Polynomials Chapter 2 NCERT Solutions for Class 9  Maths

To help students get higher marks and understand the basics of the chapter, consistent practice of the exercise given in your NCERT textbook is crucial. Don’t worry if you are stuck between the questions or have a few doubts.

SimplyAcad is providing the best NCERT Solutions for Class 9 Maths Chapter 2 Polynomials below for students to practice and revise. Go through the solutions presented in an elaborate way below to develop the mathematical skills.

Overview of the exercise of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

• Exercise 2.1: The first exercise focuses on explaining the basic concept of a polynomial and its components.
• Exercise 2.2: The second exercise deals with the concept of zero of the polynomial.
• Exercise 2.3: The third exercise questions on the remainder theorem.
• Exercise 2.4: The fourth exercise deals with the concepts of polynomial factorization.
• Exercise 2.5: The final exercise is based on algebraic entities.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1

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

\((i)~4x^2 – 3x + 7\)

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

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

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

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

Solution:

(i) \((i)~4x^2 – 3x + 7\)

The given expression is \(~4x^2 – 3x + 7\).

The exponents of the variable in the expression are 2, 1 and 0, which are whole numbers.

Also, there is only one variable, x, in the expression.

Since the expression contains only one variable, and the exponents of the variable are whole numbers, it is a polynomial in one variable.

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

The given expression is \(~y^2 + \sqrt{2}\).

The exponent of the variable in the expression is 2, which is a whole number. Also, the expression contains only one variable, y.

Therefore, the above expression is a polynomial in one variable.

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

The given expression is \(3 \sqrt{t} + t\sqrt{2}\).
\(\Rightarrow 3\sqrt t+t\sqrt 2\\
~=3t^{\frac{1}{2}}+t\sqrt2\)

Here, the power exponents of the variable are \(\frac{1}{2}\) and \(1\).

As \(\frac12\) is not a whole number, the expression is not a polynomial.

The expression has only one variable, t.

Hence, it is an expression of one variable but not a polynomial.

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

The given expression is \(~y + \frac{2}{y}\).
\(\Rightarrow y + \frac{2}{y}\\
= y + 2y^{-1}\).

The above expression has –1 as an exponent, which is not a whole number. Therefore, it is not a polynomial.

The expression contains only one variable, y.

Hence, it is an expression of one variable but not a polynomial.

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

The given expression is \(~x^{10} + y^3 + t^{50}\).

The exponents of the variables are 10, 3 and 50.

As all the powers are whole numbers, it is a polynomial.

Also, the expression possesses three variables, x, y and t.

Therefore, the expression is a polynomial but not a polynomial in one variable.

Question 2 : Write the coefficients of \(x^2\) in each of the following:

(i) \(2 + x^2 + x\)

(ii) \(2 – x^2 + x^3\)

(iii) \(\frac{\pi}{2} x^2 + x\)

(iv) \(\sqrt{2x} – 1\)

Solution:

(i) In \(2 + x^2 + x,\) coefficient of \(x^2\) is 1.

(ii) In \(2 – x^2 + x,\) coefficient of \(x^2\) is – 1.

(iii) \(\frac{\pi}{2}x^2 + x\), coefficient of \(x^2\) is \(\frac{\pi}{2}\).

(iv) \(\sqrt{2x} – 1, x^2\) is not present. Hence, coefficient is 0.

Question 3 : Give one example for each, a binomial of degree 35, and a monomial of degree 100.

Solution:

A monomial of degree 100 is \(x^{100}\).

A binomial of degree of 35 is \(y^{35} + 45\).

Question 4 : Write the degree of each of the following polynomials:

(i) \(5x^3 + 4x^2 + 7x\)

(ii) \(4-y^2\)

(iii) \(5t – \sqrt{7}\)

(iv) 3

Solution:

(i) Degree is 3 as \(x^3\) is the highest power.

(ii) Degree is 2 as \(y^2\) is the highest power.

(iii) Degree is 1 as t is the highest power.

(iv) Degree is 0 as \(x^0\) is the highest power.

Question 5 : Classify the following as linear, quadriatic and cubic polynomials:

(i) \(x^2 + x\)

(ii) \(x – x^3\)

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

(iv) 1 + x

(v) 3t

(vi) \(r^2\)

(viii)\(7x^3\)

Solution:

(i) \(x^2 + x\) is quadratic as \(x^2\) has highest power.

(ii) \(x – x^3\) is cubic as \(-x^3\) has highest power.

(iii) \(y + y^2 + 4\) is quadratic as \(y^2\) has highest power.

(iv) \(1 + x\) is linear.

(v) 3t is linear.

(vi) \(r^2\) is quadratic as \(r^2\) has highest power.

