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Have you noticed how your score in a video game increases with each level you pass? Or how your savings grow month after month? Well, they don't just change randomly; they follow a certain pattern, that's what the Polynomials are all about. For example, a second-order polynomial (highest exponent 2) like $2x^2+5x+2$ can represent how your score increases as you progress through a game where $x$ is the game level. In the second chapter of the NCERT Class 9 Maths, you will find Polynomials, which are algebraic expressions of variables and coefficients - sound complex? Don't worry, we will break it down in a simple and logical manner that will help you build your foundation in algebra.
This article on NCERT solutions for class 9 Maths Chapter 2 Polynomials offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 9 Maths. Students who are in need of Polynomials class 9 solutions will find this article very useful. It covers all the important Class 9 Maths Chapter 2 question answers. These Polynomials class 9 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 9 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.
The general form of a polynomial is: p(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0
where a0, a1, a2, …., an are constants, and an ≠ 0.
Every one-variable linear polynomial will contain a unique zero, which is a real number that is a zero of the zero polynomial, and a non-zero constant polynomial that does not have any zeros.
Remainder Theorem: If p(x) has a degree greater than or equal to 1, and you divide p(x) by the linear polynomial (x - a), the remainder will be p(a).
Factor Theorem: The linear polynomial $(x-a)$ will be a factor of the polynomial p(x) whenever p(a) = 0. Similarly, if $(x-a)$ is a factor of p(x), then p(a) = 0.
NCERT Polynomials class 9 solutions Exercise: 2.1
Page number: 29, Total questions: 5
Answer:
Yes, the polynomial $4x^2 - 3x + 7$ has only one variable, which is $x$.
Answer:
YES
Given polynomial has only one variable which is y.
Answer:
NO
Because we can observe that the exponent of variable t in term $3\sqrt t$ is $\frac{1}{2}$ which is not a whole number.
Therefore, this expression is not a polynomial.
Answer:
NO
Because we can observe that the exponent of variable y in term $\frac{2}{y}$ is $-1$ which is not a whole number. Therefore this expression is not a polynomial.
Answer:
NO
Because in the given polynomial $x^{10} + y^3 + t^{50}$ there are 3 variables which are x, y, t. That's why this is a polynomial in three variables, not in one variable.
Q2. (i) Write the coefficients of $x^2$ in the following: $2 + x^2 +x$
Answer:
Coefficient of $x^2$ in polynomial $2 + x^2 +x$ is 1.
Q2. (ii) Write the coefficients of $x^2$ in the following: $2 - x^2 + x^3$
Answer:
The coefficient of $x^2$ in polynomial $2 - x^2 + x^3$ is -1.
Q2. (iii) Write the coefficients of $x^2$ in the following: $\frac{\pi}{2}x^2 + x$
Answer:
Coefficient of $x^2$ in polynomial $\frac{\pi}{2}x^2 + x$ is $\frac{\pi}{2}$
Q2. (iv) Write the coefficients of $x^2$ in the following: $\sqrt2 x - 1$
Answer:
Coefficient of $x^2$ in polynomial $\sqrt2 x - 1$ is 0
Q3 Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Answer:
The degree of a polynomial is the highest power of the variable in the polynomial.
In binomial, there are two terms
Therefore, a binomial of degree 35 is
Eg: $x^{35}+1$
In a monomial, there is only one term in it.
Therefore, a monomial of degree 100 can be written as $y^{100}$
Q4. (i) Write the degree the following polynomial: $5x^3 + 4x^2 + 7x$
Answer:
The degree of a polynomial is the highest power of the variable in the polynomial.
Therefore, the degree of the polynomial $5x^3 + 4x^2 + 7x$ is 3.
Q4. (ii) Write the degree the following polynomial: $4 - y^2$
Answer:
The degree of a polynomial is the highest power of the variable in the polynomial.
Therefore, the degree of polynomial $4 - y^2$ is 2.
Q4. (iii) Write the degree the following polynomial: $5t - \sqrt7$
Answer:
The degree of a polynomial is the highest power of the variable in the polynomial.
Therefore, the degree of polynomial $5t - \sqrt7$ is 1
Q4. (iv) Write the degree the following polynomial: 3
Answer:
The degree of a polynomial is the highest power of the variable in the polynomial.
In this case, only a constant value 3 is there and the degree of a constant polynomial is always 0.
Q5. (i) Classify the following as linear, quadratic and cubic polynomial: $x^2+x$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively
Given polynomial is $x^2+x$ with degree 2
Therefore, it is a quadratic polynomial.
Q5. (ii) Classify the following as linear, quadratic and cubic polynomials: $x - x^3$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial have its degrees as 1, 2, and 3, respectively
Given polynomial is $x - x^3$ with degree 3
Therefore, it is a cubic polynomial
Q5 (iii) Classify the following as linear, quadratic and cubic polynomials: $y + y^2 + 4$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial have its degrees as 1, 2, and 3, respectively
Given polynomial is $y + y^2 + 4$ with degree 2.
Therefore, it is a quadratic polynomial.
Q5. (iv) Classify the following as linear, quadratic and cubic polynomials: $1+x$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively
Given polynomial is $1 +x$ with degree 1
Therefore, it is linear polynomial
Q5. (v) Classify the following as linear, quadratic and cubic polynomial: $3t$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively
Given polynomial is $3t$ with degree 1
Therefore, it is a linear polynomial
Q5. (vi) Classify the following as linear, quadratic and cubic polynomials: $r^2$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial have their degrees as 1, 2, and 3, respectively
Given polynomial is $r^2$ with degree 2
Therefore, it is a quadratic polynomial
Q5. (vii) Classify the following as linear, quadratic and cubic polynomials: $7x^3$
Answer:
Linear polynomial, quadratic polynomial, and cubic polynomial have their degrees as 1, 2, and 3, respectively
Given polynomial is $7x^3$ with degree 3
Therefore, it is a cubic polynomial.
Polynomials class 9 NCERT solutions Exercise: 2.2
Page number: 31-32, Total questions: 4
Q1. (i) Find the value of the polynomial $5x - 4x^2 +3$ at $x = 0$
Answer:
Given polynomial is $5x - 4x^2 +3$
Now, at $x = 0$ value is
$\Rightarrow 5(0)-4(0)^2+3 = 0 - 0 + 3 = 3$
Therefore, value of polynomial $5x - 4x^2 +3$ at x = 0 is 3
Q1. (ii) Find the value of the polynomial $5x - 4x^2 +3$ at $x = -1$
Answer:
Given polynomial is $5x - 4x^2 +3$
Now, at $x = -1$ value is
$\Rightarrow 5(-1)-4(-1)^2+3 = -5 - 4 + 3 = -6$
Therefore, value of polynomial $5x - 4x^2 +3$ at x = -1 is -6
Q1. (iii) Find the value of the polynomial $5x - 4x^2 +3$ at $x = 2$
Answer:
Given polynomial is $5x - 4x^2 +3$
Now, at $x = 2$ value is
$\Rightarrow 5(2)-4(2)^2+3 = 10 - 16 + 3 = -3$
Therefore, value of polynomial $5x - 4x^2 +3$ at x = 2 is -3
Q2. (i) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(y)= y^2 - y + 1$
Answer:
Given polynomial is
$p(y)= y^2 - y + 1$
Now,
$p(0)= (0)^2 - 0 + 1= 1$
$p(1)= (1)^2 - 1 + 1 = 1$
$p(2)= (2)^2 - 2 + 1 = 3$
Therefore, values of p(0) , p(1) and p(2) are 1 , 1 and 3 respectively .
Q2. (ii) Find p(0) , p(1) and p(2) for each of the following polynomials: $p(t) = 2 + t + 2t^2 - t^3$
Answer:
Given polynomial is
$p(t) = 2 + t + 2t^2 - t^3$
Now,
$p(0) = 2 + 0 + 2(0)^2 - (0)^3 = 2$
$p(1) = 2 + 1 + 2(1)^2 - (1)^3 = 4$
$p(2) = 2 + 2 + 2(2)^2 - (2)^3 = 4$
Therefore, values of p(0) , p(1) and p(2) are 2 , 4 and 4 respectively
Q2. (iii) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x) = x^3$
Answer:
Given polynomial is
$p(x) = x^3$
Now,
$p(0) = (0)^3 =0$
$p(1) = (1)^3=1$
$p(2)=(2)^3=8$
Therefore, values of p(0) , p(1) and p(2) are 0 , 1 and 8 respectively
Q2. (iv) Find p(0), p(1) and p(2) for each of the following polynomials: $p(x)= (x-1)(x+ 1)$
Answer:
Given polynomial is
$p(x)=(x-1)(x+1)=x^2-1$
Now,
$p(0) = (0)^2-1 = -1$
$p(1) = (1)^2-1 = 0$
$p(2) = (2)^2-1 = 3$
Therefore, values of p(0) , p(1) and p(2) are -1 , 0 and 3 respectively
Answer:
Given polynomial is $p(x) = 3x + 1$
Now, at $x = -\frac{1}{3}$ it's value is
$p\left ( -\frac{1}{3} \right )=3\times \left ( -\frac{1}{3} \right )+1 = -1+1=0$
Therefore, yes $x = -\frac{1}{3}$ is a zero of polynomial $p(x) = 3x + 1$
Answer:
Given polynomial is $p(x) = 5x - \pi$
Now, at $x =\frac{4}{5}$ it's value is
$p\left ( \frac{4}{5} \right )=5\times \left ( \frac{4}{5} \right ) -\pi = 4-\pi \neq 0$
Therefore, no $x =\frac{4}{5}$ is not a zero of polynomial $p(x) = 5x - \pi$
Answer:
Given polynomial is $p(x) = x^2-1$
Now, at x = 1 its value is
$p(1) = (1)^2-1 = 1 -1 =0$
And at x = -1
$p(-1) = (-1)^2-1 = 1 -1 =0$
Therefore, yes x = 1 , -1 are zeros of polynomial $p(x) = x^2-1$
Answer:
Given polynomial is $p(x) = (x+1)(x-2)$
Now, at x = 2 it's value is
$p(2) = (2+1)(2-2) = 0$
And at x = -1
$p(-1) = (-1+1)(-1-2) = 0$
Therefore, yes x = 2 , -1 are zeros of polynomial $p(x) = (x+1)(x-2)$
Q3. (v) Verify whether the following are zeroes of the polynomial, indicated against it. $p(x) = x^2. x =0$
Answer:
Given polynomial is $p(x) = x^2$
Now, at x = 0 it's value is
$p(0) = (0)^2=0$
Therefore, yes x = 0 is a zeros of polynomial $p(x) = (x+1)(x-2)$
Answer:
Given polynomial is $p(x) = lx+m$
Now, at $x = -\frac{m}{l}$ it's value is
$p\left ( -\frac{m}{l} \right )= l \times \left ( -\frac{m}{l} \right )+m = -m+m =0$
Therefore, yes $x = -\frac{m}{l}$ is a zeros of polynomial $p(x) = lx+m$
Answer:
Given polynomial is $p(x) = 3x^2-1$
Now, at $x = -\frac{1}{\sqrt3}$ it's value is
$p\left ( -\frac{1}{\sqrt3} \right )= 3 \times \left ( -\frac{1}{\sqrt3} \right )^2-1 = 1-1 =0$
And at $x = \frac{2}{\sqrt3}$
$p\left ( \frac{2}{\sqrt3} \right )= 3 \times \left ( \frac{2}{\sqrt3} \right )^2-1 = 4-1 =3\neq 0$
Therefore, $x = -\frac{1}{\sqrt3}$ is a zeros of polynomial $p(x) = 3x^2-1$ .
