Polynomials - Class 9 Mathematics - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Exercise 2.1 - Polynomials | NCERT | Mathematics | Class 9
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) $4 x^{2}-3 x+7$
(ii) $y^{2}+\sqrt{2}$
(iii) $3 \sqrt{t}+t \sqrt{2}$
(iv) $y+\frac{2}{y}$
(v) $x^{10}+y^{3}+t^{50}$
Here's an analysis for each expression to determine if it's a polynomial in one variable or not, alongside reasons:
(i) $4x^2 - 3x + 7$
This is a polynomial in one variable ($x$) because:
- It is expressed in terms of a single variable, $x$.
- All the powers of $x$ are whole numbers (2 and 1).
- It consists of a finite number of terms, each a product of a constant and a power of $x$.
(ii) $y^2 + \sqrt{2}$
This is a polynomial in one variable ($y$) because:
- It is expressed in terms of a single variable, $y$.
- The power of $y$ is a whole number (2).
- $\sqrt{2}$ is a constant term, which is allowed in polynomials.
(iii) $3\sqrt{t} + t\sqrt{2}$
This is not a polynomial for several reasons:
- Although it's in terms of a single variable, $t$, the term $3\sqrt{t}$ involves a square root of the variable, which means the power of $t$ is $\frac{1}{2}$, not a whole number.
(iv) $y + \frac{2}{y}$
This is not a polynomial because:
- It includes a term with $y$ in the denominator ($\frac{2}{y}$), which implies a negative exponent (-1) for $y$. Polynomials require all powers of the variable to be non-negative whole numbers.
(v) $x^{10} + y^3 + t^{50}$
This is not a polynomial in one variable because:
- It involves three different variables ($x$, $y$, and $t$), making it a polynomial in three variables, not one.
In summary:
- Expressions (i) and (ii) are polynomials in one variable.
- Expressions (iii) and (iv) are not polynomials due to the presence of non-whole number and negative exponents, respectively.
- Expression (v) is not a polynomial in one variable due to involving more than one variable.
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) $\sqrt{2} x-1$
To identify the coefficients of $x^2$ in each expression, we look for the term that includes $x^2$ and note the number (coefficient) that multiplies $x^2$. If the term $x^2$ is not present, the coefficient is 0.
- For the expression $2 + x^2 + x$, the coefficient of $x^2$ is 1.
- For the expression $2 - x^2 + x^3$, the coefficient of $x^2$ is -1.
- For the expression $\frac{\pi}{2} x^2 + x$, the coefficient of $x^2$ is $\frac{\pi}{2}$.
- For the expression $\sqrt{2} x - 1$, there is no $x^2$ term, so the coefficient of $x^2$ is 0.
Give one example each of a binomial of degree 35 , and of a monomial of degree 100 .
A binomial is an algebraic expression containing two terms, and its degree is determined by the highest degree of its terms when it's fully expanded. An example of a binomial of degree 35 could be:
$$x^{35} + 1$$
This is a binomial because it contains two terms, $x^{35}$ and $1$, and its highest degree is 35.
A monomial is an algebraic expression containing only one term. An example of a monomial of degree 100 could be:
$$x^{100}$$
This is a monomial because it only contains one term, and its degree is 100.
Write the degree of each of the following polynomials:
(i) $5 x^{3}+4 x^{2}+7 x$
(ii) $4-y^{2}$
(iii) $5 t-\sqrt{7}$
(iv) 3
The degree of a polynomial is the highest power of the variable in the polynomial. Let's see the degree of each given polynomial:
(i) $5x^{3} + 4x^{2} + 7x$
The highest power of $x$ in this polynomial is $3$, so its degree is $3$.
(ii) $4 - y^{2}$
The highest power of $y$ in this polynomial is $2$, so its degree is $2$.
(iii) $5t - \sqrt{7}$
The highest power of $t$ in this polynomial is $1$, so its degree is $1$.
(iv) $3$
This is a constant polynomial, and the degree of a constant polynomial (apart from the zero polynomial) is $0$. So, its degree is $0$.
Classify the following as linear, quadratic and cubic polynomials:
(i) $x^{2}+x$
(ii) $x-x^{3}$
(iii) $y+y^{2}+4$
(iv) $1+x$
(v) $3 t$
(vi) $r^{2}$
(vii) $7 x^{3}$
Based on the highest power of the variable present in each expression, we can classify the given polynomials as follows:
- Linear Polynomials: These are polynomials with the highest power of the variable being 1.