(vii) \(7x^3\) is cubic \(7x^3\) has the highest power.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Question 1 : Find the value of the polynomial \(5x -4x^2 + 3\) at

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

\(p(x) = 5x – 4x^2 + 3\)

(i) \(At~ x = 0, p(0) = 5 \times 0 – 4 \times 0^2 + 3 = 3\)

(ii) At x = – 1, p(-1) = \( 5 \times (-1) – 4 \times (-1)^2 + 3 = – 5 – 4 + 3 = – 6\)

(iii) At x = 2, p(2) = \(5 \times 2 – 4 \times (2)^2 + 3 = 10 – 16 + 3 = – 3\)

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

(i) \(p(y) = y^2 – y + 1\)

(ii) \(p(t) = 2 + t + 2t^2 – t^3\)

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

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

Solution: (i) \(p(y) = y^2 – y + 1\\
p(0) = 0^2 – 0 + 1 = 1\\
p(1) = 1^2 – 1 + 1 = 1\)
\(p(2) = 2^2 – 2 + 1 = 3.\)

(ii) \(p(t) = 2 + t + 2t^2 – t^3\\
p(0) = 2 + 0 + 2 \times 0^2 – 0^3 = 2\\
p(1) = 2 + 1 + 2 \times 1^2 – 1^3 = 4\)
\(p(2) = 2 + 2 + 2 \times 2^2 – 2^3 = 4 + 8 – 8 = 4.\)

(iii) \(p(x) = x^3\\
p(0) = 0\\
p(1) = 1\)
\(p(2) = 8.\)

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

Question 3 : Find p(0), p(1) and p(2) for each of the following polynomials:

(i) \(p(y) = y^2 – y + 1\)

(ii) \(p(t) = 2 + t + 2t^2 – t^3\)

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

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

Solution: (i) \(p(y) = y^2 – y + 1\\
p(0) = 0^2 – 0 + 1 = 1\\
p(1) = 1^2 – 1 + 1 = 1\)
\(p(2) = 2^2 – 2 + 1 = 3.\)

(ii) \(p(t) = 2 + t + 2t^2 – t^3\\
p(0) = 2 + 0 + 2 \times 0^2 – 0^3 = 2\\
p(1) = 2 + 1 + 2 \times 1^2 – 1^3 = 4\)
\(p(2) = 2 + 2 + 2 \times 2^2 – 2^3 = 4 + 8 – 8 = 4.\)

(iii) \(p(x) = x^3\\
p(0) = 0\\
p(1) = 1\)
\(p(2) = 8.\)

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

Question 4 : Find the zero of the polynomial in each of the following cases:

(i) p(x) = x + 5

(ii) p(x) = x – 5

(iii) p(x) = 2x + 5

(iv) p(x) = 3x – 2

(v) p(x) = 3x

(vi) p(x) = ax, a \(\neq\) 0

(vii) p(x) = cx + d, c \(\neq\) 0, c, d are real numbers.

Solution: (i) x + 5 = 0, x = – 5, so, – 5 is the zero of x + 5.

(ii) x – 5 = 0, x = – 5 so, 5 is the zero of x – 5.

(iii) \(2x + 5 = 0, \Rightarrow 2x = -5, \Rightarrow x = \frac{-5}{2}, so – \frac{5}{2}\) is the zero of 2x + 5.

(iv) \(3x – 2 = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}, so \frac{2}{3}\) is the zero of 3x – 2.

(v) \(3x = 0, \Rightarrow x = 0,\) so 0 is the zero of 3x.

(vi) \(ax = 0 (a \neq 0) \Rightarrow x = \frac{0}{a} = 0,\) so, is the zero of ax.

(vii) \(cx + d = 0 (c \neq 0) \Rightarrow cx = – d \Rightarrow x = \frac{-d}{c}, so, \frac{-d}{c}\) is the 0 of cx + d.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3

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

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

(ii) \(x^4 + x^3 + x^2 + x + 1\)

(iii) \(x^4 + 3x^3 + 3x^2 + x + 1\)

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

Solution: To have (x + 1) as a factor, substituting x = – 1 must give p(-1) = 0.

(i) \(x^3 + x^2 + x + 1\\
= (-1)^3 + (-1)^2 + (-1) + 1 = – 1 + 1 – 1 + 1 = 0\)
Therefore, \(x + 1\) is a factor \(x^3 + x^2+x + 1\).

(ii) \(x^4 + x^3 + x^2 + x + 1\\
= (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 – 1 + 1 – 1 + 1 = 1\)
Remainder is not 0. Therefore (x + 1) is not its factor.

(iii) \(x^4 + 3x^3 + 3x^2 + x + 1\\
= (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1\\
= 1 – 3 + 3 – 1 + 1 = 1.\) Remainder is not 0.
Therefore, (x + 1) is not a factor.

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

Remainder not 0.

Therefore, (x + 1) is not a factor.

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

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

\((ii) ~p(x) = x^3 + 3x^2 + 3x + 1, ~g(x) = x + 2\)

\((iii) ~p(x) = x^3 – 4x^2 + x + 6, ~g(x) = x – 3\)

Solution:

\((i)\) Determining whether \(g(x)\) is a factor of \(p(x)\).

