+ x + 1\)

 

(ii) x4 + x3 + x2 + x + 1

Answer:

Let p(x)= \(x^{4} + x^{3}+x^{2} +x + 1\)

The zero of x+1 is -1.

\(p(-1)=(-1)^{4} + (-1)^{3}+ (-1)^{2} + (-1)+1\\ =1-1+1-1+1=1\neq 0\) \(∴\)By factor theorem, x+1 is a factor of \(x^{4} + x^{3}+x^{2} +x + 1\)

 

 (iii) x4 + 3x3 + 3x2 + x + 1 

Answer:

Let p(x)= \(x^{4} +3 x^{3}+3x^{2} +x + 1\)

The zero of x+1 is -1.

\(p(-1)=(-1)^{4} + 3(-1)^{3}+ 3(-1)^{2} + (-1)+1\\ =1-3+3-1+1=1\neq 0\) \(∴\)By factor theorem, x+1 is a factor of \(x^{4} + 3x^{3}+3x^{2} +x + 1\)

 

(iv) \(x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}\)

Answer:

Let p(x)= \(x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}\)

The zero of x+1 is -1.

\(p(-1)=(-1)^{3} – (-1)^{2} – (2 + \sqrt{2})(-1) + \sqrt{2}\)

\(∴\)By factor theorem, x+1 is not a factor of \(x^{3} – x^{2} – (2 + \sqrt{2})x + \sqrt{2}\)

 

Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:


(i)
\(p(x) = 2x^{3} + x^{2} – 2x – 1\), g(x) = x + 1

Answer: \(p(x) = 2x^{3} + x^{2} – 2x – 1\), g(x) = x + 1

g(x)=0

\(\Rightarrow x+1=0\Rightarrow x=-1\)

\(∴\)Zero of g(x) is -1.

Now, \(p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1\\ =-2+1+2-1=0\)

\(∴\)By factor theorem, g(x) is a factor of p(x).
(ii) \( p(x) = x^{3}+ 3x^{2} + 3x + 1\), g(x) = x + 2

Answer: \(p(x) = x^{3} + 3x^{2} +3x + 1\), g(x) = x + 2

g(x)=0

\(\Rightarrow x+2=0\Rightarrow x=-2\)

\(∴\)Zero of g(x) is -2.

Now, \(p(-2) = (-2)^{3} + 3(-2)^{2} +3(-2) + 1\\ =-8+12-6+1=-1\neq=0\)

\(∴\)By factor theorem, g(x) is not a factor of p(x).
(iii) \( p(x) = x^{3} – 4 x^{2} + x + 6\), g(x) = x – 3

Answer: \(p(x) = x^{3} – 4x^{2} +x + 6\), g(x) = x -3

g(x)=0

\(\Rightarrow x-3=0\Rightarrow x=3\)

\(∴\)Zero of g(x) is 3.

Now, \(p(3) = (3)^{3} -4(3)^{2} +(3) + 6\\ =27-36+3+6=0\)

\(∴\)By factor theorem, g(x) is a factor of p(x).

 

Q.3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

(i)\( p(x) = x^{2}+ x + k\)

Answer: If x-1 is a factor of p(x), then p(1)=0   |By Factor Theorem \( \Rightarrow (1)^{2}+ (1) + k=0\Rightarrow 1+1+k=0\)

\( \Rightarrow2+k=0\Rightarrow k=-2\)

 

(ii) \( p(x) = 2x^{2} + kx + \sqrt{2}\)

Answer: If x-1 is a factor of p(x), then p(1)=0

\( \Rightarrow 2(1)^{2}+ k(1) +\sqrt{2} =0\Rightarrow 2+k+\sqrt{2}=0\Rightarrow k=-(2+\sqrt{2})\)

 

(iii) \( p(x) = kx^{2} – \sqrt{2}x + 1\)

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

\( \Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0\\ \Rightarrow k=\sqrt{2}-1\)
(iv) \( p(x) = kx^{2} – 3x + k\)