- Quadratic Polynomials: These are polynomials with the highest power of the variable being 2.
- Cubic Polynomials: These are polynomials with the highest power of the variable being 3.
Now, classifying the given polynomials:
(i) $x^{2}+x$ - Quadratic Polynomial (highest power of $x$ is 2)
(ii) $x-x^{3}$ - Cubic Polynomial (highest power of $x$ is 3)
(iii) $y+y^{2}+4$ - Quadratic Polynomial (highest power of $y$ is 2)
(iv) $1+x$ - Linear Polynomial (highest power of $x$ is 1)
(v) $3t$ - Linear Polynomial (highest power of $t$ is 1)
(vi) $r^{2}$ - Quadratic Polynomial (highest power of $r$ is 2)
(vii) $7x^{3}$ - Cubic Polynomial (highest power of $x$ is 3)
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Ask Chatterbot AIExercise 2.2 - Polynomials | NCERT | Mathematics | Class 9
Find the value of the polynomial $5 x-4 x^{2}+3$ at
(i) $x=0$
(ii) $x=-1$
(iii) $x=2$
The value of the polynomial $5x - 4x^2 + 3$ at the given points are:
(i) At $x=0$, the value is $3$.
(ii) At $x=-1$, the value is $-6$.
(iii) At $x=2$, the value is $-3$.
Find $p(0), p(1)$ and $p(2)$ for each of the following polynomials:
(i) $p(y)=y^{2}-y+1$
(ii) $p(t)=2+t+2 t^{2}-t^{3}$
(iii) $p(x)=x^{3}$
(iv) $p(x)=(x-1)(x+1)$
Here are the values of $p(0), p(1)$, and $p(2)$ for each of the given polynomials:
For $p(y)=y^{2}-y+1$:
$p(0) = 1$
$p(1) = 1$
$p(2) = 3$
For $p(t)=2+t+2t^{2}-t^{3}$:
$p(0) = 2$
$p(1) = 4$
$p(2) = 4$
For $p(x)=x^{3}$:
$p(0) = 0$
$p(1) = 1$
$p(2) = 8$
For $p(x)=(x-1)(x+1)$:
$p(0) = -1$
$p(1) = 0$
$p(2) = 3$
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p(x)=3 x+1, x=-\frac{1}{3}$
(ii) $p(x)=5 x-\pi, x=\frac{4}{5}$
(iii) $p(x)=x^{2}-1, x=1,-1$
(iv) $p(x)=(x+1)(x-2), x=-1,2$
(v) $p(x)=x^{2}, x=0$
(vi) $p(x)=l x+m, x=-\frac{m}{l}$
(vii) $p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
(viii) $p(x)=2 x+1, x=\frac{1}{2}$
Here are the evaluations of the given polynomials at the specified points:
-
For $p(x) = 3x + 1$ at $x = -\frac{1}{3}$, the result is 0. Hence, $x = -\frac{1}{3}$ is a zero of the polynomial.
-
For $p(x) = 5x - \pi$ at $x = \frac{4}{5}$, the result is $4 - \pi$. Hence, $x = \frac{4}{5}$ is not a zero of the polynomial.
-
For $p(x) = x^2 - 1$ at $x = 1$ and $x = -1$, the result is 0 in both cases. Hence, $x = 1$ and $x = -1$ are zeros of the polynomial.
-
For $p(x) = (x + 1)(x - 2)$ at $x = -1$ and $x = 2$, the result is 0 in both cases. Hence, $x = -1$ and $x = 2$ are zeros of the polynomial.
-
For $p(x) = x^2$ at $x = 0$, the result is 0. Hence, $x = 0$ is a zero of the polynomial.
-
For $p(x) = lx + m$ at $x = -\frac{m}{l}$, the result is 0. Hence, $x = -\frac{m}{l}$ is a zero of the polynomial.
-
For $p(x) = 3x^2 - 1$ at $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{2}{\sqrt{3}}$, the result is 0 at $x = -\frac{1}{\sqrt{3}}$ and 3 at $x = \frac{2}{\sqrt{3}}$. Hence, $x = -\frac{1}{\sqrt{3}}$ is a zero of the polynomial, but $x = \frac{2}{\sqrt{3}}$ is not.
-
For $p(x) = 2x + 1$ at $x = \frac{1}{2}$, the result is 2. Hence, $x = \frac{1}{2}$ is not a zero of the polynomial.