Given, \(p(x) = 2x^3 + x^2 – 2x -1\)

\(g(x) = x + 1\)

Zero of \(g(x)\) is given by:

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

By Factor Theorem

\(\Rightarrow ~p(-1) = 2(-1)^3 + (-1)^2 – 2(-1) – 1\)
\(= -2 + 1 + 2 – 1 = 0\)
Hence, by factor theorem \(g(x)\) is a factor of \(p(x)\).

\((ii)\) Determining whether \(g(x)\) is a factor of \(p(x)\).
Given, \(p(x) = x^3 + 3x^2 +3x +1\)
\(g(x) = x + 2\)
Zero of \(g(x)\) is given by:
\(x + 2 = 0 ~\Rightarrow ~ x = -2\)

By Factor Theorem
\(\Rightarrow ~p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 \\

= – 8 + 12 – 6 + 1

= – 1\)

Hence, by factor theorem \(g(x)\) is not a factor of \(p(x)\).

\((iii)\) Determining whether \(g(x)\) is a factor of \(p(x)\).
Given, \(p(x) = x^3 -4 x^2 + x +6\)
\(g(x) = x – 3\)

Zero of \(g(x)\) is given by:
\(x -3 = 0 ~\Rightarrow ~ x = 3\)

By Factor Theorem

\(\Rightarrow ~p(3) = (3)^3 – 4(3)^2 + 3 + 6\\
~~~~~~~~~~~~~= 27 – 36 + 3 + 6
= 0\)

Hence, by factor theorem \(g(x)\) is a factor of \(p(x)\).

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

\((i)~p(x) = x^2 + x + k\)

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

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

\((iv)~p(x) = kx^2 – 3x + k\)

Solution: \(\textbf {Solution:}\)

\((x -1)\) is a factor, so we substitute \(x = 1\) in each case and solve for \(k\) by making \(p(1)\) equal to \(0\).

\((i)\) Find the value of \(k\)

Given : \(p(x) =x^2 + x + k\)

If \((x – 1)\) is factor of \(p(x)\), then \(p(1) = 0\)

By Factor Theorem

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

\(= 1 + 1 + k = 0 \\ \Rightarrow ~k = – 2\)

Hence, the value of \(k = – 2 \)

\((ii)\) Find the value of \(k\)

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

If \((x – 1)\) is factor of \(p(x)\), then \(p(1) = 0\)

By Factor Theorem

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

\(\Rightarrow ~ 2 + k + \sqrt{2} = 0\)

\(\Rightarrow ~k = – ( 2 + \sqrt{2})\)

Hence, the value of \(k = – (2 + \sqrt{2})\)

\((iii)\) Find the value of \(k\)

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

If \((x – 1)\) is factor of \(p(x)\), then \(p(1) = 0\)

By Factor Theorem

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

\( \Rightarrow k – \sqrt{2} + 1 = 0\)

\(\Rightarrow k = \sqrt{2} – 1\)

Hence, the value of \(k = \sqrt{2} – 1\)

\((iv)\) Find the value of \(k\)

Given : \(p(x) = kx^2 – 3x + k\)

If \((x – 1)\) is factor of \(p(x)\), then \(p(1) = 0\)

By Factor Theorem

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

\( \Rightarrow k – 3 + k = 0 \\ \Rightarrow ~2k – 3 = 0\)

\(\Rightarrow~ k = \frac{3}{2}\)

Hence, the value of \(k = \frac{3}{2}\)

Question 4 : Factorise:

\((i)~12x^2 – 7x + 1\)

\((ii)~2x^2 + 7x + 3\)

\((iii)~6x^2 + 5x – 6\)

\((iv)~3x^2 – x – 4\)

Solution: \((i)\) Factorize the given expression

Given expression \(12x^2 – 7x + 1\)

Using the splitting the middle term method.

We get \(-4\) and \(-3\) as the numbers \([(-4) + (-3) = (-7) ~ \text{and} ~(-4) \times (-3) = 12]\)

\(~=12x^2 – 7x + 1\)

\(= 12x^2 – 4x – 3x+ 1\)

\(= 4x (3x-1)-1 (3x-1)\)

\( = (4x-1) (3x-1)\)

Hence, \(12x^2 – 7x + 1\) \( = (4x-1) (3x-1)\)

\((ii)\) Factorize the given expression

Given expression \(2x^2 + 7x+ 3\)

Using the splitting the middle term method.

We get \(6\) and \(1\) as the numbers \([6 + 1 = 7~ \text{and} ~6 \times 1 = 6]\)

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

\(\Rightarrow~ 2x^2 + 6x + x + 3\)

\(= 2x (x+3) + 1 (x+3) \)

\(= (2x+ 1) (x+3)\)

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

\((iii)\) Factorize the given expression

Given expression \(~6x^2 + 5x – 6\)

Using the splitting the middle term method.

We get \(9\) and \((-4)\) as the numbers \([9 + (-4) = 5~ \text{and} ~9 \times (-4) = -36]\)

\(=~6x^2 + 5x – 6\)

\(= 6x^2 + 9x – 4x – 6\)

\(= 3x (2x+3) – 2 (2x+ 3)\)

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

Hence, \(~6x^2 + 5x – 6\) \(= (3x-2) (2x+3)\)

\((iv)\) Factorize the given expression

Given expression \(~3x^2 – x – 4\)

Using the splitting the middle term method.

We get \(-4\) and \(3\) as the numbers \([-4 + 3 = -1 ~ \text{and} ~-4 \times 3 = -12]\)

\(=~3x^2 – x – 4\)

\(= 3x^2 – 4x + 3x – 4 \)

\(= x (3x – 4) + 1 (3x-4)\)

\( = (x+1) (3x-4)\)

Question 5 : Factorise:

\((i)~x^3 – 2x^2 – x + 2\)

\((ii)~x^3 – 3x^2- 9x – 5\)

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

\((iv)~2y^3 + y^2 – 2y – 1\)

Solution: Solution:

\((i)~Given;~x^3 – 2x^2 – x + 2\)

Factorising

\(x^3 – x – 2x^2 + 2 \)

\(= x (x^2 – 1) – 2(x^2 – 1)\)

\(= (x^2 – 1)(x – 2)\)

\(= [(x)^2 – (1)^2](x – 2)\)

\(= (x – 1)(x + 1)(x – 2)\)

\(\ [(a^2 – b^2) = (a + b)(a – b)]\)

Conclusion

\(Thus,~x^3 – 2x^2 – x + 2 \\
~~~~~~~~~~~~~~~= (x – 1)(x + 1)(x – 2) \)