whereas $x = \frac{2}{\sqrt3}$ is not a zeros of polynomial $p(x) = 3x^2-1$
Answer:
Given polynomial is $p(x) = 2x+1$
Now, at $x = \frac{1}{2}$ it's value is
$p\left ( \frac{1}{2} \right )= 2 \times \left ( \frac{1}{2} \right )+1 = 1+1=2 \neq 0$
Therefore, $x = \frac{1}{2}$ is not a zeros of polynomial $p(x) = 2x+1$
Q4. (i) Find the zero of the polynomial in each of the following cases: $p(x)= x + 5$
Answer:
Given polynomial is $p(x)= x + 5$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow x+5 = 0$
$\Rightarrow x=-5$
Therefore, x = -5 is the zero of polynomial $p(x)= x + 5$
Q4. (ii) Find the zero of the polynomial in each of the following cases: $p(x) = x - 5$
Answer:
Given polynomial is $p(x)= x - 5$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow x-5 = 0$
$\Rightarrow x=5$
Therefore, x = 5 is a zero of polynomial $p(x)= x - 5$
Q4. (iii) Find the zero of the polynomial in each of the following cases: $p(x)= 2x + 5$
Answer:
Given polynomial is $p(x)= 2x + 5$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow 2x+5 = 0$
$\Rightarrow x=-\frac{5}{2}$
Therefore, $x=-\frac{5}{2}$ is a zero of polynomial $p(x)= 2x + 5$
Q4. (iv) Find the zero of the polynomial in each of the following cases: $p(x) = 3x - 2$
Answer:
Given polynomial is $p(x) = 3x - 2$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow 3x-2 = 0$
$\Rightarrow x=\frac{2}{3}$
Therefore, $x=\frac{2}{3}$ is a zero of polynomial $p(x) = 3x - 2$
Q4. (v) Find the zero of the polynomial in each of the following cases: $p(x) = 3x$
Answer:
Given polynomial is $p(x) = 3x$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow 3x = 0$
$\Rightarrow x=0$
Therefore, $x=0$ is a zero of polynomial $p(x) = 3x$
Q4. (vi) Find the zero of the polynomial in each of the following cases: $p(x) = ax, \ a\neq 0$
Answer:
Given polynomial is $p(x) = ax$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow ax = 0$
$\Rightarrow x=0$
Therefore, $x=0$ is a zero of polynomial $p(x) = ax$
Answer:
Given polynomial is $p(x) = cx+d$
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
Now,
$p(x)=0$
$\Rightarrow cx+d = 0$
$\Rightarrow x=-\frac{d}{c}$
Therefore, $x=-\frac{d}{c}$ is a zero of polynomial $p(x) = cx+d$
Class 9 polynomials NCERT solutions Exercise: 2.3
Page number: 35-36, Total questions: 5
Q1. (i) Determine which of the following polynomials has $(x + 1)$ a factor : $x^3 + x^2 +x + 1$
Answer:
Zero of polynomial $(x + 1)$ is -1.
If $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$
Then, $p(-1)$ must be equal to zero
Now,
$\Rightarrow p(-1)=(-1)^3+(-1)^2-1+1$
$\Rightarrow p(-1)=-1+1-1+1 = 0$
Therefore, $(x + 1)$ is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$
Q1. (ii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + x^3 + x^2 +x + 1$
Answer:
Zero of polynomial $(x + 1)$ is -1.
If $(x + 1)$ is a factor of polynomial $p(x)=x^4 + x^3 + x^2 +x + 1$
Then, $p(-1)$ must be equal to zero
Now,
$\Rightarrow p(-1)=(-1)^4+(-1)^3+(-1)^2-1+1$
$\Rightarrow p(-1)=1-1+1-1+1 = 1\neq 0$
Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^4 + x^3 + x^2 +x + 1$
Q1. (iii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + 3x^3 + 3x^2 +x + 1$
Answer:
Zero of polynomial $(x + 1)$ is -1.
If $(x + 1)$ is a factor of polynomial $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$
Then, $p(-1)$ must be equal to zero
Now,
$\Rightarrow p(-1)=(-1)^4+3(-1)^3+3(-1)^2-1+1$
$\Rightarrow p(-1)=1-3+3-1+1 = 1\neq 0$
Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$
Answer:
Zero of polynomial $(x + 1)$ is -1.
If $(x + 1)$ is a factor of polynomial $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$
Then, $p(-1)$ must be equal to zero
Now,
$\Rightarrow p(-1)=(-1)^3-(-1)^2-(2+\sqrt2)(-1)+\sqrt2$
$\Rightarrow p(-1)=-1-1+2+\sqrt2+\sqrt2 = 2\sqrt2\neq 0$
Therefore, $(x + 1)$ is not a factor of polynomial $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$
Answer:
Zero of polynomial $g(x)=x+1$ is $-1$
If $g(x)=x+1$ is factor of polynomial $p(x) = 2x^3 + x^2 - 2x - 1$
Then, $p(-1)$ must be equal to zero
Now,
$\Rightarrow p(-1)= 2(-1)^3+(-1)^2-2(-1)-1$
$\Rightarrow p(-1)= -2+1+2-1 = 0$
Therefore, $g(x)=x+1$ is factor of polynomial $p(x) = 2x^3 + x^2 - 2x - 1$
Answer:
Zero of polynomial $g(x)=x+2$ is $-2$
If $g(x)=x+2$ is factor of polynomial $p(x) = x^3 + 3x^2 + 3x + 1$
Then, $p(-2)$ must be equal to zero
Now,
$\Rightarrow p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$
$\Rightarrow p(-2)= -8+12-6+1 = -1\neq 0$
Therefore, $g(x)=x+2$ is not a factor of polynomial $p(x) = x^3 + 3x^2 + 3x + 1$
Answer:
Zero of polynomial $g(x)=x-3$ is $3$
If $g(x)=x-3$ is factor of polynomial $p(x) = x^3 - 4x^2 + x + 6$
Then, $p(3)$ must be equal to zero
Now,
$\Rightarrow p(3) = (3)^3 - 4(3)^2 + 3 + 6$
$\Rightarrow p(3) = 27-36+3+6=0$
Therefore, $g(x)=x-3$ is a factor of polynomial $p(x) = x^3 - 4x^2 + x + 6$
Q3. (i) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = x^2 + x + k$
Answer:
Zero of polynomial $x - 1$ is $1$
If $x - 1$ is factor of polynomial $p(x) = x^2 + x + k$
Then, $p(1)$ must be equal to zero
Now,
$\Rightarrow p(1) = (1)^2 + 1 + k$
$\Rightarrow p(1) =0$
$\Rightarrow 2+k = 0$
$\Rightarrow k = -2$
Therefore, the value of k is $-2$
Answer:
Zero of the polynomial $x - 1$ is $1$
If $x - 1$ is factor of polynomial $p(x) = 2x^2 + kx + \sqrt2$
Then, $p(1)$ must be equal to zero
Now,
$\Rightarrow p(1) = 2(1)^2 + k(1) + \sqrt2$
$\Rightarrow p(1) =0$
$\Rightarrow 2+k +\sqrt2= 0$
$\Rightarrow k = -(2+\sqrt2)$
Therefore, the value of k is $-(2+\sqrt2)$
Answer:
Zero of polynomial $x - 1$ is $1$
If $x - 1$ is factor of polynomial $p(x) = kx^2-\sqrt2 x + 1$
Then, $p(1)$ must be equal to zero
Now,
$\Rightarrow p(1) = k(1)^2 -\sqrt2(1) + 1$
$\Rightarrow p(1) =0$
$\Rightarrow k -\sqrt2 +1= 0$
$\Rightarrow k = -1+\sqrt2$
Therefore, the value of k is $-1+\sqrt2$
Q3. (iv) the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = kx^2 -3 x + k$
Answer:
Zero of polynomial $x - 1$ is $1$
If $x - 1$ is factor of polynomial $p(x) = kx^2 -3 x + k$
Then, $p(1)$ must be equal to zero
Now,
$\Rightarrow p(1) = k(1)^2 -3(1) + k$
$\Rightarrow p(1) =0$
$\Rightarrow k -3+k= 0$
$\Rightarrow k = \frac{3}{2}$
Therefore, value of k is $\frac{3}{2}$
Q4. (i) Factorise : $12x^2 - 7x + 1$
Answer:
Given polynomial is $12x^2 - 7x + 1$
We need to factorise the middle term into two terms such that their product is equal to $12 \times 1 = 12$ and their sum is equal to $-7$
We can solve it as
$\Rightarrow 12x^2 - 7x + 1$
$\Rightarrow 12x^2-3x-4x+1$ $(\because -3 \times -4 = 12 \ \ and \ \ -3+(-4) = -7)$
$\Rightarrow 3x(4x-1)-1(4x-1)$
$\Rightarrow (3x-1)(4x-1)$
Q4. (ii) Factorise : $2x^2 + 7x + 3$
Answer:
Given polynomial is $2x^2 + 7x + 3$
We need to factorise the middle term into two terms such that their product is equal to $2 \times 3 = 6$ and their sum is equal to $7$
We can solve it as
$\Rightarrow 12x^2 - 7x + 1$
$\Rightarrow 2x^2+6x+x+3$ $(\because 6 \times 1 = 6 \ \ and \ \ 6+1 = 7)$
$\Rightarrow 2x(x+3)+1(x+3)$
$\Rightarrow (2x+1)(x+3)$
Q4. (iii) Factorise : $6x^2 +5x - 6$
Answer:
Given polynomial is $6x^2 +5x - 6$
We need to factorise the middle term into two terms such that their product is equal to $6 \times -6 =-3 6$ and their sum is equal to $5$
We can solve it as
$\Rightarrow 6x^2 +5x - 6$
$\Rightarrow 6x^2 +9x -4x - 6$ $(\because 9 \times -4 = -36 \ \ and \ \ 9+(-4) = 5)$
$\Rightarrow 3x(2x+3)-2(2x+3)$
$\Rightarrow (2x+3)(3x-2)$
Q4. (iv) Factorise : $3x^2 - x - 4$
Answer:
Given polynomial is $3x^2 - x - 4$
We need to factorise the middle term into two terms such that their product is equal to $3 \times -4 =-12$ and their sum is equal to $-1$
We can solve it as
$\Rightarrow 3x^2 - x - 4$
$\Rightarrow 3x^2 -4 x+3x - 4$ $(\because 3 \times -4 = -12 \ \ and \ \ 3+(-4) = -1)$
$\Rightarrow x(3x-4)+1(3x-4)$
$\Rightarrow (x+1)(3x-4)$
Q5. (i) Factorise : $x^3 - 2x^2 - x +2$
Answer:
Given polynomial is $x^3 - 2x^2 - x +2$
Now, by hit and trial method we observed that $(x+1)$ is one of the factors of the given polynomial.