Find the zero of the polynomial in each of the following cases:
(i) $p(x)=x+5$
(ii) $p(x)=x-5$
(iii) $p(x)=2 x+5$
(iv) $p(x)=3 x-2$
(v) $p(x)=3 x$
(vi) $p(x)=a x, a \neq 0$
(vii) $p(x)=c x+d, c \neq 0, c, d$ are real numbers.
Here are the zeros of the polynomial in each of the given cases:
(i) For $p(x) = x+5$, the zero is $x = -5$.
(ii) For $p(x) = x-5$, the zero is $x = 5$.
(iii) For $p(x) = 2x+5$, the zero is $x = -\frac{5}{2}$.
(iv) For $p(x) = 3x-2$, the zero is $x = \frac{2}{3}$.
(v) For $p(x) = 3x$, the zero is $x = 0$.
(vi) For $p(x) = ax, a \neq 0$, the zero is $x = 0$ for any nonzero $a$.
(vii) For $p(x) = cx+d, c \neq 0, c, d$ are real numbers, the zero is $x = -\frac{d}{c}$ for any real numbers $c, d$, with $c \neq 0$.
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Ask Chatterbot AIExercise 2.3 - Polynomials | NCERT | Mathematics | Class 9
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) $x^{4}+3 x^{3}+3 x^{2}+x+1$
(iv) $x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$
For a polynomial to have $(x + 1)$ as a factor, plugging $x = -1$ into the polynomial must result in $0$.
- For the polynomial $x^{3}+x^{2}+x+1$, substituting $x = -1$ does not result in $0$, hence $(x+1)$ is not a factor.
- For $x^{4}+x^{3}+x^{2}+x+1$, substituting $x = -1$ results in $0$. Thus, $(x+1)$ is a factor.
- For $x^{4}+3 x^{3}+3 x^{2}+x+1$, substituting $x = -1$ also results in $0$, indicating that $(x+1)$ is a factor.
- Finally, for $x^{3}-x^{2}-(2+\sqrt{2}) x+\sqrt{2}$, substituting $x = -1$ does not yield a result of $0$, which means $(x+1)$ is not a factor.
Therefore, the polynomials which have $(x+1)$ as a factor are:
(ii) $x^{4}+x^{3}+x^{2}+x+1$ and (iii) $x^{4}+3 x^{3}+3 x^{2}+x+1$.
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x)=2 x^{3}+x^{2}-2 x-1, g(x)=x+1$
(ii) $p(x)=x^{3}+3 x^{2}+3 x+1, g(x)=x+2$
(iii) $p(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3$
Using the Factor Theorem, we can determine whether $g(x)$ is a factor of $p(x)$ by evaluating $p(x)$ at the root of $g(x)$. If $p(x) = 0$ when $g(x) = 0$, then $g(x)$ is a factor of $p(x)$. Here are the results for each case:
-
For $p(x)=2x^{3}+x^{2}-2x-1$ and $g(x)=x+1$, we evaluate $p(x)$ at $x = -1$. The result is $0$, which means $g(x)=x+1$ is a factor of $p(x)$.
-
For $p(x)=x^{3}+3x^{2}+3x+1$ and $g(x)=x+2$, we evaluate $p(x)$ at $x = -2$. The result is $-1$, which means $g(x)=x+2$ is not a factor of $p(x)$.
-
For $p(x)=x^{3}-4x^{2}+x+6$ and $g(x)=x-3$, we evaluate $p(x)$ at $x = 3$. The result is $0$, which means $g(x)=x-3$ is a factor of $p(x)$.
In summary:
- $g(x)=x+1$ is a factor of $p(x)=2x^{3}+x^{2}-2x-1$.
- $g(x)=x+2$ is not a factor of $p(x)=x^{3}+3x^{2}+3x+1$.
- $g(x)=x-3$ is a factor of $p(x)=x^{3}-4x^{2}+x+6$.
Find the value of $k$, if $x-1$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x)=x^{2}+x+k$
(ii) $p(x)=2 x^{2}+k x+\sqrt{2}$
(iii) $p(x)=k x^{2}-\sqrt{2} x+1$
(iv) $p(x)=k x^{2}-3 x+k$
The values of $k$ for each case when $x-1$ is a factor of $p(x)$ are as follows:
(i) For $p(x)=x^{2}+x+k$, the value of $k$ is $-2$.