\((ii)~Given:~x^3 – 3x^2 – 9x – 5\)

Factorising

\(x^3 – 3x^2 – 9x – 5\)

\( = x^3 + x^2 – 4x^2 – 4x – 5x – 5\)

\(= x^2 (x + 1) – 4x (x + 1) – 5(x + 1)\)

\(= (x + 1)(x^2 – 4x – 5)\)

\(= (x + 1)(x^2 – 5x + x – 5)\)

\(= (x + 1)[(x(x – 5) + 1(x – 5))]\)

\(= (x+1)(x-5)(x+1)\)

Conclusion

\(Thus,~x^3 – 3x^2 – 9x – 5 \\
~~~~~~~~~~~~~~~= (x+1)(x-5)(x+1) \)

\((iii)~Given:~ x^3 + 13x^2 + 32x + 20\)

Factorising

\(x^3 + 13x^2 + 32x + 20 \)

\(= x^3 + x^2 + 12x^2 +12x+ 20x + 20\)

\(= x^2 (x+1) + 12x (x + 1) + 20 (x +1)\)

\(= (x + 1) (x^2 + 12x + 20)\)

\(= (x + 1) (x^2 + 2x + 10x + 20)\)

\(= (x +1) {[x(x + 2) + 10(x + 2)}]\)

\(= (x + 1) (x + 2) (x+10)\)

Conclusion

\(Thus,~x^3 + 13x^2 + 32x + 20 \\
~~~~~~~~~~~~~~~= (x + 1) (x + 2) (x+10) \)

\((iv)~ Given:~2y^3 + y^2 – 2y – 1\)

Factorising

\(2y^3 + y^2 – 2y – 1\)

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

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

\(= (y – 1) (2y^{2} + 3y +1)\)

\(= (y – 1) (2y^{2} + 2y + y +1)\)

\(= (y – 1)[(2y(y + 1) + 1( y + 1)]\)

\(= (y – 1) (y + 1) (2y + 1)\)

Conclusion:

\(Thus, 2y^3 + y^2 – 2y – 1\)
\(~~~~~~~~~~~~~~~~~= (y – 1) (y + 1) (2y + 1)\)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4

Question 1 : Use suitable identities to find the following products:

\((i)~(x + 4) (x + 10)\)

\((ii)~(x+8) (x-10)\)

\((iii)~(3x + 4) (3x-5)\)

\((iv)~\left ( y^2 + \frac{3}{2} \right ) \left ( y^2 – \frac{3}{2} \right )\)

\((v)~(3-2x) (3+2x)\)

Solution: \((i)\) Given : \((x + 4) (x+10)\)

Applying the formula

Using identity, \((x + a)(x + b) = x^{2} + (a + b)x + ab\)

We have, \((x + 4) (x+10)\)

\(= x^2 + (4 + 10) x + 4 \times 10\)

\(= x^2 + 14x + 40\)

\((ii)\) Given : \((x+8) (x-10)\)

Applying the formula

Using identity, \((x + a)(x + b) = x^{2} + (a + b)x + ab\)

We have, \((x + 8) (x – 10)\)

\(= x^2 + (8-10) x+8 (-10)\)

\(= x^2 – 2x – 80\)

\((iii)\) Given : \((3x + 4) (3x-5)\)

