Polynomials - Class 9 - Mathematics
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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) $\sqrt{2} x-1$
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Sign up nowGive one example each of a binomial of degree 35 , and of a monomial of degree 100 .
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Sign up nowWrite 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
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Sign up nowClassify 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) $7 x^{3}$
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Sign up nowExercise 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) $p(x)=(x-1)(x+1)$
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Sign up nowVerify 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) $p(x)=2 x+1, x=\frac{1}{2}$
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Sign up nowFind 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.
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Sign up nowExercise 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) $p(x)=x^{3}-4 x^{2}+x+6, g(x)=x-3$
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Sign up nowFind 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) $p(x)=k x^{2}-3 x+k$
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Sign up nowFactorise :
(i) $12 x^{2}-7 x+1$
(ii) $2 x^{2}+7 x+3$
(iii) $6 x^{2}+5 x-6$
(iv) $3 x^{2}-x-4$
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Sign up nowFactorise :
(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) $2 y^{3}+y^{2}-2 y-1$
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Sign up nowExercise 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) $104 \times 96$
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Sign up nowFactorise the following using appropriate identities:
(i) $9 x^{2}+6 x y+y^{2}$
(ii) $4 y^{2}-4 y+1$
(iii) $x^{2}-\frac{y^{2}}{100}$
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Sign up nowExpand 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) $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$
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Sign up nowFactorise:
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
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Sign up nowWrite 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) $\left[x-\frac{2}{3} y\right]^{3}$
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Sign up nowEvaluate the following using suitable identities:
(i) $(99)^{3}$
(ii) $(102)^{3}$
(iii) $(998)^{3}$
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Sign up nowFactorise 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) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$
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Sign up nowVerify: (i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
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Sign up nowFactorise each of the following:
(i) $27 y^{3}+125 z^{3}$
(ii) $64 m^{3}-343 n^{3}$
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Sign up nowFactorise : $27 x^{3}+y^{3}+z^{3}-9 x y z$
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Sign up nowVerify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$
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Sign up nowIf $x+y+z=0$, show that $x^{3}+y^{3}+z^{3}=3 x y z$.
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Sign up nowWithout actually calculating the cubes, find the value of each of the following:
(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
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Sign up nowGive possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
$\text { Area: } 25 a^{2}-35 a+12$
$\text { Area: } 35 y^{2}+13 y-12$
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Sign up nowWhat are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Volume : $3 x^{2}-12 x$
Volume : $12 k y^{2}+8 k y-20 k$
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Sign up nowExtra 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
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