NCERT Solutions Class 9 Maths Chapter 2: Polynomials

Exercise 2.1

Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
4x^2-3x+7
Answer: 4x^2-3x+7 can also be written as 4x^2-3x^1+7x^0 .
We can call the expression 4x^2-3x+7 a polynomial in one variable because x is the only variable in the equation and the powers of x (i.e., 2, 1, and 0) are all whole numbers.

y^2+\sqrt2
Answer: y^2+\sqrt2 can also be written as y^2+\sqrt2{\ y}^0 .
We can call the expression y^2+\sqrt2\ a polynomial in one variable because y is the only variable in the equation and the powers of x (i.e., 2 and 0) are all whole numbers.

3\sqrt t+t\sqrt2
Answer: 3\sqrt t+t\sqrt2 can also be written as 3t^\frac{1}{2}+\sqrt2\ t^1 .
Despite the fact that t is the only variable in the equation, the powers of t (i.e.,1/2) is not whole numbers. As a result, we can say that the equation 3\sqrt t+t\sqrt2 is not a one-variable polynomial.

y+\frac{2}{y}
Answer: y+\frac{2}{y} can also be written as y^1+2y^{-1} .
Despite the fact that y is the only variable in the equation, the powers of t (i.e., -1) is not whole numbers. As a result, we can say that the equation y+\frac{2}{y}\ is not a one-variable polynomial.

x^{10}+y^3+t^{50}
Answer: In this case, the equation is x^{10}+y^3+t^{50}
Despite the fact that the powers 10, 3, and 50 are all whole numbers, the expression has three variables. As a result, it is not a one-variable polynomial.

Q2. Write the coefficients of x2 in each of the following:
2+x^2+x
Answer: 2+x^2+x can also be written as 2+\left(1\right)x^2+x.
The number that multiplies the variable is known as the coefficient.
The variable x2 is multiplied by 1 in this case.
In 2+x^2+x, the coefficients of x2 is 1.

2-x^2+x^3
Answer: 2-x^2+x^3
can also be written as 2+(-1)x^2+x^3.
The number (together with its sign, i.e., – or +) that multiplies the variable is known as the coefficient.
The variable x2 is multiplied by -1 in this case.
In 2-x^2+x^3, the coefficients of x2 is -1.

\left(\frac{\pi}{2}\right)x^2+x
Answer: \left(\frac{\pi}{2}\right)x^2+x is a form of the equation \left(\frac{\pi}{2}\right)x^2+x.
The number (together with its sign, i.e., – or +) that multiplies the variable is known as the coefficient.
The variable x2 is multiplied by the number \frac{\pi}{2} in this case.
The x2 coefficients in \left(\frac{\pi}{2}\right)x^2+x is \frac{\pi}{2}.

\sqrt2x-1
Answer: Since 0x^2 is 0, the equation \sqrt2x-1 can be written as 0x^2+\sqrt2\ x-1.
The number (together with its sign, i.e., – or +) that multiplies the variable is known as the coefficient.
The variable x2 is multiplied by 0 in this case. In \sqrt2x-1, the x2 coefficients in \sqrt2x-1 is 0.

Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Answer: Binomial of degree 35: A binomial of degree 35 is a polynomial with two terms and the greatest degree 35.
Example: 10y^{35}+10 .
A polynomial with one term and the greater degree 100 is referred to as a monomial of degree 100.
Example: 15t^{100}+3 .

Q4. Write the degree of each of the following polynomials:
5x^3+4x^2+7x
Answer: The degree of a polynomial is the largest power of the variable in the polynomial.
5x^3+4x^2+7x=5x^3+4x^2+7x^1
x has the following powers: 3, 2, 1
Because 3 is the largest power of x in the equation, the degree of 5x^3+4x^2+7x is 3.

4-y^2
Answer: The degree of a polynomial is the largest power of the variable in the polynomial.
4-y^2=4y^0-y^2
The highest power of the variable y is 2. Therefore, the degree of the given polynomial is 2.

5t-\sqrt7
Answer: The degree of a polynomial is the largest power of the variable in the polynomial.
5t^1-\sqrt7
The highest power of the variable t is 1. Therefore, the degree of the polynomial is 1.

3
Answer: The degree of a polynomial is the largest power of the variable in the polynomial.
Since there are no variables, we can assume that the power of any variable would be 0 here. Therefore, the degree of the polynomial is 0.

Q5. Classify the following as linear, quadratic and cubic polynomials:
x^2+x
Answer: The degree of the polynomial is 2. Therefore, it is a quadratic polynomial.

x-x^3
Answer: The degree of the polynomial is 3. Therefore, it is a cubic polynomial.

y+y^2+4
Answer: The degree of the polynomial is 2. Therefore, it is a quadratic polynomial.

1+x
Answer: The degree of the polynomial is 1. Therefore, it is a linear polynomial.

3t
Answer: The degree of the polynomial is 1. Therefore, it is a linear polynomial.

r^2
Answer: The degree of the polynomial is 2. Therefore, it is a quadratic polynomial.

7x^3
Answer: The degree of the polynomial is 3. Therefore, it is a cubic polynomial.

Exercise 2.2:

Q1. Find the value of the polynomial 5x-4x^2+3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Answer:
Let us assume that f\left(x\right)=5x-4x^2+3
Substituting x=0, we get:
f\left(0\right)=5\left(0\right)-4\left(0\right)^2+3
=3

Substituting x=-1, we get:
f\left(-1\right)=5\left(-1\right)-4\left(-1\right)^2+3
=-6

Substituting x=2, we get:
f\left(-1\right)=5\left(2\right)-4\left(2\right)^2+3
=-3

Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
p\left(y\right)=y^2-y+1
Answer:
p\left(0\right)=\left(0\right)^2-\left(0\right)+1=1
p\left(1\right)=\left(1\right)^2-\left(1\right)+1=1
p\left(2\right)=\left(2\right)^2-\left(2\right)+1=3

p\left(t\right)=2+t+2t^2-t^3
Answer:
p\left(0\right)=2+\left(0\right)+2\left(0\right)^2-\left(0\right)^3=2
p\left(1\right)=2+\left(1\right)+2\left(1\right)^2-\left(1\right)^3=4
p\left(0\right)=2+\left(2\right)+2\left(2\right)^2-\left(2\right)^3=4

p\left(x\right)=x^3
Answer:
p\left(0\right)=\left(0\right)^3=0
p\left(1\right)=\left(1\right)^3=1
p\left(2\right)=\left(2\right)^3=8

p\left(x\right)=(x-1)(x+1)
Answer:
p\left(0\right)=\left(\left(0\right)-1\right)\left(\left(0\right)+1\right)=-1
p\left(1\right)=\left(\left(1\right)-1\right)\left(\left(1\right)+1\right)=0
p\left(2\right)=\left(\left(2\right)-1\right)\left(\left(2\right)+1\right)=3

Q3. Verify whether the following are zeroes of the polynomial, indicated against them.
p\left(x\right)=3x+1,\ x=-1/3
Answer: For the value of x=-1/3, the output of the polynomial will be:
p\left(-\frac{1}{3}\right)=3\left(-\frac{1}{3}\right)+1=0
Therefore, the value satisfies the polynomial, thus it is a zero of the polynomial.

p\left(x\right)=5x-\pi,\ x=4/5
Answer: For the value of x=4/5 , the output of the polynomial will be:
p\left(4/5\right)=5\left(\frac{4}{5}\right)-\pi=4-\pi
Therefore, it does not satisfy the polynomial and is not a zero of it.

p\left(x\right)=x^2-1,\ x=1,-1
Answer: For the value of x=1, the output of the polynomial will be:
p\left(1\right)=\left(1\right)^2-1=0
For the value of x=-1, the output of the polynomial will be:
p\left(-1\right)=\left(-1\right)^2-1=0
Therefore, 1\ and-1 both are the zeroes of the polynomial.

p\left(x\right)=\left(x+1\right)\left(x-2\right),\ x=-1,2
Answer: For the value of x=-1, the output of the polynomial will be:
p\left(-1\right)=\left(-1+1\right)\left(-1-2\right)=0
For the value of x=2, the output of the polynomial will be:
p\left(2\right)=\left(2+1\right)\left(2-2\right)=0
Therefore, -1 and 2 both are the zeroes of the polynomial.

p\left(x\right)=x^2,\ x=0
Answer: For the value of x=0, the output of the polynomial will be:
p\left(0\right)=0^2=0
Therefore, 0 is the zero of the polynomial.

p\left(x\right)=lx+m,\ x=-m/l
Answer: For the value of x=-m/l , the output of the polynomial will be:
p\left(-m/l\right)=l\left(-\frac{m}{l}\right)+m=0
Therefore, -\frac{m}{l} is the zero of the polynomial.

p\left(x\right)=3x^2-1,\ x=-\frac{1}{\sqrt3},\frac{2}{\sqrt3}
Answer: For the value of x=-\frac{1}{\sqrt3} , the output of the polynomial will be:
p\left(-\frac{1}{\sqrt3}\right)=3\left(-\frac{1}{\sqrt3}\right)^2-1=0
For the value of x=2/\sqrt3 , the output of the polynomial will be:
p\left(\frac{2}{\sqrt3}\right)=3\left(\frac{2}{\sqrt3}\right)^2-1=3
Therefore, -\frac{1}{\sqrt3} is the zero of the polynomial.

p\left(x\right)=2x+1,\ x=\frac{1}{2}
Answer: For the value of x=1/2 , the output of the polynomial will be:
p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1=2
Therefore, \frac{1}{2} is not the zero of the polynomial.

Q4. Find the zero of the polynomials in each of the following cases:
p\left(x\right)=x+5
Answer:
p(x)\ =\ x+5
=>\ x+5\ =\ 0
=>x\ =\ -5
Therefore, -5 is a zero of the given polynomial.

p\left(x\right)=x-5
Answer:
p(x)\ =\ x-5
=>\ x-5\ =\ 0
=>x\ =\ 5
Therefore, 5 is a zero of the given polynomial.
p\left(x\right)=2x+5
Answer:
p(x)\ =\ 2x+5
=>2x+5\ =\ 0
=>2x\ =\ -5
=>\ x\ =\ -5/2
Therefore, -5/2 is a zero of the given polynomial.

p\left(x\right)=3x-2
Answer:
p(x)\ =\ 3x–2
=>3x-2\ =\ 0
=>3x\ =\ 2
=>x\ =\ 2/3
Therefore, -2/3 is a zero of the given polynomial.

p\left(x\right)=3x
Answer:
p(x)\ =\ 3x
=>\ 3x\ =\ 0
=>x\ =\ 0
Therefore, 0 is a zero of the given polynomial.

p\left(x\right)=ax,\ a\neq0
Answer:
p(x)\ =\ ax
=>\ ax\ =\ 0
=>\ x\ =\ 0
Therefore, 0 is a zero of the given polynomial.

p\left(x\right)=cx+d,\ c\neq0,c,d\ are\ real\ numbers
Answer:
p(x)\ =\ cx\ +\ d
=>\ cx+d\ =0
=>\ x\ =\ -d/c
Therefore, -d/c is a zero of the given polynomial.

Exercise 2.3:

Q1. Find the remainder when x^3+3x^2+3x+1 is divided by
x+1
Answer:
x+1=\ 0
\Rightarrow x\ =\ -1
∴The Remainder will be:
p(-1)\ =\ \left(-1\right)^3+3\left(-1\right)^2+3(-1)+1

=\ -1+3-3+1
=\ 0

x-\frac{1}{2}
Answer:
x-1/2\ =\ 0
\Rightarrow\ x\ =\ ½
∴The Remainder will be:
p\left(\frac{1}{2}\right)\ =\ \left(\frac{1}{2}\right)^3+3\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)+1
=\ \left(\frac{1}{8}\right)+\left(\frac{3}{4}\right)+\left(\frac{3}{2}\right)+1
=\ 27/8

x
Answer:
x\ =\ 0
∴The Remainder will be:
p(0)\ =\ \left(0\right)^3+3\left(0\right)^2+3(0)+1
=\ 1

x+\pi
Answer:
x+\pi\ =\ 0
\Rightarrow\ x\ =\ -\pi
∴The Remainder will be:
p(0)\ =\ \left(-\pi\right)^3\ +3\left(-\pi\right)^2+3(-\pi)+1
=\ -\pi^3+3\pi^2-3\pi+1

5+2x
Answer:
5+2x=0
\Rightarrow\ 2x\ =\ -5
\Rightarrow\ x\ =\ -5/2
∴The Remainder will be:
\left(-\frac{5}{2}\right)^3+3\left(-\frac{5}{2}\right)^2+3\left(-\frac{5}{2}\right)+1\
=\ \left(-\frac{125}{8}\right)+\left(\frac{75}{4}\right)-\left(\frac{15}{2}\right)+1
=-\frac{27}{8}

Q2. Find the remainder when x^3-ax^2+6x-a is divided by x-a.
Answer:
Let us assume
p(x)\ =\ x^3-ax^2+6x-a
x-a\ =\ 0
x\ =\ a
The Remainder will be:
p(a)\ =\ \left(a\right)^3-a(a^2)+6(a)-a
=\ a^3-a^3+6a-a\
=\ 5a

Q3. Check whether 7+3x is a factor of 3x^3+7x.
Answer:
7+3x\ =\ 0
\Rightarrow\ 3x\ =\ -7
\Rightarrow\ x\ =\ -7/3
∴The Remainder will be:
3\left(-\frac{7}{3}\right)^3+7\left(-\frac{7}{3}\right)\
=\ -\left(\frac{343}{9}\right)+\left(-\frac{49}{3}\right)
=\frac{-343-\left(49\right)3}{9}
=\frac{-343-147}{9}
=\ -\frac{490}{9}\ \neq\ 0
Therefore, 7+3x\ is not a factor of \ 3x^3+7x .

Exercise 2.4

Q1. Determine which of the following polynomials has (x + 1) a factor:
x^3+x^2+x+1
Answer:
Assuming a function:
f\left(x\right)=x^3+x^2+x+1
x+1=0
=>x=-1
Substituting x=-1 in the polynomial.
f(-1)\ =\ \left(-1\right)^3+\left(-1\right)^2+(-1)+1
=\ -1+1-1+1
=\ 0
Therefore, according to factor theorem, x+1 is a factor of the given polynomial.

x^4+x^3+x+1
Answer:
Assuming a function:
f\left(x\right)=x^4+x^3+x^2+x+1
x+1=0
=>x=-1
Substituting x=-1 in the polynomial.
f(-1)\ =\ \left(-1\right)^4+\left(-1\right)^3+\left(-1\right)^2+(-1)+1
=\ 1-1+1-1+1
=\ 1\ \neq\ 0
Therefore, according to the factor theorem, x+1 is not a factor of the given polynomial.

x^4+3x^3+3x^2+x+1
Answer:
Assuming a function:
f\left(x\right)=x^4+{3x}^3+3x^2+x+1
x+1=0
=>x=-1
Substituting x=-1 in the polynomial.
f(-1)=\left(-1\right)^4+3\left(-1\right)^3+3\left(-1\right)^2+(-1)+1
=1-3+3-1+1
=1\ \neq\ 0
Therefore, according to the factor theorem, x+1 is not a factor of the given polynomial.

x^3-x^2-\left(2+\sqrt2\right)x+\sqrt2
Answer:
Assuming a function:
f\left(x\right)=x^3+x^2-(2+\sqrt2)x+\sqrt2
x+1=0
=>x=-1
Substituting x=-1 in the polynomial.
f\left(-1\right)=\ \left(-1\right)^3–(-1)2–(2+√2)(-1)+ √2
=\ -1-1+2+\sqrt2+\sqrt2
=\ 2\sqrt2\ \neq\ 0
Therefore, according to the factor theorem, x+1 is not a factor of the given polynomial.

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
\ p\left(x\right)=\ 2x^3+x^2–2x–1, g(x) = x+1
Answer:
Firstly, we will calculate the zero of g(x) then we will substitute the zero of g(x) in p(x) to see if g(x) is a factor of p(x) or not.
g\left(x\right)=0
x+1=0
x=-1
Substituting the zero of g(x) in p(x).
p(-1)\ =\ 2\left(-1\right)^3+\left(-1\right)^2–2(-1)–1
=\ -2+1+2-1
=\ 0
Therefore, according to the factor theorem g(x) is a factor of p(x).

p\left(x\right)=x^3+3x^2+3x+1,\ \ g\left(x\right)=\ x+2
Answer:
Firstly, we will calculate the zero of g(x) then we will substitute the zero of g(x) in p(x) to see if g(x) is a factor of p(x) or not.
g(x)\ =\ 0
=>\ x+2\ =\ 0
=>\ x\ =\ -2
Substituting the zero of g(x) in p(x).
p(-2)\ =\ \left(-2\right)^3+3\left(-2\right)^2+3(-2)+1
=\ -8+12-6+1
=\ -1\ \neq\ 0
Therefore, according to the factor theorem g(x) is not a factor of p(x).

p\left(x\right)=x^3–4x^2+x+6, g(x)= x–3
Answer:
Firstly, we will calculate the zero of g(x) then we will substitute the zero of g(x) in p(x) to see if g(x) is a factor of p(x) or not.
g(x)\ =\ 0
=>x-3\ =\ 0
=>x\ =\ 3
Substituting the zero of g(x) in p(x).
p(3)\ =\ \left(3\right)^3-4\left(3\right)^2+(3)+6
=\ 27-36+3+6
=\ 0
Therefore, according to the factor theorem g(x) is a factor of p(x).

Q3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:
p(x)\ =\ x^2+x+k
Answer:
If x-1 is a p(x) factor, then p(1) = 0.
According to Factor Theorem:
p\left(1\right)=\left(1\right)^2+(1)+k\ =\ 0
\Rightarrow\ 1+1+k\ =\ 0
\Rightarrow\ 2+k\ =\ 0
\Rightarrow\ k\ =\ -2
p(x)\ =\ 2x^2+kx+\sqrt2
Answer:
If x-1 is a p(x) factor, then p(1) = 0.
According to Factor Theorem:
p\left(1\right)=\ 2\left(1\right)^2+k(1)+\sqrt2\ =\ 0
\Rightarrow\ 2+k+\sqrt2\ =\ 0
\Rightarrow\ k\ =\ -(2+\sqrt2)

p(x)\ =\ kx^2–√2 x+1
Answer:
If x-1 is a p(x) factor, then p(1) = 0.
According to Factor Theorem:
p\left(1\right)=\ k\left(1\right)^2-\sqrt2(1)+1=0
\Rightarrow\ k\ =\ \sqrt2-1

p(x)=kx^2–3x+k
Answer: If x-1 is a p(x) factor, then p(1) = 0.
According to Factor Theorem:
p\left(1\right)=k\left(1\right)^2–3(1)+k = 0
\Rightarrow\ k-3+k\ =\ 0
\Rightarrow\ 2k-3\ =\ 0
\Rightarrow\ k=\ 3/2

Q4. Factorize:
12x^2–7x+1
Answer:
Using the procedure of breaking the middle term, we must discover a number with a total of -7 and a product of 12.
The integers -3 and -4 are obtained [-3+-4=-7 and -3×-4 = 12].
12x^2–7x+1= 12x^2-4x-3x+1
=\ 4x(3x-1)-1(3x-1)
=\ (4x-1)(3x-1)

2x^2+7x+3
Answer:
Using the procedure of breaking the middle term, we must discover a number with a total of 7 and a product of 6.
The integers 6 and 1 are obtained [6+1=7 and 6×1 = 6].
2x^2+7x+3\ =\ 2x^2+6x+1x+3
=\ 2x\ (x+3)+1(x+3)
=\ (2x+1)(x+3)

6x^2+5x-6
Answer:
Using the procedure of breaking the middle term, we must discover a number with a total of 5 and a product of -36.
The integers -4 and -9 are obtained [-4+9 = 5 and -4×9 = -36].
6x^2+5x-6\ =\ 6x^2+9x–4x–6
=\ 3x(2x+3)–2(2x+3)
=\ (2x+3)(3x–2)

3x^2–x–4
Answer:
Using the procedure of breaking the middle term, we must discover a number with a total of -1 and a product of -12.
The integers -4 and 3 are obtained [-4+3 = -1 and -4×3 = -12].
3x^2–x–4 = 3x^2–4x+3x–4
=\ x(3x–4)+1(3x–4)
=\ (3x–4)(x+1)

Q5. Factorize:
x^3-2x^2-x+2
Answer:
Let us assume a function p\left(x\right)=x^3-2x^2-x+2
Now, the factors of 2 are: \pm1\ and\pm2
p(x)\ =\ x^3–2x^2–x+2
p(-1)\ =\ \left(-1\right)^3–2(-1)^2–(-1)+2
=\ -1-2+1+2
=\ 0
So, (x+1) can be said as a factor of the function p(x).
The factorized polynomial will be:
\left(x+1\right)(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)

x^3-3x^2-9x-5
Answer:
Let us assume a function p\left(x\right)=x^3-3x^2-9x-5
Now, the factors of 5 are: \pm1\ and\pm5.
Using the trial method, we can find that p(5)=0.
Therefore, x-5 is a zero of p(x).
p(x)\ =\ x^3–3x^2–9x–5
p(5)\ =\ \left(5\right)^3–3(5)^2–9(5)–5
=\ 125-75-45-5
=\ 0
The factorized polynomial will be:
\left(x-5\right)\left(x^2+2x+1\right)
=\ (x-5)(x^2+x+x+1)
=\ (x-5)(x(x+1)+1(x+1))
=\ (x-5)(x+1)(x+1)

x^3+13x^2+32x+20
Answer:
Let us assume a function p\left(x\right)=x^3+13x^2+32x+20
Now, the factors of 20 are: \pm1,\pm2,\pm4,\pm5,\pm10\ \ and\pm20.
Using the trial method, we can find that p(-1)=0
Therefore, x+1 is a zero of p(x).
p(x)=\ x^3+13x^2+32x+20
p(-1)\ =\ \left(-1\right)^3+13\left(-1\right)^2+32(-1)+20
=\ -1+13-32+20
=\ 0
The factorized polynomial will be:
\left(x+1\right)\left(x^2+12x+20\right)
=\ (x+1)(x2+2x+10x+20)
=\ (x+1)x(x+2)+10(x+2)
=\ (x+1)(x+2)(x+10)

2y^3+y^2-2y-1
Answer:
Let us assume a function p\left(y\right)=2y^3+y^2-2y-1
Now, the factors of 2×(−1)= -2 are ±1 and ±2.
Using the trial method, we can find that p(1)=0.
Therefore, y-1 is a zero of p(y)
p(y)\ =\ 2y^3+y^2–2y–1
p(1)\ =\ 2\left(1\right)^3+\left(1\right)^2–2(1)–1
=\ 2+1-2
=\ 0
The factorized polynomial will be:
\left(y-1\right)\left(2y^2+3y+1\right)
=\ (y-1)(2y^2+2y+y+1)
=\ (y-1)(2y(y+1)+1(y+1))
=\ (y-1)(2y+1)(y+1)

Exercise 2.5:

Q1. Use suitable identities to find the following products:
(x+4)(x+10)
Answer:
We will be using the identity: \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab
Now, we will use: a=4\ and\ b=10 :
\left(x+4\right)\left(x+10\right)
=\ x^2+(4+10)x+(4\times10)
=\ x^2+14x+40

(x+8)(x-10)
Answer:
We will be using the identity: \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab
Now, we will use: a=8\ and\ b=-10 :
\left(x+8\right)\left(x-10\right)
=\ x^2+(8+(-10))x+(8\times(-10))
=\ x^2+(8-10)x–80
=\ x^2-2x-80

(3x+4)(3x-5)
Answer:
We will be using the identity: \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab
Now, we will use: x=3x,\ a=4\ and\ b=-5 :
\left(3x+4\right)\left(3x-5\right)
=\ \left(3x\right)^2+[4+(-5)]3x+4×(-5)
=\ 9x^2+3x(4–5)–20
=\ 9x^2–3x–20

\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)
Answer:
We will be using the identity: \left(x+y\right)\left(x-y\right)=x^2-y^2.
Now, we will use: x=y^2\ and\ y=3/2 .
\left(y^2+\frac{3}{2}\right)(y^2–3/2)
=\ \left(y^2\right)^2–(3/2)^2
=\ y^4–9/4

Q2. Evaluate the following products without multiplying directly:
103\times107
Answer:
103\times107=\ (100+3)\times(100+7)
We will be using the identity: \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab
Now, we will use: x=100,\ a=3\ and\ b=7.
103\times107\ =\ (100+3)\times(100+7)
=\ \left(100\right)^2+(3+7)100+(3\times7)
=\ 10000+1000+21
=\ 11021

95\times96
Answer:
95\times96\ =\ (100-5)\times(100-4)
Now, we will use: x=100,\ a=-5\ and\ b=-4.
95\times96\ =\ (100-5)\times(100-4)
=\ \left(100\right)^2+100(-5+(-4))+(-5\times-4)
=\ 10000-900+20
=\ 9120

104\times96
Answer:
104\times96\ =\ (100+4)\times(100–4)
We will be using the identity: \left(x+y\right)\left(x-y\right)=x^2-y^2.
Now, we will use: a=100\ and\ b=4 .
104\times96\ =\ (100+4)\times(100–4)
=\ \left(100\right)^2–(4)^2
=\ 10000–16
=\ 9984

Q3. Factorize the following using appropriate identities:
9x^2+6xy+y^2
Answer: 9x^2+6xy+y^2\ =\ \left(3x\right)^2+(2\times3x\times y)+y^2
We will be using the identity: x^2+2xy+y^2=\left(x+y\right)^2
Now, we will use: x=3x\ and\ y=y .
9x^2+6xy+y^2\
=\ \left(3x\right)^2+(2\times3x\times y)+y^2
=\ \left(3x+y\right)^2
=\ (3x+y)(3x+y)

4y^2-4y+1
Answer:
4y^2-4y+1\ =\ \left(2y\right)^2–(2×2y×1)+1
We will be using the identity: x^2-2xy+y^2=\left(x-y\right)^2
Now, we will use: x=2y\ and\ y=1 .
4y^2-4y+1\
=\ \left(2y\right)^2–(2×2y×1)+12
=\ (2y–1)^2
=\ (2y–1)(2y–1)

x^2-\frac{y^2}{100}
Answer:
x^2–y^2/100 = x^2–(y/10)^2
We will be using the identity: x^2-y^2=(x-y)(x+y)
Now, we will use: x=x\ and\ y=\frac{y}{10} .
x^2–y^2/100
=\ x^2–(y/10)^2
=\ (x–y/10)(x+y/10)

Q4. Expand each of the following, using suitable identities:
\left(x+2y+4z\right)^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx
Now, we will use: x=x,\ y=2y\ and\ z=4z .
\left(x+2y+4z\right)^2\
=\ x^2+\left(2y\right)^2+\left(4z\right)^2+(2\times x\times2y)+(2\times2y\times4z)+(2\times4z\times x)
=\ x^2+4y^2+16z^2+4xy+16yz+8xz

\left(2x-y+z\right)^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
Now, we will use: x=2x,\ y=-y\ and\ z=z .
\left(2x-y+z\right)^2\
=\left(2x\right)^2+\left(-y\right)^2+z^2+(2\times2x\times-y)+(2\times-y\times z)+(2\times z\times2x)
=\ 4x^2+y^2+z^2–4xy–2yz+4xz

\left(-2x+3y+2z\right)^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
Now, we will use: x=-2x,\ y=3y\ and\ z=2z .
\left(-2x+3y+2z\right)^2
=\ \left(-2x\right)^2+\left(3y\right)^2+\left(2z\right)^2+(2\times-2x\times3y)+(2\times3y\times2z)+(2\times2z\times-2x)
=\ 4x^2+9y^2+4z^2–12xy+12yz–8xz

\left(3a-7b-c\right)^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
Now, we will use: x=3a,\ y=-7b\ and\ z=-c .
(3a –7b– c)^2
=\ \left(3a\right)^2+(– 7b)^2+(– c)^2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
=\ 9a^2\ +\ 49b^2\ +\ c^2– 42ab+14bc–6ca

(–2x+5y–3z)^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
Now, we will use: x=-2x,\ y=5y\ and\ z=-3z .
(–2x+5y–3z)^2
=\ (–2x)^2+(5y)^2+(–3z)^2+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)
=\ 4x^2+25y^2\ +9z^2– 20xy–30yz+12zx

\left[\frac{1}{4}a-\frac{1}{2}b+1\right]^2
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
Now, we will use: x=\frac{1}{4}a,\ y=-\frac{1}{2}b\ and\ z=1 .
\left[\frac{1}{4}a-\frac{1}{2}b+1\right]^2
=\left(\frac{1}{4}a\right)^2+\left(-\frac{1}{2}b\right)^2+\left(1\right)^2+\left(2\times\frac{1}{4}a\times\frac{1}{2}b\right)+\left(2\times-\frac{1}{2}b\times1\right)+\left(2\times1\times\frac{1}{4}a\right)
=\frac{1}{16}a^2+\frac{1}{4}b^2+1^2-\frac{2}{8}ab-\frac{2}{2}b+\frac{2}{4}a
=\frac{1}{16}a^2+\frac{1}{4}b^2+1-\frac{1}{4}ab-b+\frac{1}{2}a

Q5. Factorize:
4x^2+9y^2+16z^2+12xy–24yz–16xz
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\
4x^2+9y^2+16z^2+12xy–24yz–16xz
=\ \left(2x\right)^2+\left(3y\right)^2+\left(-4z\right)^2+(2\times2x\times3y)+(2\times3y\times-4z)+(2\times-4z\times2x)
=\ (2x+3y–4z)^2
=\ (2x+3y–4z)(2x+3y–4z)
2x^2+y^2+8z^2–2√2 xy+4√2 yz–8xz
Answer:
We will be using the identity: \left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\ .
2x^2+y^2+8z^2–2√2 xy+4√2 yz–8xz
=\ \left(-\sqrt2x\right)^2+\left(y\right)^2+\left(2\sqrt2z\right)^2+(2\times-\sqrt2x\times y)+(2\times y\times2\sqrt2z)+(2\times2\sqrt2\times-\sqrt2x)
=\ \left(-\sqrt2x+y+2\sqrt{2z}\right)^2
=\ (-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)

Q6. Write the following cubes in expanded form:
\left(2x+1\right)^3
Answer:
We will be using the identity: \left(x+y\right)^3=x^3+y^3+3xy(x+y)
\left(2x+1\right)^3
=\ \left(2x\right)^3+13+(3\times2x\times1)(2x+1)
=\ 8x^3+1+6x(2x+1)
=\ 8x^3+12x^2+6x+1

\left(2a-3b\right)^3
Answer:
We will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y)
\left(2a-3b\right)^3\
=\ \left(2a\right)^3-\left(3b\right)^3–(3×2a×3b)(2a–3b)
=\ 8a^3–27b^3–18ab(2a–3b)
=\ 8a^3–27b^3–36a^2 b+54ab^2

\left(\left(\frac{3}{2}\right)x+1\right)^3
Answer:
We will be using the identity: \left(x+y\right)^3=x^3+y^3+3xy(x+y)
\left(\left(\frac{3}{2}\right)x+1\right)^3
=\left(\left(\frac{3}{2}\right)x\right)^3+13+\left(3\times\left(\frac{3}{2}\right)x\times1\right)\left(\left(\frac{3}{2}\right)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+1+\frac{27}{4}x^2+\frac{9}{2}x
=\frac{27}{8}x^3+\frac{27}{4}x^2+\frac{9}{2}x+1

\left(x-\left(\frac{2}{3}\right)y\right)^3
Answer:
We will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y)
\left(x-\frac{2}{3}y\right)^3
=\left(x\right)^3-\left(\frac{2}{3}y\right)^3-\left(3\times x\times\frac{2}{3}y\right)\left(x-\frac{2}{3}y\right)
=\left(x\right)^3-\frac{8}{27}y^3-2xy\left(x-\frac{2}{3}y\right)
=\left(x\right)^3-\frac{8}{27}y^3-2x^2y+\frac{4}{3}xy^2

Q7. Evaluate the following using suitable identities:
\left(99\right)^3
Answer:
The number 99 can be written as 100-1.
Now, we will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y)
\left(99\right)^3\
=\ (100–1)^3
=\ \left(100\right)^3–13–(3×100×1)(100–1)
=\ 1000000\ –1–300(100 – 1)
=\ 1000000–1–30000+300
=\ 970299

\left(102\right)^3
Answer:
The number 102 can be written as 100+2.
Now, we will be using the identity: \left(x+y\right)^3=x^3+y^3+3xy(x+y)
\left(100+2\right)^3\
=\left(100\right)^3+2^3+(3\times100\times2)(100+2)
=\ 1000000\ +\ 8\ +\ 600(100\ +\ 2)
=\ 1000000\ +\ 8\ +\ 60000\ +\ 1200
=\ 1061208

\left(998\right)^3
Answer:
The number 998 can be written as 1000-2.
Now, we will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y).
=\left(1000\right)^3–2^3–(3×1000×2)(1000–2)
=\ 1000000000–8–6000(1000– 2)
=\ 1000000000–8- 6000000+12000
=\ 994011992

Q8. Factorize each of the following:
8a^3+b^3+12a^2b+6ab^2
Answer:
8a^3+b^3+12a^2b+6ab^2 may be written as \left(2a\right)^3+b^3+3\left(2a\right)^2b+3\left(2a\right)\left(b\right)^2
Now, we will be using the identity: \left(x+y\right)^3=x^3+y^3+3xy(x+y)
8a^3+b^3+12a^2b+6ab^2\
=\ \left(2a\right)^3+b^3+3\left(2a\right)^2b+3\left(2a\right)\left(b\right)^2
=\ \left(2a+b\right)^3
=\ (2a+b)(2a+b)(2a+b)

8a^3–b^3–12a^2 b+6ab^2
Answer:
8a^3–b^3–12a^2 b+6ab^2 may be written as: \left(2a\right)^3–b^3–3(2a)^2 b+3(2a) (b)^2
Now, we will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y).
8a^3–b^3-12a^2 b+6ab^2
=\ \left(2a\right)^3–b^3–3(2a)^2 b+3(2a) (b)^2
=\ (2a–b)^3
=\ (2a–b)(2a–b)(2a–b)

27–125a^3–135a+225a^2
Answer:
27–125a^3–135a+225a^2 may be written as 3^3–(5a)^3–3(3)^2 (5a)+3(3) (5a)^2
Now, we will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y).
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)

64a^3–27b^3–144a^2 b+108ab^2
Answer:
64a^3–27b^3–144a^2 b+108ab^2 may be written as \left(4a\right)^3–(3b)^3–3(4a)^2 (3b)+3(4a) (3b)^2 .
Now, we will be using the identity: \left(x-y\right)^3=x^3-y^3-3xy(x-y).
64a^3–27b^3–144a^2 b+108ab^2
=\left(4a\right)^3–(3b)^3–3(4a)^2 (3b)+3(4a) (3b)^2
=(4a–3b)^3
=(4a–3b)(4a–3b)(4a–3b)

7p^3– (1/216)-(9/2) p^2+(1/4)p
Answer:
7p^3– (1/216)-(9/2) p^2+(1/4)p may be written as \left(3p\right)^3–(1/6)^3–3(3p)^2 (1/6)+3(3p) (1/6)^2
27p^3–(1/216)-(9/2) p^2+(1/4)p
=\left(3p\right)^3–(1/6)^3–3(3p)^2 (1/6)+3(3p) (1/6)^2
=\ (3p–16)^3
=\ (3p–16)(3p–16)(3p–16)

Q9. Verify:
x^3+y^3=\left(x+y\right)(x^2–xy+y^2 )
Answer:

We will be using the identity: \left(x+y\right)^3=x^3+y^3+3xy(x+y)
\Rightarrow\ x^3+y^3\ =\ \left(x+y\right)^3–3xy(x+y)
\Rightarrow\ x^3+y^3\ =\ \left(x+y\right)[(x+y)2–3xy]
We will take (x+y) as 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)

x^3–y^3 = (x–y)(x^2+xy+y^2 )
Answer:

we will be using the identity: \left(x-y\right)^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]
We will take (x+y) as 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)

Q10. Factorize each of the following:
27y^3+125z^3
Answer:

27y^3+125z^3 may be written as \left(3y\right)^3+\left(5z\right)^3
We will be using the identity x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)
27y^3+125z^3\
=\ \left(3y\right)^3+\left(5z\right)^3
=\ \left(3y+5z\right)[(3y)^2–(3y)(5z)+(5z)^2 ]
=\ (3y+5z)(9y^2–15yz+25z^2)

64m^3–343n^3
Answer:
64m^3–343n^3 may be written as \left(4m\right)^3-\left(7n\right)^3.
We will be using the identity x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)
64m^3–343n^3
=\ \left(4m\right)^3–(7n)^3
=\ \left(4m-7n\right)\left[\left(4m\right)^2+\left(4m\right)\left(7n\right)+\left(7n\right)^2\right]
=\ \left(4m-7n\right)\left(16m^2+28mn+49n^2\right)

Q11. Factorize: 27x^3+y^3+z^3–9xyz
Answer:

27x^3+y^3+z^3–9xyz may be written as \left(3x\right)^3+y^3+z^3–3(3x)(y)(z)
We will be using the identity: x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
27x^3+y^3+z^3–9xyz
=\ \left(3x\right)^3+y^3+z^3–3(3x)(y)(z)
=\ (3x+y+z)[\left(3x\right)^2+y^2+z^2–3xy–yz–3xz]
=\ (3x+y+z)(9x^2+y^2+z^2–3xy–yz–3xz)

Q12. Verify that: x^3+y^3+z^3–3xyz = (1/2) (x+y+z)[(x–y)^2+(y–z)^2+(z–x)^2 ]
Answer:

We will be using the identity: x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
\Rightarrow\ x^3+y^3+z^3–3xyz = (1/2)(x+y+z)[2(x^2+y^2+z^2–xy–yz–xz)]
=\ \left(\frac{1}{2}\right)(x+y+z)(2x^2+2y^2\ +2z^2–2xy–2yz–2xz)
=\ \left(\frac{1}{2}\right)\left(x+y+z\right)[(x^2+y^2-2xy)+(y^2+z^2–2yz)+(x^2+z^2–2xz)]
=\ \left(\frac{1}{2}\right)\left(x+y+z\right)[(x–y)^2+(y–z)^2+(z–x)^2 ]

Q13. If x+y+z\ =\ 0, show that x^3+y^3+z^3=3xyz.
Answer:

We will be using the identity: x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
As mentioned in the question: \left(x+y+z\right)=0
then,
\ 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
\Rightarrow\ x^3+y^3+z^3\ =\ 3xyz

Q14. Without actually calculating the cubes, find the value of each of the following:
\left(-12\right)^3+\left(7\right)^3+\left(5\right)^3
Answer:

We already know: if x+y+z=0 then x^3+y^3+z^3=3xyz
now, in this case: -12+7+5=0
If we consider, x=-12,\ y=7\ and\ z=5 then:
\left(-12\right)^3+\left(7\right)^3+\left(5\right)^3=3xyz
=3\times-12\times7\times5
=-1260

\left(28\right)^3+\left(-15\right)^3+\left(-13\right)^3
Answer:

We already know: if x+y+z=0 then x^3+y^3+z^3=3xyz
now, in this case: 28-15-13=0
\left(28\right)^3+\left(-15\right)^3+\left(-13\right)^3\ =\ 3xyz
=\ 0+3(28)(-15)(-13)
=\ 16380

Q15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
Area: 25a^2-35a+12
Answer:
Using the technique of separating the middle term, we must discover a number with a sum of -35 and a product of 25×12=300.
As numbers, we obtain -15 and -20.
25a^2–35a+12
=\ 25a^2–15a-20a+12
=\ 5a(5a–3)–4(5a–3)
=\ (5a–4)(5a–3)
The expression that might account for the length will be: (5a-4)
The expression that might account for the width will be: (5a-3)

Area: 35y^2+13y–12
Answer:
Using the technique of separating the middle term, we must discover a number with a sum of 13 and a product of 35x-12=-420.
As numbers, we obtain -15 and 28.
35y^2+13y–12
=\ 35y^2–15y+28y–12
=\ 5y(7y–3)+4(7y–3)
=\ (5y+4)(7y–3)
The expression that might account for the length will be: (5y+4)
The expression that might account for the width will be: (7y-3)

Q16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Volume: 3x^2–12x
Answer: The given expression may be represented as: 3x(x-4)
The expression that might account for the length will be: 3
The expression that might account for the width will be: x
The expression that might account for the height will be: (x-4)

Volume: 12ky^2+8ky–20k
Answer: The given expression may be represented as: 4k(3y^2+2y-5)
Now, we can write the term (3y^2+2y-5) as: (3y^2+5y-3y-5)
=4k(3y^2+5y–3y–5)
=\ 4k[y(3y+5)–1(3y+5)]
=\ 4k(3y+5)(y–1)
The expression that might account for the length will be: 4k
The expression that might account for the width will be: (3y+5)
The expression that might account for the height will be: (y-1)