Applying the formula

Using identity, \((x + a)(x + b) = x^{2} + (a + b)x + ab\)

We have, \((3x+ 4) (3x- 5)\)

\(=3(x+\frac43)3(x-\frac53)\)

\(=9[x^2+(\frac43-\frac53)x-\frac{20}{9}]\)

\(=9(x^2-\frac13 x-\frac{20}{9})\)

\(=9x^2-3x-20\)

\((iv)\) Given : \(\left ( y^2 + \frac{3}{2} \right ) \left ( y^2 – \frac{3}{2} \right )\)

Applying the formula

Using identity, \((x + y) (x-y) = x^2 – y^2\)

We have, \(\left ( y^2 + \frac{3}{2} \right ) \left ( y^2 – \frac{3}{2} \right )\)

\(= (y^2)^2 – \left ( \frac{3}{2} \right )^2\)

\( = y^4 – \frac{9}{4}\)

\((v)\) Given : \(~(3-2x) (3+2x)\)

Applying the formula

Using identity, \((x + y) (x-y) = x^2 – y^2\)

We have, \((3-2x) (3+2x) \)

\(= 3^2 – (2x)^2\)

\(= 9 – 4x^2\)

Question 2 : Evaluate the following products without multiplying directly:

(i) 103 \(\times\) 107

(ii) 95 \(\times\) 96

(iii) 104 \(\times\) 96

Solution: (i) \(103 \times 107 = (100 + 3) (100+7)\)

\(=(100)^2 + (3+7) \times 100 + 3 \times 7 \)

\(= 10000 + 1000 + 21 = 11021\)

(ii) \(95 \times 96 = (100-5) (100-4)\)

\( = (100)^2 – (5+4) \times 100 + 5 \times 4\)

\(= 10000-900 + 20 = 9120\)

(iii) Using identity: \((a+b)(a-b)=a^2-b^2\)

\(\therefore 104 \times 96\)

\( = (100+4) (100-4)\)

\( = (100^2 – 4^2)\)

\(= 10000 – 16 = 9984\)

Question 3: Factorise the following using appropriate identities:

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

(ii) \(4y^2 – 4y + 1\)

(iii) \(x^2 – \frac{y^2}{100}\)

Solution:

(i) \(9x^2 + 6xy + y^2 = (3x)^2 + 2(3x) y + (y)^2\\
= (3x+ y)^2~~~[Using~a^2 + 2ab + b^2 = (a+b)^2]\\
= (3x+y) (3x+y)\)

(ii) \(4y^2 – 4y +1\\
= (2y^2) – 2(2y) (1) + (1)^2\\
= (2y – 1)^2 = (2y – 1) (2y – 1) [Using~a^2 – 2ab + b^2 = (a – b)^2]\)

(iii) \(x^2 – \frac{y^2}{100} = x^2 – \left ( \frac{y}{10} \right )^2\\
= \left ( x+\frac{y}{10} \right ) \left ( x-\frac{y}{10} \right )~[Using~a^2 – b^2 = (a+b)(a-b)]\)

Question 4: Expand each of the following, using suitable identities:

\((i)~(x + 2y + 4z)^2\)

\((ii)~(2x-y+z)^2\)

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

\((iv)~(3a-7b-c)^2\)

\((v)~(-2x+5y – 3z)^2\)

\((vi)~\left [ \frac{1}{4} a – \frac{1}{2} b +1 \right ]^2\)

Solution:

\((i)~(x+2y + 4z)^2\)

\([\text{Using the identity}~(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca]\)

\(= x^2 + (2y)^2 + (4z)^2 + 2x \times 2y + 2 \times 2y \times 4z + 2 \times 4z \times x\)

\(= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx\)

[0.5 marks]

\((ii)~(2x-y+z)^2\)

\([\text{Using the identity}~(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc+2ca]\)

\(= (2x)^2 + (-y)^2 + (z)^2 + 2 \times (2x) (-y) + 2 (-y) (z)+2 \times 2xz\)

\(= 4x^2 + y^2 + z^2 – 4xy – 2yz + 4zx\)

[0.5 marks]

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

\([\text{Using the identity}~(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca]\)

\(= (-2x)^2 + (3y)^2 + (2z)^2 + 2(-2x) (3y) + 2 (3y) (2z) + 2 (2z) (-2x)\)

\(= 4x^2 + 9y^2 + 4z^2 – 12xy + 12yz – 8zx\)

[0.5 marks]

\((iv)\) Given: \((3a-7b-c)^2\)

Applying formula

\((3a-7b-c)^2 = (3a)^{2} + (-7b)^{2} + (-c)^{2} + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)\)

\( = 9a^{2} + 49b^{2} + c^{2} – 42ab + 14bc – 6ac\)

[0.5 marks]

\((v)~(-2x +5y -3z)^2\)

\([\text{Using the identity}~ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca]\)

\(= (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2 (5y)(-3z)+2(-3z) (-2x)\)