Answer: If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

\( \Rightarrow k(1)^{2} – 3(1) + k=0\Rightarrow k-3+k=0\\ \Rightarrow 2k-3=0\\ \Rightarrow k=\frac{3}{2}\)

 

Q.4. Factorize:

(i) \( 12x^{2} – 7x + 1\)

Answer:

\( 12x^{2} – 7x + 1=12x^{2} – 4x-3x + 1\\ =4x(3x-1)-1(3x-1) =(4x-1)(3x-1)\)

Let p(x)= \( 12x^{2} – 7x + 1\)

Then p(x)= \( 12(x^{2} – \frac{7}{12}x + \frac{1}{12})=12q(x)\)

Where q(x)= \( x^{2} – \frac{7}{12}x + \frac{1}{12}\)

By trial, we find that

\( q(\frac{1}{3})=(\frac{1}{3})^{2} – \frac{7}{12}(\frac{1}{3}) + \frac{1}{12}\\ =\frac{4-7+3}{36}=\frac{0}{36} =0\)

\( ∴\)By Factor Theorem,

\( (x-\frac{1}{3})\) is a factor of q(x).

Similarly, by trial, we find that

\( q(\frac{1}{4})=(\frac{1}{4})^{2} – \frac{7}{12}(\frac{1}{4}) + \frac{1}{12}\\ =\frac{3-7+4}{18}=\frac{0}{48} =0\)

\( ∴\)By Factor Theorem,

\( (x-\frac{1}{4})\) is a factor of q(x).

Therefore, \( 12x^{2}-7x+1=12(x-\frac{1}{3})(x-\frac{1}{4})\\ =12(\frac{3x-1}{3})(\frac{4x-1}{4})=(3x-1)(4x-1)\)

 

(ii)\( 2x^{2} + 7x + 3\)

Answer: 

\( 2x^{2} + 7x + 3=2x^{2} + 6x+x + 3\\ =2x(x+3) +1 (x + 3)\\ =(x+3)(2x+1)\)

Let p(x)= \( 2x^{2} + 7x + 3\)

Then p(x)= \( 2(x^{2} + \frac{7}2{}x + \frac{3}{2})=2q(x)\)
Where q(x)= \( x^{2} + \frac{7}2{}x + \frac{3}{2}\)

By trial, we find that

\(q(-3)=(-3)^{2} – \frac{7}{2}(-3) + \frac{3}{2}\\ =9-\frac{21}{2}+\frac{3}{2}=0\)

\( ∴\)By Factor Theorem,

\((x-3),i.e.(x+3)\) is a factor of q(x).

Similarly, by trial, we find that

\(q(-\frac{1}{2})=(-\frac{1}{2})^{2} +\frac{7}{12}(-\frac{1}{2}) + \frac{3}{2}\\ =\frac{1}{3}-\frac{7}{4}+\frac{3}{4} =0\)

\( ∴\)By Factor Theorem,

\((x-(-\frac{1}{2})),i.e.(x+\frac{1}{2}) \;is\; a\; factor\; of\; q(x).\\ Therefore, 2x^{2}+7x+3=2(x+3)(x+\frac{1}{2})\\ =2(x+3)(\frac{2x+1}{2})\\ =(x+3)(2x+1)\)

By Simpler method,

(iii) \(6x^{2}+5x-6 =6x^{2}+ 9x – 4x – 6

=3x (2x + 3) – 2 (2x + 3) = (2x + 3) (3x – 2)\)

 

(iv)3x2 – x – 4 =\(3x^{2} – x – 4 \\ =3×2 – 4x + 3x – 4 \\ = x (3x – 4) + 1 (3x – 4)\\ = (3x – 4) (x + 1)\)

 

Q.5. Factorize:


(i)
\(x^{3} – 2x^{2} – x + 2\)

Answer: Let \(p(x) = x^{3} – 2x^{2} – x + 2\)
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
\(p(x) = x^{3} – 2x^{2} – x + 2\)

\(p(-1) = (-1)^{3} – 2(-1)^{2} – (-1) + 2=-1-1+1+2=0\)

Therefore, (x+1) is the factor of  p(x)

1

Now, Dividend = Divisor × Quotient + Remainder

\((x+1) (x^{2} – 3x + 2)\\ = (x+1) (x^{2} – x – 2x + 2)\\ = (x+1) (x(x-1) -2(x-1))\\ = (x+1) (x-1) (x+2)\)

 

 (ii) \(x^{3} – 3x^{2} – 9x – 5\)
Answer: Let p(x) = \(x^{3} – 3x^{2} – 9x – 5\)
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = \(x^{3} – 3x^{2} – 9x – 5\)

p(5) = \((5)^{3} – 3(5)^{2} – 9(5) – 5=125-75-45-5=0\)

Therefore, (x-5) is the factor of  p(x)

2
Now, Dividend = Divisor × Quotient + Remainder

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

 

(iii) \(x^{3} + 13x^{2} + 32x + 20\)

Answer: Let p(x) = \(x^{3} + 13x^{2} + 32x + 20\)

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = \(x^{3} + 13x^{2} + 32x + 20\)

p(-1) = \((-1)^{3} + 13(-1)^{2} + 32(-1) + 20=-1+13-32+20=0\)

 Therefore, (x+1) is the factor of  p(x)

3

Now, Dividend = Divisor × Quotient + Remainder

\((x+1) (x^{2} + 12x + 20)\\ = (x+1) (x^{2} + 2x + 10x + 20)\\ = (x-5) {x(x+2) +10(x+2)}\\ = (x-5) (x+2) (x+10)\)

 

(iv) \(2y^{3} + y^{2} – 2y – 1\)
Answer: Let p(y) = \(2y^{3} + y^{2} – 2y – 1\)

Factors of ab = \(2\times (-1)\)= -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) = \(2y^{3} + y^{2} – 2y – 1\)

p(1) = \(2(1)^{3} + (1)^{2} – 2(1) – 1=2+1-2=0\)

Therefore, (y-1) is the factor of  p(y)

4

Now, Dividend = Divisor × Quotient + Remainder

\((y-1) (2y^{2} + 3y + 1)\\ = (y-1) (2y^{2} + 2y + y + 1)\\ = (y-1) (2y(y+1) +1(y+1))\\ = (y-1) (2y+1) (y+1)\)

 

Exercise – 5

Q.1.Use suitable identities to find the following products:

 

(i)(x + 4) (x + 10) 

Answer: \((x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 \times 10)\\ = x^{2} + 14x + 40\)
(ii)(x + 8) (x – 10)     

Answer: \((x + 8) (x -10 ) = x^{2} + (8+(- 10))x + (8\times(-10))\\ = x^{2} +(8-10)x – 80\\ =x^{2}-2x-80\)

 

(iii)(3x + 4) (3x – 5)

Answer: \((3x + 4)(3x -5) = (3x) ^{2} + {4 + (-5)}3x + {4×(-5)}\\ = 9x^{2} + 3x(4 – 5) – 20\\ = 9x^{2} – 3x – 20\)
(iv)\((y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2})\)

Answer: \((y^{2} + \frac{3}{2}) (y^{2} – \frac{3}{2}) = (y^{2})^{2} – (\frac{3}{2})^{2}\\ = y^{4} – \frac{9}{4}\)

 

Q.2. Evaluate the following products without multiplying directly:

(i)103 × 107

Answer:   \(103 \times 107 = (100 + 3)\times (100 + 7)\\ = (100)^{2} + (3 + 7)(100 + (3 \times 7))\\ = 10000 + 1000 + 21 \\ = 11021\)

 

(ii) 95 × 96  

Answer: \(95 \times 96 = (90 + 5) \times (90 + 6)\\ = (90)^{2} + 90(5 + 6) + (5 \times 6) \\ = 8100 + 990 + 30 = 9120\)

 

(iii) 104 × 96
Answer: \(104 \times 96 = (100 + 4) \times (100 – 4)\\ = (100)^{2} – (4)^{2}\\ = 10000 – 16 \\ = 9984\)

 

Q.3. Factorize the following using appropriate identities:

(i)\(9x^{2} + 6xy + y^{2}\)

Answer: \(9x^{2} + 6xy + y^{2} \\ = (3x) ^{2} + (2\times 3x\times y) + y^{2}\\ = (3x + y)^{2} \\ = (3x + y) (3x + y)\)

 

(ii) \(4y^{2}- 4y + 1\)

Answer: \(4y^{2}- 4y + 1 \\ =(2y)^{2} – (2\times 2y\times 1) + 1^{2}\\ = (2y – 1)^{2}\\ = (2y – 1) (2y – 1)\)

 

 (iii) \(x2 – \frac{y^{2}}{100}\)

Answer: \(x^{2} – (\frac{y}{10})^{2} = (x – \frac{y}{10}) (x + \frac{y}{10})\)

 

Q.4. Expand each of the following, using suitable identities:

(i) \((x + 2y + 4z)^{2}\)

(ii) \((2x- y + z)^{2}\)

(iii) \((-2x+3y+2z)^{2}\)

 (iv) \([\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}\)

Answer:

(i) \((x + 2y + 4z)^{2}\)

Using identity,

\((x + 2y + 4z)^{2} \\ = x^{2} + (2y)^{2} + (4z)^{2} + (2\times x\times 2y) + (2\times 2y\times 4z) + (2\times 4z\times x)\\ = x^{2} + 4y^{2} + 16z2 + 4xy + 16yz + 8xz\)
(ii)  \((2x- y + z)^{2}\)
Using identity,

\((2x-y + z)^{2}\\ = (2x)^{2} + (-y)^{2}+ z^{2} + (2\times 2x\times -y) + (2\times -y\times z) + (2\times z\times 2x) \\ = 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz\)

 

(iii) \((-2x+3y+2z)^{2}\)

Using identity,

\((-2x+ 3y + 2z)^{2} \\ =(-2x)^{2} + (3y)^{2} + (2z)^{2} + (2\times -2x\times 3y) + (2\times 3y\times 2z) + (2\times 2z\times -2x)\\ = 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz – 8xz\)
(iv) \([\frac{1 }{4}a – \frac{1}{2} b + 1]^{2}\)
 Using identity,
\([\frac{1}{4} a -\frac{ 1}{2} b + 1]^{2} = (\frac{1}{4} a)^{2} + (-\frac{1}{2} b)^{2} + 1^{2} + (2\times \frac{1}{4} a\times -\frac{1}{2} b) + (2\times -\frac{1}{2} b\times 1) + (2\times 1\times \frac{1}{4} a) \\ = \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 – \frac{1}{4} ab – b + \frac{1}{2} a\)

 

Q.5. Factorize:
(i) \(4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz\)

(ii) \(2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz\)
Answer:

(i) \(4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz\)

Answer:
\(4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz\)

\(= (2x)^{2} + (3y)^{2} + (-4z)^{2} + (2\times 2x\times 3y) + (2\times 3y\times -4z) + (2\times -4z\times 2x)\\ = (2x + 3y – 4z)^{2}\\ = (2x + 3y – 4z) (2x + 3y – 4z)\)
(ii) \(2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz\)
Answer:

\(2x^{2} + y^{2} + 8z^{2} – 2\sqrt{2} xy + 4\sqrt{2} yz – 8xz\)
\(= (-\sqrt{2}x)^{2} + (y)^{2} + (2\sqrt{2}z)^{2} + (2\times -\sqrt{2}x\times y) + (2\times y\times 2\sqrt{2}z) + (2\times 2\sqrt{2}z\times -\sqrt{2}x)\\ = (-\sqrt{2}x + y + 2\sqrt{2}z)^{2}\\ = (-\sqrt{2}x + y + 2\sqrt{2}z) (-\sqrt{2}x + y + 2\sqrt{2}z)\)

 

Q.6. Write the following cubes in expanded form:

(i)\((2x + 1)^{3}\)

(ii) \((2a -3b)^{3}\)

(iii) \([\frac{3}{2} x + 1]^{3}\)

(iv) \([ x -\frac{2}{3} y]^{3}\)

Answer:

(i)\((2x + 1)^{3}\)

\((2x + 1)^{3} = (2x)^{3} + 1^{3} + (3\times 2x\times 1)(2x + 1)\\ = 8x^{3} + 1 + 6x(2x + 1)\\ = 8x^{3} + 12x^{3} + 6x + 1\)

 

(ii) \((2a -3b)^{3}\)

\((2a-3b)^{3}=(2a)^{3}-(3b)^{3} – (3\times 2a\times 3b)(2a – 3b)\\ = 8a^{3} – 27b^{3} – 18ab(2a – 3b)\\ = 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}\)

 

(iii) \([\frac{3}{2} x + 1]^{3}\)

\([\frac{3}{2} x + 1]^{3} = (\frac{3}{2} x)^{3} + 1^{3} + (3\times\frac{ 3}{2} x\times 1)(\frac{3}{2} x + 1)\\ = \frac{27}{8} x^{3}+ 1 + \frac{9}{2}x (\frac{3}{2} x + 1)\\ = \frac{27}{8} x3 + 1 + \frac{27}{4}^{2} + \frac{9}{2} x\\ = \frac{27}{8} x3 + \frac{27}{4} ^{2} + \frac{9}{2} x + 1\)
(iv) \([ x -\frac{2}{3} y]^{3}\)
\([x – \frac{2}{3} y]^{3} = (x)^{3} – (\frac{2}{3} y)^{3} – (3\times x\times \frac{2}{3} y)(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y^{3} – 2xy(x – \frac{2}{3} y)\\ = (x)^{3} – \frac{8}{27}y3 – 2x^{2}y + \frac{4}{3}xy^{2}\)

 

Q.7. Evaluate the following using suitable identities: 

(i)\((99)^{3}\)

(ii)\((102)^{3}\)

(iii) \((998)^{3}\)

Answer:

(i) \((99)^{3} = (100 – 1)^{3}\)

\((100 – 1)^{3} = (100)^{3} – 1^{3} – (3\times 100\times 1)(100 – 1)\)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300 = 970299


(ii) \((102)^{3} = (100 + 2)^{3}\)
\((100 + 2)^{3} = (100)^{3} + 2^{3}+ (3\times 100\times 2)(100 + 2)\)
= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200= 1061208

 

(iii) \((998)^3= (1000 – 2)^{3}\)

\(= (1000)^{3} – 2^{3} – (3\times 1000\times 2)(1000 – 2)\)

= 1000000000 – 8 – 6000(1000 – 2)

= 1000000000 – 8- 6000000 + 12000 = 994011992

 

Q.8. Factorise each of the following:
(i)
\(8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}\)

(ii) \(8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}\)

(iii)27 – 125a3 – 135a + 225a2   

(iv) \(64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}\)

(v) \(27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p\)

Answer:

(i) \(8a^{3}+ b^{3} + 12a^{2}b + 6ab^{2}\)

\(8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}\\ = (2a)^{3} + b^{3} + 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a + b)^{3}\\ = (2a + b)(2a + b)(2a + b)\)
(ii) \(8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}\)

\(8a^{3} – b^{3}- 12a^{2}b + 6ab^{2}\\ = (2a)^{3} – b^{3} – 3(2a)^{2}b + 3(2a)(b)^{2}\\ = (2a – b)^{3}\\ = (2a – b)(2a – b)(2a – b)\)
(iii)27 – 125a3 – 135a + 225a2   

\(27 – 125a^{3} – 135a + 225a^{2}\\ = 3^{3} – (5a)^{3} – 3(3)^{2}(5a) + 3(3)(5a)^{2}\\ = (3 – 5a)^{3}\\ = (3 – 5a)(3 – 5a)(3 – 5a)\)

 

 (iv) \(64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}\)

\(64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}\\ = (4a)^{3} – (3b)^{3} – 3(4a)^{2}(3b) + 3(4a)(3b)^{2}\\ = (4a – 3b)^{3}\\ = (4a – 3b)(4a – 3b)(4a – 3b)\)
(v) \(27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p\) 

\(27p^{3} – \frac{1}{216}- \frac{9}{2} p^{2} + \frac{1}{4} p\)

\(= (3p)^{3} – (\frac{1}{6})^{3} – 3(3p)^{2}(\frac{1}{6}) + 3(3p)(\frac{1}{6})^{2}\\ = (3p – \frac{1}{6})^{3}\\ = (3p – \frac{1}{6})(3p – \frac{1}{6})(3p – \frac{1}{6})\)

 

Q.9. Verify:

(i) \(x^{3} + y^{3}\\ = (x + y) (x^{2} – xy + y^{2})\)

(ii) \(x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})\)

Answer:

(i) \(x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})\\ We \: know\: that, \\ (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)^{3} – 3xy(x + y)\\ \Rightarrow x^{3} +y^{3} = (x + y)[(x + y)^{2} – 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} +y^{3}= (x + y)[(x^{2} + y^{2} + 2xy) – 3xy] \\ \Rightarrow x^{3} +y^{3} = (x + y)(x^{2} + y^{2} – xy)\)
(ii) \(x^{3} – y^{3}\\ = (x – y) (x^{2} + xy + y^{2})\)
\(x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})\\ We \: know\: that, \\ (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)^{3} + 3xy(x – y)\\ \Rightarrow x^{3} -y^{3} = (x – y)[(x – y)^{2} + 3xy] {Taking\: (x+y) \: common}\\ \Rightarrow x^{3} -y^{3}= (x – y)[(x^{2} + y^{2} – 2xy) + 3xy] \\ \Rightarrow x^{3} +y^{3} = (x – y)(x^{2} + y^{2} + xy)\)

 

Q.10. Factorize each of the following:

 (i) \(27y^{3} + 125z^{3}\)

 (ii) \(64m^{3} – 343n^{3}\)

Answer:

(i) \(27y^{3} + 125z^{3}\)

\(27y^{3} + 125z^{3}\\ = (3y)^{3} + (5z)^{3}\\ = (3y + 5z) [{(3y)^{2} – (3y)(5z) + (5z)^{2}}]\\ = (3y + 5z) (9y^{2} – 15yz + 25z)^{2}\)
(ii) \(64m^{3} – 343n^{3}\)

\(64m^{3} – 343n^{3} \\ = (4m)^{3} – (7n)^{3}\\ = (4m + 7n) [{(4m)^{2} + (4m)(7n) + (7n)^{2}}]\\ = (4m + 7n) (16m^{2} + 28mn + 49n)^{2}\)

 

Q.11. Factorise : \(27x^{3} + y^{3} + z^{3} – 9xyz\)

Answer:

\(27x^{3} + y^{3} + z^{3} – 9xyz \\ = (3x)^{3} + y^{3} + z^{3} – 3 (3x)(y)(z)\\ = (3x + y + z) {(3x)^{2} + y^{2} + z^{2} – 3xy – yz – 3xz}\\ = (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)\)
Q.12. Verify that:

\(x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]\)

Answer:

We know that,
\(x^{3} + y^{3} + z^{3} -3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)\\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = \frac{1}{2}\times (x + y + z) [2(x^{2} + y^{2} + z^{2} – xy – yz – xz)]\\ = \frac{1}{2}(x + y + z) (2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2xz) \\ = \frac{1}{2}(x + y + z) [(x^{2} + y^{2} -2xy) + (y^{2} + z^{2} – 2yz) + (x^{2} + z^{2} – 2xz)] \\ = \frac{1}{2}(x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]\)

 

Q.13. If x + y + z = 0, show that \(x^{3} + y^{3} + z^{3} = 3xyz.\)

Answer:

We know that,
\(x^{3} + y^{3} + z^{3} = 3xyz\)= (x + y + z)(x2 + y2 + z2 – xy – yz – xz)
Now put (x + y + z) = 0,  
\(x^{3} + y^{3} + z^{3} = 3xyz= (0)(x^{2} + y^{2}+ z^{2} – xy – yz – xz) \\ \Rightarrow x^{3} + y^{3} + z^{3} – 3xyz = 0\)

 

Q.14. Without actually calculating the cubes, find the value of each of the following:
(i) \((-12)^{3} + (7)^{3} + (5)^{3}\)

(ii) \((28)^{3} +(-15)^{3}+(-13)^{3}\)

Answer:

(i) \((-12)^{3} + (7)^{3} + (5)^{3}\)
\((-12)^{3} + (7)^{3} + (5)^{3}= 0+3(-12)(7)(5)………………(Since. (-12)+(7)+(5)=0)Using\; Identity\\ = -1260\)
(ii) \((28)^{3} +(-15)^{3}+(-13)^{3}\)
x + y + z = 28 – 15 – 13 = 0

\((28)^{3}+ (-15)^{3} + (-13)^{3}=0+ 3(28)(-15)(-13) = 16380\)

 

Q.15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : \(25a^{2} – 35a + 12\)

(ii) Area : \(35 y^{2} + 13y – 12\)

Answer:

(i) Area : \(25a^{2} – 35a + 12\)
\(25a^{2} – 35a + 12\\ = 25a^{2} – 15a -20a + 12\\ = 5a(5a – 3) – 4(5a – 3)\\ = (5a – 4)(5a – 3)\)

Possible expression for length = 5a – 4
Possible expression for breadth = 5a – 3

 

(ii) Area : \(35 y^{2} + 13y – 12\)

\(35 y^{2} + 13y – 12\\ = 35y^{2} – 15y + 28y – 12\\ = 5y(7y – 3) + 4(7y – 3)\\ = (5y + 4)(7y – 3)\)

Possible expression for length = (5y + 4)
Possible expression for breadth = (7y – 3)

 

Q.16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? 

(i) Volume : \(3x^{2} – 12x\)
(ii) Volume : \(12ky^{2} + 8ky – 20k\)

Answer:

(i) Volume : \(3x^{2} – 12x\)
\(3x^{2} – 12x\)
= 3x(x – 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x – 4)

 

(ii) Volume : \(12ky^{2} + 8ky – 20k\)

\(12ky^{2} + 8ky – 20k\\ = 4k(3y^{2} + 2y – 5)\\ = 4k(3y^{2} +5y – 3y – 5)\\ = 4k[y(3y +5) – 1(3y + 5)]\\ = 4k (3y +5) (y – 1)\)

Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1) 

 

Non-polynomial equations give more challenges when doing mathematical questions. Monomials are the building blocks of polynomials. It is a polynomial that comprises variables and coefficients without subtraction or addition. Monomials are treated as the part of a larger polynomial.
In conclusion, NCERT Solutions For Class 9 Maths Chapter 2 talks about Non-polynomial equations. Monomials are the building blocks of polynomials. It is a polynomial that comprises variables and coefficients without subtraction or addition. Monomials are treated as the part of a larger polynomial.

Related Links
Ncert Solutions For Class 10 Science Pdf Free Download 11Th Ncert Biology Pdf
Ncert Chemistry Class 11 Pdf Free Download Part 2 Ncert Solution For Class 11Th Maths
Ncert 12 Biology Pdf Ncert Solutions For Class 12 Chemistry Pdf Download
Ncert 12Th Math Solution Pdf Ncert Solutions For Class 7 Mathematics
Science 7Th Class Ncert Ncert 8Th Maths Solutions Pdf
Ncert Class 8 Science Solutions Pdf 10Th Ncert