(ii) For $p(x)=2x^{2}+kx+\sqrt{2}$, the value of $k$ is $-2-\sqrt{2}$.
(iii) For $p(x)=kx^{2}-\sqrt{2}x+1$, the value of $k$ is $\sqrt{2}-1$.
(iv) For $p(x)=kx^{2}-3x+k$, the value of $k$ is $\frac{3}{2}$.
Factorise :
(i) $12 x^{2}-7 x+1$
(ii) $2 x^{2}+7 x+3$
(iii) $6 x^{2}+5 x-6$
(iv) $3 x^{2}-x-4$
Here are the factorizations for each of the given expressions:
(i) For $12x^{2}-7x+1$, the factorization is: $$(4x - 1)(3x - 1)$$
(ii) For $2x^{2}+7x+3$, the factorization is: $$(2x + 1)(x + 3)$$
(iii) For $6x^{2}+5x-6$, the factorization is: $$(3x - 2)(2x + 3)$$
(iv) For $3x^{2}-x-4$, the factorization is: $$(x + 1)(3x - 4)$$
Factorise :
(i) $x^{3}-2 x^{2}-x+2$
(ii) $x^{3}-3 x^{2}-9 x-5$
(iii) $x^{3}+13 x^{2}+32 x+20$
(iv) $2 y^{3}+y^{2}-2 y-1$
The factorized forms of the given expressions are:
(i) $x^{3}-2 x^{2}-x+2 = (-2 + x)(-1 + x)(1 + x)$
(ii) $x^{3}-3 x^{2}-9 x-5 = (-5 + x)(1 + x)^2$
(iii) $x^{3}+13 x^{2}+32 x+20 = (1 + x)(2 + x)(10 + x)$
(iv) $2 y^{3}+y^{2}-2 y-1 = (-1 + y)(1 + y)(1 + 2y)$
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Ask Chatterbot AIExercise 2.4 - Polynomials | NCERT | Mathematics | Class 9
Use suitable identities to find the following products:
(i) $(x+4)(x+10)$
(ii) $(x+8)(x-10)$
(iii) $(3 x+4)(3 x-5)$
(iv) $\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)$
(v) $(3-2 x)(3+2 x)$
To find the products for each given pair, we can use the algebraic identities:
$$ (a + b)(a + c) = a^2 + a(b + c) + bc $$
$$ (a + b)(a - b) = a^2 - b^2 $$
Let's compute each of the products using these identities:
(i) $(x+4)(x+10)$:
Using the first identity: $$ x^2 + x(4 + 10) + 4\cdot10 = x^2 + 14x + 40 $$
(ii) $(x+8)(x-10)$:
Using the second identity: $$ x^2 - 8\cdot10 = x^2 - 80 $$
(iii) $(3x+4)(3x-5)$:
Using the first identity but notice the coefficients in front of x are the same in both terms, so it's also applicable to treat it like the difference of two squares with an additional middle term. $$ (3x)^2 + 3x(4 - 5) + 4\cdot-5 = 9x^2 - 3x - 20 $$
(iv) $\left(y^2+\frac{3}{2}\right)\left(y^2-\frac{3}{2}\right)$:
Using the second identity: $$ \left(y^2\right)^2 - \left(\frac{3}{2}\right)^2 = y^4 - \frac{9}{4} $$
(v) $(3-2x)(3+2x)$:
Again, using the second identity: $$ 3^2 - (2x)^2 = 9 - 4x^2 $$
So the products are:
(i) $x^2 + 14x + 40$
(ii) $x^2 - 80$
(iii) $9x^2 - 3x - 20$
(iv) $y^4 - \frac{9}{4}$
(v) $9 - 4x^2$
Evaluate the following products without multiplying directly:
(i) $103 \times 107$
(ii) $95 \times 96$
(iii) $104 \times 96$
Here are the evaluations for the given products without multiplying directly:
(i) For $103 \times 107$, the result is 11,021.
(ii) For $95 \times 96$, the result is 9,120.
(iii) For $104 \times 96$, the result is 9,984.
Factorise the following using appropriate identities:
(i) $9 x^{2}+6 x y+y^{2}$
(ii) $4 y^{2}-4 y+1$
(iii) $x^{2}-\frac{y^{2}}{100}$
To factorize the given expressions, we will use the following algebraic identities:
$(a+b)^2 = a^2 + 2ab + b^2$
$(a-b)^2 = a^2 - 2ab + b^2$
$a^2 - b^2 = (a-b)(a+b)$
Let's factorize each expression according to these identities:
(i) $9x^2 + 6xy + y^2$
This can be factorized using the first identity, where $a = 3x$ and $b = y$:
$$(3x + y)^2$$
(ii) $4y^2 - 4y + 1$
This can be factorized using the second identity, where $a = 2y$ and $b = 1$:
$$(2y - 1)^2$$
(iii) $x^2 - \frac{y^2}{100}$
This can be factorized using the third identity, where $a = x$ and $b = \frac{y}{10}$:
$$\left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right)$$
Therefore, the factorized forms are:
(i) $(3x + y)^2$
(ii) $(2y - 1)^2$
(iii) $\left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right)$
Expand each of the following, using suitable identities:
(i) $(x+2 y+4 z)^{2}$
(ii) $(2 x-y+z)^{2}$
(iii) $(-2 x+3 y+2 z)^{2}$
(iv) $(3 a-7 b-c)^{2}$
(v) $(-2 x+5 y-3 z)^{2}$
(vi) $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
Here are the expanded forms using suitable identities:
(i) $$(x+2y+4z)^2 = x^2 + 4xy + 4y^2 + 8xz + 16yz + 16z^2$$
(ii) $$(2x-y+z)^2 = 4x^2 - 4xy + y^2 + 4xz - 2yz + z^2$$
(iii) $$(-2x+3y+2z)^2 = 4x^2 - 12xy + 9y^2 - 8xz + 12yz + 4z^2$$
(iv) $$(3a-7b-c)^2 = 9a^2 - 42ab + 49b^2 - 6ac + 14bc + c^2$$
(v) $$(-2x+5y-3z)^2 = 4x^2 - 20xy + 25y^2 + 12xz - 30yz + 9z^2$$
(vi) $$\left(\frac{1}{4}a - \frac{1}{2}b + 1\right)^2 = 1 + \frac{1}{2}a + \frac{1}{16}a^2 - b - \frac{1}{4}ab + \frac{1}{4}b^2$$
Factorise:
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
The factorizations for the given expressions are as follows:
(i) $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz = (2x + 3y - 4z)^2$
(ii) $2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$ could not be further factorised into simpler polynomials, indicating it's already in its simplest form or does not factor over the rationals.
Write the following cubes in expanded form:
(i) $(2 x+1)^{3}$
(ii) $(2 a-3 b)^{3}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}$
(iv) $\left[x-\frac{2}{3} y\right]^{3}$
Here are the expansions in expanded form:
(i) $(2 x+1)^{3}$ expands to $$8x^3 + 12x^2 + 6x + 1$$
(ii) $(2 a-3 b)^{3}$ expands to $$8a^3 - 36a^2b + 54ab^2 - 27b^3$$
(iii) $\left[\frac{3}{2} x+1\right]^{3}$ expands to $$\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$$
(iv) $\left[x-\frac{2}{3} y\right]^{3}$ expands to $$x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$$
Evaluate the following using suitable identities:
(i) $(99)^{3}$
(ii) $(102)^{3}$
(iii) $(998)^{3}$
To evaluate these expressions using suitable identities, we can apply the identity for the cube of a binomial, which is $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.$$ Let's apply this identity to each expression:
(i) For $(99)^3$, we can view this as $(100-1)^3$. Applying the identity, we get: $$ (100-1)^3 = 100^3 - 3 \cdot 100^2 \cdot 1 + 3 \cdot 100 \cdot 1^2 - 1^3. $$
(ii) For $(102)^3$, we can view this as $(100+2)^3$. Applying the identity, we get: $$ (100+2)^3 = 100^3 + 3 \cdot 100^2 \cdot 2 + 3 \cdot 100 \cdot 2^2 + 2^3. $$
(iii) For $(998)^3$, we can view this as $(1000-2)^3$. Applying the identity, we get: $$ (1000-2)^3 = 1000^3 - 3 \cdot 1000^2 \cdot 2 + 3 \cdot 1000 \cdot 2^2 - 2^3. $$
Now, let's compute each of these explicitly:
For $(99)^3$: $$ 100^3 - 3 \cdot 100^2 \cdot 1 + 3 \cdot 100 \cdot 1^2 - 1^3 = 1000000 - 3 \cdot 10000 + 3 \cdot 100 - 1. $$
For $(102)^3$: $$ 100^3 + 3 \cdot 100^2 \cdot 2 + 3 \cdot 100 \cdot 2^2 + 2^3 = 1000000 + 3 \cdot 20000 + 3 \cdot 400 + 8. $$
For $(998)^3$: $$ 1000^3 - 3 \cdot 1000^2 \cdot 2 + 3 \cdot 1000 \cdot 2^2 - 2^3 = 1000000000 - 3 \cdot 2000000 + 3 \cdot 4000 - 8. $$
Let's calculate each of these. Here are the evaluated results using the identities:
(i) $(99)^{3} = 970299$
(ii) $(102)^{3} = 1061208$
(iii) $(998)^{3} = 994011992$
Factorise each of the following:
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
(iii) $27-125 a^{3}-135 a+225 a^{2}$
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$
The factorisations for each of the given expressions are:
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$ can be factorised as $(2a + b)^3$.
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$ can be factorised as $(2a - b)^3$.
(iii) $27-125 a^{3}-135 a+225 a^{2}$ can be factorised as $-(-3 + 5a)^3$.
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$ can be factorised as $(4a - 3b)^3$.
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$ can be factorised as $\frac{(-1 + 18p)^3}{216}$.
Verify: (i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
To verify the given identities, we'll expand the right-hand side of each equation and see if we can arrive back at the left-hand side.
For the first identity:
$$ (x+y)(x^{2}-xy+y^{2}) $$
Expansion:
Distribute $x$ over the second term: $x(x^2) - x(xy) + x(y^2)$
Distribute $y$ over the second term: $y(x^2) - y(xy) + y(y^2)$
Combine the steps: $x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$
Simplification:
The terms $- x^2y$ and $x^2y$ cancel out; the terms $xy^2$ and $-xy^2$ also cancel out. Thus, we're left with:
$$ x^3 + y^3 $$
So, we have shown that:
$$ x^{3}+y^{3} = (x+y)(x^{2}-xy+y^{2}) $$
For the second identity:
$$ (x-y)(x^{2}+xy+y^{2}) $$
Expansion:
Distribute $x$ over the second term: $x(x^2) + x(xy) + x(y^2)$
Distribute $-y$ over the second term: $-y(x^2) - y(xy) - y(y^2)$
Combine the steps: $x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3$
Simplification:
The terms $x^2y$ and $-x^2y$ cancel out; the terms $xy^2$ and $-xy^2$ also cancel out. Thus, we're left with:
$$ x^3 - y^3 $$
So, we have shown that:
$$ x^{3}-y^{3}=(x-y)(x^{2}+x y+y^{2}) $$
Both identities are thus verified as true.
Factorise each of the following:
(i) $27 y^{3}+125 z^{3}$
(ii) $64 m^{3}-343 n^{3}$
The factorisation for each of the given expressions is as follows:
(i) For $27 y^{3}+125 z^{3}$, the factorised form is: $$(3y + 5z)(9y^2 - 15yz + 25z^2)$$
(ii) For $64 m^{3}-343 n^{3}$, the factorised form is: $$(4m - 7n)(16m^2 + 28mn + 49n^2)$$
Factorise : $27 x^{3}+y^{3}+z^{3}-9 x y z$
The given expression $27x^3 + y^3 + z^3 - 9xyz$ can be factorised as: $$(3x + y + z)(9x^2 - 3xy + y^2 - 3xz - yz + z^2)$$
Verify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$
The expression $$x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$$ is verified to be true. The simplification of the difference between the left-hand side and the right-hand side of the equation results in $0$, indicating that both sides of the equation are equal for any values of $x$, $y$, and $z$.
If $x+y+z=0$, show that $x^{3}+y^{3}+z^{3}=3 x y z$.
To show that if $x+y+z=0$, then $x^3+y^3+z^3=3xyz$, follow the steps below:
Express $z$ in terms of $x$ and $y$:Given $x+y+z=0$, we can express $z$ as $z=-(x+y)$.
Cubing both sides of $z$:Taking the cube of both sides, we get $z^3 = -(x+y)^3$.
Expanding the right side using the binomial theorem:Expand $-(x+y)^3$ to get $-(x^3 + 3x^2y + 3xy^2 + y^3)$.
Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)$ with $a=x$, $b=y$, and $c=z$:We substitute $z$ with $-(x+y)$ and simplify to match both sides.
Substitute value of $z^3$ and simplify:$x^3 + y^3 + z^3 = x^3 + y^3 - (x^3 + 3x^2y + 3xy^2 + y^3)$.
Simplify the expression further and relate it to $3xyz$:We find that the expression simplifies to $3xyz$ after substituting $z$ with $-(x+y)$ and carrying out the operations.
Now, let's go into detail for some of these steps, particularly the crucial parts of the simplification that leads to $3xyz$.
Substituting $z = -(x+y)$, we expand $x^3+y^3+z^3$ and use the fact that $x+y+z=0$,
[ x^3 + y^3 + (-(x+y))^3 = 3xyz ]
We can rewrite the given condition $x+y+z=0$ as $z=-(x+y)$. When we substitute this into the equation $x^3+y^3+z^3$, we aim to simplify the expression to eventually show the equality. Simplifying the terms involves expanding the cubed terms and grouping like terms, applying algebraic identities where necessary.
However, direct expansion and substitution followed by simplification will lead us towards the identity $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - xz - yz)$. Given $x + y + z = 0$, this simplifies further. Let's illustrate the key steps to reach our conclusion.
Start with $x^3+y^3+(-(x+y))^3$.
Expand it and use $z=-(x+y)$ to simplify, noting how terms cancel or combine.
Apply the identity if necessary, and use $x+y+z=0$ to simplify the resulting expressions.
The direct substitution and simplification involve algebraic manipulations that will ultimately show that $x^3+y^3+z^3 = 3xyz$. This process demonstrates the beauty of algebraic identities and how they can be employed to prove such statements.
Without actually calculating the cubes, find the value of each of the following:
(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
To solve these problems without directly calculating the cubes, we can utilize the algebraic identity for the sum of cubes:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)$$
When $a + b + c = 0$, this simplifies to:
$$a^3 + b^3 + c^3 = 3abc$$
Let's check whether $a + b + c = 0$ applies to each case:
(i) For $(-12) + 7 + 5$:
$$-12 + 7 + 5 = 0$$
Thus, $a^3 + b^3 + c^3 = 3abc$ applies.
(ii) For $(28) + (-15) + (-13)$:
$$28 - 15 - 13 = 0$$
Thus, $a^3 + b^3 + c^3 = 3abc$ applies here as well.
Calculating the values:
(i) For $(-12)^3 + (7)^3 + (5)^3$:
$$3 \cdot (-12) \cdot 7 \cdot 5$$
(ii) For $(28)^3 + (-15)^3 + (-13)^3$:
$$3 \cdot 28 \cdot (-15) \cdot (-13)$$
We can now proceed to calculate these simplified expressions. Here are the values for each of the expressions:
(i) $(-12)^{3}+(7)^{3}+(5)^{3} = -1260$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3} = 16380$
These results were obtained using the algebraic identity for the sum of cubes and the condition when the sum of the elements equals zero.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
$\text { Area: } 25 a^{2}-35 a+12$
$\text { Area: } 35 y^{2}+13 y-12$
The possible expressions for the length and breadth of each of the given rectangles, where their areas are given, can be factored as follows:
For the area $25 a^{2}-35 a+12$, the possible expressions are: $$(-4 + 5a) \text{ and } (-3 + 5a)$$
For the area $35 y^{2}+13 y-12$, the possible expressions are: $$(4 + 5y) \text{ and } (-3 + 7y)$$
These expressions represent the length and breadth (interchangeably) of each rectangle.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Volume : $3 x^{2}-12 x$
Volume : $12 k y^{2}+8 k y-20 k$
To find possible expressions for the dimensions of the cuboids given their volumes, we'll factorize each polynomial volume expression. This will break down the volume into a product of expressions, each of which can represent a dimension of the cuboid.
For the Volume: $3x^{2} - 12x$
We factor out the common factor:
$$3x^{2} - 12x = 3x(x - 4)$$
This gives two possible expressions for the dimensions of the cuboid: $3x$ and $(x - 4)$. We might consider a third dimension as $1$ to account for all three dimensions, making the volume formula $length \times width \times height$ valid. So, the dimensions could be $3x$, $x - 4$, and $1$.
For the Volume: $12k y^{2} + 8ky - 20k$
First, we factor out the common factor $4k$:
$$12k y^{2} + 8ky - 20k = 4k(3y^{2} + 2y - 5)$$
Then, we look to factor the quadratic expression $(3y^{2} + 2y - 5)$. However, this expression does not easily factor into products of simpler linear terms using integers and might require further sophisticated factorization or might remain prime if such factorization isn’t straightforward or possible with integer coefficients. Let's check if this quadratic expression is easily factorizable. The quadratic expression factors into $(-1 + y)(5 + 3y)$. Therefore, the factorized form of the second volume expression is:
$$12k y^{2} + 8ky - 20k = 4k(-1 + y)(5 + 3y)$$
This provides possible expressions for the dimensions of the cuboid: $4k$, $(-1 + y)$, and $(5 + 3y)$.
Summary
For the volume $3x^{2} - 12x$, the possible expressions for the dimensions of the cuboid are $3x$, $x - 4$, and an implied dimension of $1$.
For the volume $12k y^{2} + 8ky - 20k$, the possible expressions for the dimensions of the cuboid are $4k$, $-1 + y$, and $5 + 3y$.
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Ask Chatterbot AIExtra Questions - Polynomials | NCERT | Mathematics | Class 9
State whether the following expression is a polynomial or not: $ 5a^{-2} + \sqrt{a + 4} ? $
A polynomial is defined as an expression composed of variables (also called indeterminates) and coefficients, employing operations like addition, subtraction, multiplication, and entire non-negative exponents of variables.
In the expression $$ 5a^{-2} + \sqrt{a + 4}, $$ the term $5a^{-2}$ involves an exponent of $-2$, which is not a non-negative integer, and $\sqrt{a + 4}$ is equivalent to $(a + 4)^{1/2}$, which involves an exponent of $\frac{1}{2}$, also not a non-negative integer.
Conclusion: The given expression, $$ 5a^{-2} + \sqrt{a + 4}, $$ is not a polynomial due to the inclusion of negative and fractional exponents.
If $a + b = 1$ and $a - b = 7$, find:
(i) $\left(a^{2} + b^{2}\right)$
(ii) $ab$
A) (i) 25 (ii) -12
B) (i) 51 (ii) 12
C) (i) 26 (ii) -13
D) (i) -25 (ii) 12
The correct option is A
(i) 25
(ii) -12
Given equations are:
$a + b = 1 $
$a - b = 7$
To find $\left(a^{2} + b^{2}\right):$Square both equations:
Squaring $a+b=1$, we have: $$ (a+b)^2 = a^2 + 2ab + b^2 \ 1^2 = a^2 + 2ab + b^2 \ 1 = a^2 + 2ab + b^2 $$
Squaring $a-b=7$, we get: $$ (a-b)^2 = a^2 - 2ab + b^2 \ 7^2 = a^2 - 2ab + b^2 \ 49 = a^2 - 2ab + b^2 $$
Adding the results of these square terms to eliminate $ab$: $$ \left( a^2 + 2ab + b^2 \right) + \left( a^2 - 2ab + b^2 \right) = 1 + 49 \ 2a^2 + 2b^2 = 50 \ a^2 + b^2 = 25 $$
To find $ab$:Substitute $a^2 + b^2 = 25$ back into the first squared equation: $$ 1 = a^2 + 2ab + b^2 \ 1 = 25 + 2ab \ 2ab = 1 - 25 \ 2ab = -24 \ ab = -12 $$
Conclusions:
(i) $\mathbf{a^2 + b^2 = 25}$
(ii) $\mathbf{ab = -12}$
Thus, the correct option is $\mathbf{A}$.
Find the value of $a$ for which $(x+2a)$ is a factor of $\left(x^5 - 4a^2x^3 + 2x + 2a + 3\right)$.
Let the polynomial be defined as: $$ f(x) = x^5 - 4a^2 x^3 + 2x + 2a + 3 $$ It is specified that $(x + 2a)$ is a factor of $f(x)$. According to the Factor Theorem, for $(x + 2a)$ to be a factor, the condition $f(-2a) = 0$ must hold.
Substituting $x = -2a$ into $f(x)$, we find: $$ f(-2a) = (-2a)^5 - 4a^2(-2a)^3 + 2(-2a) + 2a + 3 = 0 $$ Simplifying each term, $$ f(-2a) = -32a^5 + 32a^5 -4a + 2a + 3 = 0 $$ which simplifies further to: $$ -2a + 3 = 0 $$ Solving for $a$, $$ -2a = -3 \ a = \frac{3}{2} $$
Thus, the value of $a$ for which $(x+2a)$ is a factor of the polynomial is $\frac{3}{2}$.
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