\(= 4x^2 + 25y^2 + 9z^2 – 20xy – 30yz + 12zx\)

[0.5 marks]

\((vi)~\left [ \frac{1}{4}a – \frac{1}{2} b + 1 \right ]^2 \)

\([\text{Using the identity}~ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca]\)

\(\Rightarrow \left ( \frac{1}{4}a \right )^2 + \left ( \frac{-1}{2}b \right )^2 + (1)^2 + 2\left ( \frac{1}{4}a \right ) \left ( \frac{-1}{2}b \right ) + 2\left ( \frac{-1}{2}b \right )\times 1+2(1) \times \frac{1}{4}a\)

\(= \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 – \frac{1}{4}ab – b + \frac{1}{2}a\)

\(= \frac{a^2}{16} + \frac{b^2}{4} + 1- \frac{1}{4}ab-b +\frac{1}{2}a\)

\(= \frac{a^2}{16} + \frac{b^2}{4} + 1 – \frac{ab}{4}-b+\frac{a}{2}\)

[0.5 marks]

Question 5: Factorise:

\((i)~~~4x^2+9y^2 + 16z^2 + 12xy – 24yz – 16xz\\
(ii)~~~2x^2 +y^2 +8z^2 – 2\sqrt{2} xy + 4\sqrt{2}yz – 8xz\)

Solution:

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

\(Using ~identity ~(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y)+2(3y)(-4z)+2(-4z)(2x)\\
=(2x+3y-4z)^2 = (2x+3y-4z)(2x+3y-4z)\)

\((ii)~~~2x^2 +y^2 +8z^2 – 2\sqrt{2} xy + 4\sqrt{2}yz – 8xz\\
Using ~the ~identity~(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\\
=(\sqrt{2}x)^2+(-y)^2+(-2\sqrt{2}z)^2 + 2(\sqrt{2}x)(-y)+2(-y)(-2\sqrt{2}z) + 2 (\sqrt{2}x)(-2\sqrt{2}z)\\
=(\sqrt{2}x-y-2\sqrt{2}z)^2\\
=(\sqrt{2}x-y-2\sqrt{2}z) (\sqrt{2}x-y-2\sqrt{2}z)\)

Question 6: Write the following cubes in expanded form:

\((i) ~(2x+1)^3\\
(ii) ~(2a-3b)^3\\
(iii) \left [ \frac{3}{2}x+1 \right ]^3\\
(iv) \left [ x-\frac{2}{3}y \right ]^3 \\\)

Solution:

\((i)~~~~~~(2x+1)^3 \\Using~ identity~(a+b)^3=a^3+b^3+3ab(a+b)\\
= (2x)^3+1^3+3(2x)(1)(2x+1)\\
=8x^3 + 1 + 6x (2x+1) \\
= 8x^3 + 12x^2 + 6x+1\)

\((ii)~~~~~~(2a-3b)^3\\ Using~ identity~(a-b)^3=a^3-b^3-3ab(a-b)\\ = (2a)^3-(3b)^3 – 3 \times 2a\times 3b(2a-3b)\\
= 8a^3 – 27b^3 – 18ab (2a-3b)\\
= 8a^3 – 27b^3 – 36a^2b + 54ab^2\)

\((iii)~~~~~~\left [ \frac{3}{2}x+1 \right ]^3\\Using~ identity~(a+b)^3=a^3+b^3+3ab(a+b)\\ = \left ( \frac{3}{2}x \right )^3 + 1^3 + 3\left ( \frac{3}{2}x \right )(1)\left ( \frac{3}{2}x+1 \right )\\
= \frac{27}{8}x^3 + 1 + \frac{9}{2}x \left ( \frac{3}{2}x+1 \right )\\
= \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x+1\)

\((iv)~~~~~~~\left [ x-\frac{2}{3}y \right ]^3 \\ Using~ identity~(a-b)^3=a^3-b^3-3ab(a-b)\\= x^3 – \left ( \frac{3}{2}y \right )^3 -3(x)\left ( \frac{2}{3}y \right )\left (x- \frac{2}{3}y \right )\\
= x^3 – \frac{8}{27}y^3 -2xy \left ( x-\frac{2}{3}y \right )\\
= x^3 – \frac{8}{27}y^3 – 2x^2 y + \frac{4}{3}xy^2\)

Question 7: Evaluate the following using suitable identities:

(i) \((99)^3\)

(ii) \((102)^3\)

(iii) \((998)^3\)

Solution:

\((i) ~(99)^3~[ Using~ identity~(a-b)^3 =a^3-b^3-3ab(a-b)]\\
\Rightarrow (100-1)^3 \\
= (100)^3 + (-1)^3 + 3(100)(-1)(100-1)\\
= 1000000 – 1 – 300 (100-1)\\
= 1000000 – 1 – 30000 + 300 = 970299\)