By the long division method, we will get
We know that Dividend = (Divisor × Quotient) + Remainder
$x^3 - 2x^2 - x +2 = (x+1)(x^2-3x+2)+0$
$= (x+1)(x^2-2x-x+2)$
$= (x+1)(x-2)(x-1)$
Therefore, on factorization of $x^3 - 2x^2 - x +2$ we will get $(x+1)(x-2)(x-1)$
Q5. (ii) Factorise : $x^3 - 3x^2 -9x -5$
Answer:
Given polynomial is $x^3 - 3x^2 -9x -5$
Now, by the hit-and-trial method, we observed that $(x+1)$ is one of the factors of the given polynomial.
By the long division method, we will get
We know that Dividend = (Divisor $\times$ Quotient) + Remainder
$x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)$
$= (x+1)(x^2-5x+x-5)$
$= (x+1)(x-5)(x+1)$
Therefore, on factorization of $x^3 - 3x^2 -9x -5$ we will get $(x+1)(x-5)(x+1)$
Q5. (iii) Factorise : $x^3 + 13x^2 + 32x + 20$
Answer:
Given polynomial is $x^3 + 13x^2 + 32x + 20$
Now, by hit and trial method we observed that $(x+1)$ is one of the factors of given polynomial.
By long division method, we will get
We know that Dividend = (Divisor $\times$ Quotient) + Remainder
$x^3 + 13x^2 + 32x + 20=(x+1)(x^2+12x+20)$
$= (x+1)(x^2+10x+2x+20)$
$= (x+1)(x+10)(x+2)$
Therefore, on factorization of $x^3 + 13x^2 + 32x + 20$ we will get $(x+1)(x+10)(x+2)$
Q5. (iv) Factorise : $2y^3 + y^2 - 2y - 1$
Answer:
Given polynomial is $2y^3 + y^2 - 2y - 1$
Now, by hit and trial method we observed that $(y-1)$ is one of the factors of the given polynomial.
By long division method, we will get
We know that Dividend = (Divisor $\times$ Quotient) + Remainder
$2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)$
$= (y-1)(2y^2+2y+y+1)$
$= (y-1)(2y+1)(y+1)$
Therefore, on factorization of $2y^3 + y^2 - 2y - 1$ we will get $(y-1)(2y+1)(y+1)$.
Class 9 maths chapter 2 question answer Exercise: 2.4
Page number: 40-42, Total questions: 16
Q1. (i) Use suitable identities to find the following product: $(x + 4) ( x + 10)$
Answer:
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = 4 \ \ and \ \ b = 10$
$(x+4)(x+10)= x^2+(10+4)x+10\times 4$
$= x^2+14x+40$
Therefore, $(x + 4) ( x + 10)$ is equal to $x^2+14x+40$
Q1. (ii) Use suitable identities to find the following product: $(x+8)(x-10)$
Answer:
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = 8 \ \ and \ \ b = -10$
$(x+8)(x-10)= x^2+(-10+8)x+8\times (-10)$
$= x^2-2x-80$
Therefore, $(x+8)(x-10)$ is equal to $x^2-2x-80$
Q1. (iii) Use suitable identities to find the following product: $(3x+4)(3x - 5)$
Answer:
We can write $(3x+4)(3x - 5)$ as
$(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}$
$9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )$
$=9x^2-3x-20$
Therefore, $(3x+4)(3x - 5)$ is equal to $9x^2-3x-20$
Q1. (iv) Use suitable identities to find the following product: $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$
Answer:
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $x=y^2 \ \ and \ \ a = \frac{3}{2}$
$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2$
$= y^4-\frac{9}{4}$
Therefore, $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$ is equal to $y^4-\frac{9}{4}$
Q1. (v) Use suitable identities to find the following product: $(3 - 2x) (3 + 2x)$
Answer:
We can write $(3 - 2x) (3 + 2x)$ as
$(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )$
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $a = \frac{3}{2}$
$-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )$
$=9-4x^2$
Therefore, $(3 - 2x) (3 + 2x)$ is equal to $9-4x^2$
Q2. (i) Evaluate the following product without multiplying directly: $103 \times 107$
Answer:
We can rewrite $103 \times 107$ as
$\Rightarrow 103 \times 107= (100+3)\times (100+7)$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $x =100 , a=3 \ \ and \ \ b = 7$
$(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7$
$=10000+1000+21= 11021$
Therefore, value of $103 \times 107$ is $11021$
Q2. (ii) Evaluate the following product without multiplying directly: $95 \times 96$
Answer:
We can rewrite $95 \times 96$ as
$\Rightarrow 95 \times 96= (100-5)\times (100-4)$
We will use identity
$(x+a)(x+b)=x^2+(a+b)x+ab$
Put $x =100 , a=-5 \ \ and \ \ b = -4$
$(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)$
$=10000-900+20= 9120$
Therefore, value of $95 \times 96$ is $9120$
Q2. (iii) Evaluate the following product without multiplying directly: $104 \times 96$
Answer:
We can rewrite $104 \times 96$ as
$\Rightarrow 104 \times 96= (100+4)\times (100-4)$
We will use identity
$(x+a)(x-a)=x^2-a^2$
Put $x =100 \ \ and \ \ a=4$
$(100+4)\times (100-4)= (100)^2-(4)^2$
$=10000-16= 9984$
Therefore, value of $104 \times 96$ is $9984$
Q3. (i) Factorise the following using appropriate identities: $9x^2 + 6xy + y^2$
Answer:
We can rewrite $9x^2 + 6xy + y^2$ as
$\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2$
Using identity $\Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2$
Here, $a= 3x \ \ and \ \ b = y$
Therefore,
$9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)$
Q3. (ii) Factorise the following using appropriate identities: $4y^2 - 4y + 1$
Answer:
We can rewrite $4y^2 - 4y + 1$ as
$\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2$
Using identity $\Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2$
Here, $a= 2y \ \ and \ \ b = 1$
Therefore,
$4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)$
Q3. (iii) Factorise the following using appropriate identities: $x^2 - \frac{y^2}{100}$
Answer:
We can rewrite $x^2 - \frac{y^2}{100}$ as
$\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2$
Using identity $\Rightarrow a^2-b^2 = (a-b)(a+b)$
Here, $a= x \ \ and \ \ b = \frac{y}{10}$
Therefore,
$x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )$
Q4. (i) Expand each of the following, using suitable identities: $(x + 2y+4z)^2$
Answer:
Given is $(x + 2y+4z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = x , b = 2y \ \ and \ \ c = 4z$
Therefore,
$(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x$
$= x^2+4y^2+16z^2+4xy+16yz+8zx$
Q4. (ii) Expand each of the following, using suitable identities: $(2x - y + z)^2$
Answer:
Given is $(2x - y + z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = 2x , b = -y \ \ and \ \ c = z$
Therefore,
$(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x$
$= 4x^2+y^2+z^2-4xy-2yz+4zx$
Q4. (iii) Expand each of the following, using suitable identities: $(-2x + 3y + 2z)^2$
Answer:
Given is $(-2x + 3y + 2z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a =- 2x , b = 3y \ \ and \ \ c = 2z$
Therefore,
$(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)$
$= 4x^2+9y^2+4z^2-12xy+12yz-8zx$
Q4. (iv) Expand each of the following, using suitable identities: $(3a - 7b - c)^2$
Answer:
Given is $(3a - 7b - c)^2$
We will Use identity
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$
Here, $x =3a , y = -7b \ \ and \ \ z = -c$
Therefore,
$(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c)$ $.3a$
$= 9a^2+49b^2+c^2-42ab+14bc-6ca$
Q4. (v) Expand each of the following, using suitable identities: $(-2x + 5y -3z)^2$
Answer:
Given is $(-2x + 5y -3z)^2$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a =- 2x , b = 5y \ \ and \ \ c = -3z$
Therefore,
$(-2x +5y-3z)^2$ $= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)$
$= 4x^2+25y^2+9z^2-20xy-30yz+12zx$
Answer:
Given is $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$
We will Use identity
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$
Here, $x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1$
Therefore,
$\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$ $=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )$
$= \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$
Q5. (i) Factorise: $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$
Answer:
We can rewrite $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ as
$\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ $= (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = 2x , b = 3y \ \ and \ \ c = -4z$
Therefore,
$4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x+3y-4z)^2$
$= (2x+3y-4z)(2x+3y-4z)$
Q5. (ii) Factorise: $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$
Answer:
We can rewrite $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ as
$\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ $= (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z$
We will Use identity
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
Here, $a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z$
Therefore,
$2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2$
$=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)$
Q6 (i) Write the following cubes in expanded form: $(2x + 1)^3$
Answer:
Given is $(2x + 1)^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a= 2x \ \ and \ \ b= 1$
Therefore,
$(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2$
$= 8x^3+1+12x^2+6x$
Q6. (ii) Write the following cube in expanded form: $(2a-3b)^3$
Answer:
Given is $(2a-3b)^3$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x= 2a \ \ and \ \ y= 3b$
Therefore,
$(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2$
$= 8a^3-9b^3-36a^2b+54ab^2$
Q6. (iii) Write the following cube in expanded form: $\left[\frac{3}{2}x + 1\right ]^3$
Answer:
Given is $\left[\frac{3}{2}x + 1\right ]^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a= \frac{3x}{2} \ \ and \ \ b= 1$
Therefore,
$\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2$
$= \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}$
Q6. (iv) Write the following cube in expanded form: $\left[x - \frac{2}{3} y\right ]^3$
Answer:
Given is $\left[x - \frac{2}{3} y\right ]^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=x \ and \ \ b= \frac{2y}{3}$
Therefore,
$\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2$
$= x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}$
Q7. (i) Evaluate the following using suitable identities: $(99)^3$
Answer:
We can rewrite $(99)^3$ as
$\Rightarrow (99)^3=(100-1)^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=100 \ and \ \ b= 1$
Therefore,
$(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2$
$= 1000000-1-30000+300= 970299$
Q7. (ii) Evaluate the following using suitable identities: $(102)^3$
Answer:
We can rewrite $(102)^3$ as
$\Rightarrow (102)^3=(100+2)^3$
We will use identity
$(a+b)^3=a^3+b^3+3a^2b+3ab^2$
Here, $a=100 \ and \ \ b= 2$
Therefore,
$(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2$
$= 1000000+8+60000+1200= 1061208$
Q7. (iii) Evaluate the following using suitable identities: $(998)^3$
Answer:
We can rewrite $(998)^3$ as
$\Rightarrow (998)^3=(1000-2)^3$
We will use identity
$(a-b)^3=a^3-b^3-3a^2b+3ab^2$
Here, $a=1000 \ and \ \ b= 2$
Therefore,
$(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2$
$= 1000000000-8-6000000+12000= 994011992$
Q8. (i) Factorise the following: $8a^3 + b^3 + 12a^2 b + 6ab^2$
Answer:
We can rewrite $8a^3 + b^3 + 12a^2 b + 6ab^2$ as
$\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2$ $= (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2$
We will use identity
$(x+y)^3=x^3+y^3+3x^2y+3xy^2$
Here, $x=2a \ \ and \ \ y= b$
Therefore,
$8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3$
$=(2a+b)(2a+b)(2a+b)$
Q8. (ii) Factorise the following: $8a ^3 - b^3 - 12a^2 b + 6ab^2$
Answer:
We can rewrite $8a ^3 - b^3 - 12a^2 b + 6ab^2$ as
$\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2$ $= (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=2a \ \ and \ \ y= b$
Therefore,
$8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3$
$=(2a-b)(2a-b)(2a-b)$
Q8. (iii) Factorise the following: $27 - 125a^ 3 - 135a + 225a^2$
Answer:
We can rewrite $27 - 125a^ 3 - 135a + 225a^2$ as
$\Rightarrow 27 - 125a^ 3 - 135a + 225a^2$ $= (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=3 \ \ and \ \ y= 5a$
Therefore,
$27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3$
$=(3-5a)(3-5a)(3-5a)$
Q8. (iv) Factorise the following: $64a^3 - 27b^3 - 144a^2 b + 108ab^2$
Answer:
We can rewrite $64a^3 - 27b^3 - 144a^2 b + 108ab^2$ as
$\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2$ $= (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=4a \ \ and \ \ y= 3b$
Therefore,
$64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2$
$=(4a-3b)(4a-3b)(4a-3b)$
Q8. (v) Factorise the following: $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$
Answer:
We can rewrite $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ as
$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$
We will use identity
$(x-y)^3=x^3-y^3-3x^2y+3xy^2$
Here, $x=3p \ \ and \ \ y= \frac{1}{6}$
Therefore,
$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$
$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$
Q9. (i) Verify: $x^3 + y^3 = (x +y)(x^2 - xy + y^2)$
Answer:
We know that
$(x+y)^3=x^3+y^3+3xy(x+y)$
Now,
$\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)$
$\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )$
$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )$ $(\because (a+b)^2=a^2+b^2+2ab)$
$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )$
Hence proved.
Q9. (ii) Verify: $x^3 - y^3 = (x -y)(x^2 + xy + y^2)$
Answer:
We know that
$(x-y)^3=x^3-y^3-3xy(x-y)$
Now,
$\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)$
$\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )$ $(\because (a-b)^2=a^2+b^2-2ab)$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )$
Hence proved.
Q10. (i) Factorise the following: $27y^3 + 125z^3$
Answer:
We know that
$a^3+b^3=(a+b)(a^2+b^2-ab)$
Now, we can write $27y^3 + 125z^3$ as
$\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3$
Here, $a = 3y \ \ and \ \ b = 5z$
Therefore,
$27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )$
$27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )$
Q10. (ii) Factorise the following: $64m^3 - 343n^3$
Answer:
We know that
$a^3-b^3=(a-b)(a^2+b^2+ab)$
Now, we can write $64m^3 - 343n^3$ as
$\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3$
Here, $a = 4m \ \ and \ \ b = 7n$
Therefore,
$64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )$
$64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )$
Q11. Factorise: $27x^3 + y^3 + z^3 - 9xyz$
Answer:
Given is $27x^3 + y^3 + z^3 - 9xyz$
Now, we know that
$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Now, we can write $27x^3 + y^3 + z^3 - 9xyz$ as
$\Rightarrow 27x^3 + y^3 + z^3 - 9xyz$ $=(3x)^3+(y)^3+(z)^3-3.3x.y.z$
Here, $a= 3x , b = y \ \ and \ \ c = z$
Therefore,
$27x^3 + y^3 + z^3 - 9xyz$ $=(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )$
$=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )$.
Answer:
We know that
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Now, multiply and divide the R.H.S. by 2
$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)$
$= \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)$
$= \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )$ $\left(\because a^2+b^2-2ab=(a-b)^2 \right )$
Hence proved.
Q13. If $x + y + z = 0$ , show that $x^3 + y^3 + z^3 = 3xyz$ .
Answer:
We know that
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Now, It is given that $x + y + z = 0$
Therefore,
$x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)$
$x^3+y^3+z^3-3xyz =0$
$x^3+y^3+z^3=3xyz$
Hence proved.
Answer:
Given is $(-12)^3 + (7)^3 + (5)^3$
We know that
If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$
Here, $x = -12 , y = 7 \ \ an d \ \ z = 5$
$\Rightarrow x+y+z = -12+7+5 = 0$
Therefore,
$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12)\times 7 \times 5 = -1260$
Therefore, value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$
Answer:
Given is $(28)^3 + (-15)^3 + (-13)^3$
We know that
If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$
Here, $x = 28 , y = -15 \ \ an d \ \ z = -13$
$\Rightarrow x+y+z =28-15-13 = 0$
Therefore,
$(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380$
Therefore, value of $(28)^3 + (-15)^3 + (-13)^3$ is $16380$
$25a^2 - 35a + 12$ |
Answer:
We know that
Area of rectangle is = $length \times breadth$
It is given that area = $25a^2-35a+12$
Now, by splitting middle term method
$\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12$
$= 5a(5a-4)-3(5a-4)$
$= (5a-3)(5a-4)$
Therefore, two answers are possible
case (i) :- Length = $(5a-4)$ and Breadth = $(5a-3)$
case (ii) :- Length = $(5a-3)$ and Breadth = $(5a-4)$
$35y^2 + 13y- 12$ |
Answer:
We know that
Area of rectangle is = $length \times breadth$
It is given that area = $35y^2 + 13y- 12$
Now, by splitting the middle term method
$\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12$
$= 7y(5y+4)-3(5y+4)$
$= (7y-3)(5y+4)$
Therefore, two answers are possible
case (i) :- Length = $(5y+4)$ and Breadth = $(7y-3)$
case (ii) :- Length = $(7y-3)$ and Breadth = $(5y+4)$
Q16. (i) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?
Volume : $3x^2 - 12x$ |
Answer:
We know that
Volume of cuboid is = $length \times breadth \times height$
It is given that volume = $3x^2-12x$
Now,
$\Rightarrow 3x^2-12x=3\times x\times (x-4)$
Therefore,one of the possible answer is possible
Length = $3$ and Breadth = $x$ and Height = $(x-4)$
Q16. (ii) What are the possible expressions for the dimensions of the cuboid whose volumes is given below?
Volume : $12ky^2 + 8ky - 20k$ |
Answer:
We know that
Volume of cuboid is = $length \times breadth \times height$
It is given that volume = $12ky^2+8ky-20k$
Now,
$\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)$
$= k(12y^2+20y-12y-20)$
$= k\left(4y(3y+5)-4(3y+5) \right )$
$= k(3y+5)(4y-4)$
$= 4k(3y+5)(y-1)$
Therefore,one of the possible answer is possible
Length = $4k$ and Breadth = $(3y+5)$ and Height = $(y-1)$.
Students can practice Class 9 Maths Chapter 2 question answers using the exercise link given below.
Here are the subject-wise links for the NCERT solutions of class 9:
Given below are some useful links for NCERT books and the NCERT syllabus for class 9:
Keep Working Hard and Happy Learning!
The NCERT class 9 maths chapter 2 includes topics such as definition of a polynomial, zeroes, coefficient, degrees, and terms of a polynomial, different types of a polynomial, remainder and factor theorems, and the factorization of polynomials. students should practice these NCERT solutions to get indepth understanding of these concepts which ultimately lead to score well in the exam.
Maths chapter 2 includes five exercises covering topics such as Polynomials in one variable, Zeros of a Polynomial, Real Numbers and their Decimal Expansions, Representing Real Numbers on the Number Line, Operations on Real Numbers, and Laws of Exponents for Real Numbers. Practicing these exercises of NCERT maths class 9 chapter 2 is crucial for achieving a better understanding of the concepts and scoring well in Mathematics. To help students gain confidence, Careers360 experts have designed these solutions to provide comprehensive explanations of the concepts covered in this chapter.
NCERT Solutions for Class 9 Maths Chapter 2 use straightforward language to explain the concepts, making it accessible even for students who struggle with Mathematics. These solutions are meticulously crafted by a team of experts at Careers360 with the objective of helping students prepare for their CBSE exams effectively.
Regular practice with NCERT Solutions for Class 9 Maths Chapter 2 can enable students to excel in their CBSE exams. These solutions are created by a team of skilled Maths experts at Careers360, and by solving all the questions and comparing their answers with the solutions, students can aim for high scores in their exams.
Here, students can get detailed NCERT solutions for Class 9 Maths which includes solutions to all the exercise of each chapters.
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