\((ii)~ (102)^3~[ Using~ identity~(a+b)^3 =a^3+b^3+3ab(a+b)]\\
\Rightarrow (100+2)^3 \\
= (100)^3 + 2^3 + 3(100) (2)(100+2)\\
= 1000000 + 8 + 600 (100+2)\\
= 1000000 + 8 + 60000 + 1200 = 1061208\)

\((iii)~ (998)^3~[ Using~ identity~(a-b)^3 =a^3-b^3-3ab(a-b)]\\
\Rightarrow (1000-2)^3\\
= (1000)^3 + (-2)^3 + 3(1000) (-2)(998)\\
= (1000)^3 – 8 – 6000 (998)\\
= 1000000000 – 8 – 5988000 = 994011992\)

Question 8: Factorise each of the following:

\((i)~~~~~~~8a^3 + b^3 + 12a^2b + 6ab^2\\
(ii)~~~~~~~ 8a^3 – b^3 – 12a^2b + 6ab^2\\
(iii)~~~~~~~27-125a^3 – 135a + 225a^2\\
(vi)~~~~~~~ 64a^3 – 27b^3 – 144a^2b + 108ab^2\\
(v)~~~~~~~ 27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p\)

Solution:

\((i) ~8a^3 + b^3 + 12a^2b + 6ab^2\\
= (2a)^3 + b^3 + 3 (2a) (b) (2a+b)\\~Using~ identity(a+b)^3=a^3+b^3+3ab(a+b)\\
= (2a+b)^3 = (2a+b) (2a+b) (2a+b)\)

\((ii) ~8a^3 – b^3 – 12a^2b + 6ab^2\\
= (2a)^3 + (-b)^3 + 3(2a) (-b)(2a-b)\\\\~Using~ identity(a-b)^3=a^3-b^3-3ab(a-b)\\
= (2a-b)^3 = (2a-b)^3 = (2a-b)(2a-b)(2a-b)\)

\((iii) ~27-125a^3 – 135a + 225a^2\\
= 3^3+(-5a)^3 + 3\times (3) (-5a) (3-5a)\\\\~Using~ identity(a-b)^3=a^3-b^3-3ab(a-b)\\
= (3-5a)^3 = (3-5a) (3-5a)(3-5a)\)

\((iv)~ 64a^3 – 27b^3 – 144a^2b + 108ab^2\\
= (4a)^3 + (-3b)^3 + 3(4a) \times (-3b) (4a-3b)\\\\~Using~ identity(a-b)^3=a^3-b^3-3ab(a-b)\\
= (4a-3b)^3 = (4a-3b)(4a-3b) (4a-3b)\)

\((v) ~27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p\\
= (3p)^3 + \left ( -\frac{1}{6} \right )^3 + 3(3p) \left ( -\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )\\\\~Using~ identity(a-b)^3=a^3-b^3-3ab(a-b)\\
= \left ( 3p-\frac{1}{6} \right )^3 = \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )\)

Question 9: Verify:

\((i)~~~x^3 + y^3 = (x+y)(x^2 -xy + y^2)\\
(ii)~~~x^3 – y^3 = (x-y)(x^2 + xy + y^2)\)

Solution:

\((i) ~x^3 +y^3~[using~identity~a^3+b^3=(a+b)(a^2-ab+b^2)]\\= (x+y)(x^2- xy + y^2)\\
R.H.S. = x(x^2 – xy + y^2) + y(x^2 – xy + y^2)\\
= x^3 – x^2y + xy^2 + yx^2 – xy^2 + y^3 \\
= x^3 + y^3 \\
= L.H.S\)

\((ii) ~x^3-y^3[using~identity~a^3-b^3=(a-b)(a^2+ab+b^2)]\\ = (x-y)(x^2+xy + y^2)\\
R.H.S. = x(x^2 + xy + y^2)-y(x^2 +xy + y^2)\\
= x^3 + x^2y + xy^2 – yx^2 -xy^2- y^3 \\
=x^3 – y^3 \\
= L.H.S.\)

Question 10: Factorise each of the following:

\((i) ~27y^3 + 125z^3\)
\((ii) ~64m^3 – 343n^3\)

Solution:

(i) Factorizing

$27y^3 + 125z^3$

27y3+125z3:

This expression is a sum of cubes. We can use the identity for the sum of cubes:

 

$$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$

a3+b3=(a+b)(a2−ab+b2)

For

$27y^3 + 125z^3$

27y3+125z3:

• 
  $a = 3y$ 

  $a=3y$

• 
  $b = 5z$ 

  $b=5z$

Thus:

 

$$27y^3 + 125z^3 = (3y + 5z)\left((3y)^2 – (3y)(5z) + (5z)^2\right)$$

27y3+125z3=(3y+5z)((3y)2−(3y)(5z)+(5z)2)

Simplify the expression inside the parentheses:

 

$$= (3y + 5z)(9y^2 – 15yz + 25z^2)$$

=(3y+5z)(9y2−15yz+25z2)

So, the factorization of

$27y^3 + 125z^3$

27y3+125z3 is:

 

$$27y^3 + 125z^3 = (3y + 5z)(9y^2 – 15yz + 25z^2)$$

27y3+125z3=(3y+5z)(9y2−15yz+25z2)

(ii) Factorizing

$64m^3 – 343n^3$

64m3−343n3:

This expression is a difference of cubes. We can use the identity for the difference of cubes:

 

$$a^3 – b^3 = (a – b)(a^2 + ab + b^2)$$

a3−b3=(a−b)(a2+ab+b2)

For

$64m^3 – 343n^3$

64m3−343n3:

• 
  $a = 4m$ 

  $a=4m$

• 
  $b = 7n$ 

  $b=7n$

Thus:

 

$$64m^3 – 343n^3 = (4m – 7n)\left((4m)^2 + (4m)(7n) + (7n)^2\right)$$

64m3−343n3=(4m−7n)((4m)2+(4m)(7n)+(7n)2)

Simplify the expression inside the parentheses:

 

$$= (4m – 7n)(16m^2 + 28mn + 49n^2)$$

=(4m−7n)(16m2+28mn+49n2)

So, the factorization of

$64m^3 – 343n^3$

64m3−343n3 is:

 

$$64m^3 – 343n^3 = (4m – 7n)(16m^2 + 28mn + 49n^2)$$

64m3−343n3=(4m−7n)(16m2+28mn+49n2)

Question 11:

Factorize:

 

$$27x^3 + y^3 + z^3 – 9xyz$$

 

Solution:

We recognize that the expression can be factored using the identity:

 

$$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)$$

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)

Here, let:

• 
  $a = 3x$ 

  $a=3x$

• 
  $b = y$ 

  $b=y$

• 
  $c = z$ 

  $c=z$

So, applying the identity:

 

$$27x^3 + y^3 + z^3 – 9xyz = (3x + y + z)\left((3x)^2 + y^2 + z^2 – 3xy – yz – 3zx\right)$$

27x3+y3+z3−9xyz=(3x+y+z)((3x)2+y2+z2−3xy−yz−3zx)

Now, simplifying the expression inside the parentheses:

 

$$= (3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$$

=(3x+y+z)(9x2+y2+z2−3xy−yz−3zx)

So, the factorization of the expression is:

 

$$27x^3 + y^3 + z^3 – 9xyz = (3x + y + z)(9x^2 + y^2 + z^2 – 3xy – yz – 3zx)$$

27x3+y3+z3−9xyz=(3x+y+z)(9x2+y2+z2−3xy−yz−3zx)

Question 12: Verify that:

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

Solution:

\(x^3 + y^3 + z^3 – 3xyz = \frac{1}{2} (x+y+z) [(x-y)^2 + (y-z)^2 + (z-x)^2]\\
R.H.S. = \frac{1}{2} (x+y+z) [x^2 + y^2 – 2xy + y^2 + z^2 – 2yz + z^2 + x^2 – 2zx]\\
= \frac{1}{2} (x+y+z) [2x^2+2y^2 + 2z^2 – 2xy – 2yz-2zx]\\
= (x+y+z)[x^2 + y^2 + z^2 – xy – yz -zx]\\
= x[x^2 + y^2 + z^2 – xy – yz – zx]\\
+y(x^2 + y^2 + z^2-xy – yz-zx)\\
+z(x^2 + y^2 + z^2 – xy-yz-zx)\\
= x^3 + xy^2 + xz^2 – x^2y – xyz – zx^2 + yx^2 + y^3 + yz^2 – xy^2\\
-y^2z – zxy + zx^2 + zy^2 + z^3 – zxy – yz^2 – z^2x\\
x^3 + y^3 + z^3 – 3xyz = L.H.S.\)

Question 13: If x + y + z = 0, show that \(x^3 + y^3 + z^3 = 3xyz.\)

Solution:

\(x + y + z = 0\\
(x + y + z)^3 = x^3 + y^3 + z^3 – 3xyz = 0\\
\Rightarrow x^3 + y^3 + z^3 = 3xyz.\)

Question 14: Without actually calculating the cubes, find the value of each of the following:

\((i) (-12)^3 + (7)^3 + (5)^3\)

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

Solution:

From the above question, we have \(x^3 + y^3 + z^3 = 3xyz, if~x + y + z = 0\).

(i) Here – 12 + 7 + 5 = 0

So,

\(\Rightarrow (-12)^3 + (7)^3 + (5)^3\\
=3(-12)(7)(5) \\
= – 1260\)

(ii) Here 28 + (-15) + (-13) = 0

So,

\(\Rightarrow(28)^3 + (-15)^3 + (-13)^3\\
= 3 \times 28 (-15) (-13) \\
= 16380\)

Question 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 : $35y^2+13y-12$

Solution:

$\left(i\right)$ $Area=25a^2-35a+12$

Question 16:

What are the possible expressions for the dimensions of the cuboid whose volumes are given below?
(i) Volume: ${3x}^2-12x$

(ii) Volume: $12k{y}^2+8ky-20k$

Solution: