1 Marks Question

Question 11 Mark

Factories : 64a³ - 27b³ - 144a²b + 108ab²

Answer

64a³ - 27b³ - 144a²b + 108ab²
= (4a)³ - (3b)³ - 3(4a)(3b)(4a - 3b)
= (4a - 3b)³
(Using Identity (a – b)³= a³– b³– 3ab (a – b))
= (4a - 3b)(4a - 3b)(4a- 3b)

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Question 21 Mark

Factorise :27 - 125a³ - 135a + 225a²

Answer

27 - 125a³ - 135a + 225a²
= (3)³ - (5a)³ - 3(3)(5a)(3 - 5a)
= (3 - 5a)³
(Using Identity (a – b)³= a³– b³– 3ab (a – b))
= (3 - 5a)(3 - 5a)(3 - 5a)

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Question 31 Mark

Factorise : 8a³ - b³ - 12a²b + 6ab²

Answer

8a³ - b³ - 12a²b + 6ab²
= (2a)³ - (b)³ + 3(2a)(b)(2a - b)
= (2a- b)³
(Using Identity (a – b)³= a³– b³– 3ab (a – b))
= (2a - b)(2a - b)(2a - b)

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Question 41 Mark

Evaluate using suitable identity: ${\left( {99} \right)^3}$

Answer

${\left( {99} \right)^3}$
${\left( {99} \right)^3}{\text{ can also be written as}}{\left( {100 - 1} \right)^3}.$
Using identity, ${\left( {x - y} \right)^3} = {x^3} - {y^3} - 3xy\left( {x - y} \right)$
${\left( {100 - 1} \right)^3} = {\left( {100} \right)^3} - {\left( 1 \right)^3} - 3 \times 100 \times 1\left( {100 - 1} \right)$
= 1000000-1-300(99)
= 999999-29700
= 970299.

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Question 51 Mark

Write ${\left( {\frac{3}{2}x + 1} \right)^3}$ in expanded form.

Answer

${\left( {\frac{3}{2}x + 1} \right)^3}$
We know that ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$
${\left( {\frac{3}{2}x + 1} \right)^3} = {\left( {\frac{3}{2}x} \right)^3} + {\left( 1 \right)^3} + 3 \times \frac{3}{2}x \times 1\left( {\frac{3}{2}x + 1} \right)$
$= \frac{{27}}{8}{x^3} + 1 + \frac{9}{2}x\left( {\frac{3}{2}x + 1} \right)$$ = \frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1.$
Therefore, the expansion of the expression ${\left( {\frac{3}{2}x + 1} \right)^3}$ is $\frac{{27}}{8}{x^3} + \frac{{27}}{4}{x^2} + \frac{9}{2}x + 1$

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Question 61 Mark

Write (2a - 3b)³ in the expanded form

Answer

(2a - 3b)³
= (2a)³ - (3b)³ - 3(2a)(3b)(2a - 3b)
(Using Identity (a – b)³= a³– b³– 3ab (a – b))
= 8a³ - 27b³ - 18ab(2a - 3b)
= 8a³ - 27b³ - 36a²b + 54ab²

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Question 71 Mark

Factorise :4x² + 9y² + 16z² + 12xy - 24yz - 16xz

Answer

4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (- 4z)² + 2(2x)(3y)
+ 2(3y)(-4z) + 2(- 4z)(2x)
= {2x + 3y + (-4z)}² = (2x + 3y - 4z)²
= (2x + 3y - 4z)(2x + 3y - 4z)

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Question 81 Mark

Using suitable identity Expand: $\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$

Answer

$\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}$ = $\left[\frac{1}{4} a+\left(-\frac{1}{2} b\right)+1\right]^{2}$
(Using Identity (a + b + c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca)
= $\left(\frac{1}{4} a\right)^{2}+\left(-\frac{1}{2} b\right)^{2}$ + $+(1)^{2}+2\left(\frac{1}{4} a\right)\left(-\frac{1}{2} b\right)+2\left(-\frac{1}{2} b\right)(1)+2(1)\left(\frac{1}{4} a\right)$
= $\frac{1}{16} a^{2}+\frac{1}{4} b^{2}+1-\frac{1}{4} a b-b+\frac{1}{2} a$

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Question 91 Mark

Expand using appropriate identity: (-2x + 5y - 3z)²

Answer

(-2x + 5y - 3z)²
= {(-2x) + 5y + (-3z)}²
= (-2x)²+ (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x² + 25y² + 9z² - 20xy - 30yz + 12zx

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Question 101 Mark

Expand using appropriate identity: (3a - 7b - c)²

Answer

(3a - 7b - c)²= {3a + (- 7b) + (-c)}²
= (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a² + 49b² + c² - 42ab + 14bc - 6ca

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Question 111 Mark

Expand using suitable identity: (-2x + 3y + 2z)²

Answer

(- 2x + 3y + 2z)²
= {(-2x) + 3y + (2z)}²
= (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
(Using Identity (a + b + c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca)
= 4x² + 9y² + 4z² - 12xy+ 12yz - 8zx

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Question 121 Mark

Expand using suitable identity: (2x - y + z)²

Answer

(2x - y + z)²
= {2x + (-y) + z}²
= (2x)² + (-y)² + (z)²+ 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
(Using Identity (a + b + c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca)
= 4x² + y² + z² - 4xy - 2yz + 4zx

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Question 131 Mark

Expand using suitable identity: (x + 2y + 4z)²

Answer

(x + 2y + 4z)²
= (x)² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
(Using Identity (a + b + c)²= a²+ b²+ c²+ 2ab + 2bc + 2ca)
= x² + 4y²+ 16z² + 4xy + 16yz + 8zx

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Question 141 Mark

Factorize the using appropriate identity: ${x^2} - \frac{{{y^2}}}{{100}}$

Answer

${x^2} - \frac{{{y^2}}}{{100}}$
We can observe that we can apply the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
$ \Rightarrow {\left( x \right)^2} - {\left( {\frac{y}{{10}}} \right)^2} = \left( {x + \frac{y}{{10}}} \right)\left( {x - \frac{y}{{10}}} \right).$

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Question 151 Mark

Factorize the using appropriate identity: $4{y^2} - 4y + 1$

Answer

$4{y^2} - 4y + 1 = {\left( {2y} \right)^2} - 2 \times 2y \times 1 + {\left( 1 \right)^2}$

We can observe that we can apply the identity ${\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2} \Rightarrow {\left( {2y} \right)^2} - 2 \times 2y \times 1 + {\left( 1 \right)^2} = {\left( {2y - 1} \right)^2}$.

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Question 161 Mark

Factorize the using appropriate identity: $9{x^2} + 6xy + {y^2}$

Answer

$9{x^2} + 6xy + {y^2}{\text{ = }}{\left( {3x} \right)^2} + 2 \times 3x \times y + {\left( y \right)^2}$
We can observe that we can apply the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
$ \Rightarrow {\left( {3x} \right)^2} + 2 \times 3x \times y + {\left( y \right)^2} = {\left( {3x + y} \right)^2}.$

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Question 171 Mark

Evaluate the product without multiplying directly: $103 \times 107$

Answer

$103 \times 107$
103×107 can also be written as(100+3)(100+7).
We can observe that we can apply the identity (x+a)(x+b)=x² + (a+b)x + ab
(100+3)(100+7)=(100)² + (3 + 7)(100) + 3×7
=10000 + 1000 + 21
= 11021
Therefore, we conclude that the value of the product $103 \times 107$ is 11021.

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Question 181 Mark

What are the possible expression for the dimensions of the cuboid whose volume are: $3{x^2} - 12x$

Answer

${\text{Volume : }}3{x^2} - 12x$
${\text{The expression }}3{x^2} - 12x{\text{ can also be written as}}\,3 \times x \times \left( {x - 4} \right).$
Therefore, we can conclude that a possible expression for the dimensions of a cuboid of volume $3{x^2} - 12x$ is $3,x{\text{ and }}\left( {x - 4} \right)$

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Question 191 Mark

Without actually calculating the cube, find the value of (28)³ + (-15)³ + (-13)³

Answer

(28)³ + (-15)³ + (-13)³
= 3(28)(-15)(-13) ($\because$ (28) + (-15) + (-13) = 0)
= 16380

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Question 201 Mark

Use suitable identity to find the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$

Answer

$\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$
We know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
We need to apply the above identity to find the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$
Here $a = {y^2}$ and $b = {3 \over 2}$
$\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right) = {\left( {{y^2}} \right)^2} - {\left( {\frac{3}{2}} \right)^2}$
$= {y^4} - \frac{9}{4}.$
Therefore, we conclude that the product $\left( {{y^2} + \frac{3}{2}} \right)\left( {{y^2} - \frac{3}{2}} \right)$is $\left( {{y^4} - \frac{9}{4}} \right)$

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Question 211 Mark

If x + y + z = 0 then show that x³ + y³ + z³ = 3xyz.

Answer

We know that
x³+ y³ + z³ - 3xyz = (x+ y + z)(x² + y² + z² - xy - yz - zx)
(Using Identity a³+ b³+ c³– 3abc= (a + b + c)(a²+ b²+ c²– ab – bc – ca))
= (0) (x² + y² + z² - xy - yz - zx) (∵x + y + z = 0)
= 0
⇒ x³ + y³+ z³ = 3xyz.

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Question 221 Mark

Use suitable identity to find the product: (3x + 4)(3x - 5)

Answer

(3x + 4)(3x - 5)
= (3x + 4){3x + (-5)}
{Using identity (x + a)(x + b) = x²+ (a + b) x + ab}
= (3x)² + {4 + (-5)}(3x) + (4)(-5)
= 9x² - 3x - 20

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Question 231 Mark

Factorise : 27x³ + y³ + z³ - 9xyz.

Answer

27x³ + y³ + z³ - 9xyz.
(3x)³ + (y)³ + (z)³ - 3(3x)(y)(z)
= (3x + y + z){(3x)² + (y)² + (z)² - (3x)(y) - (y)(z) - (z)(3x)}
(Using Identity a³+ b³+ c³– 3abc= (a + b + c)(a²+ b²+ c²– ab – bc – ca))
= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx).

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Question 241 Mark

Factorise :64m³ - 343n³

Answer

64m³ - 343n³
= (4m)³ - (7n)³ = (4m - 7n){(4m)² + (4m)(7n) + (7n)²}
= (4m - 7n)(16m² + 28mn + 49n²)

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Question 251 Mark

Check whether $7 + 3x $ is a factor of $ 3{x^3} + 7x$.

Answer

We know that if the polynomial $7+3x$ is a factor of $ 3{x^3} + 7x$, then on dividing the polynomial $3{x^3} + 7x$ by $7+3x$, we must get the remainder as 0.
We need to find the zero of the polynomial 7 + 3x
$\begin{gathered} 7 + 3x = 0 \hfill \\ \Rightarrow {\text{ }}x = - \frac{7}{3} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $7+3x$ in the polynomial $ 3{x^3} + 7x$,to get
$\,\,p\left( x \right) = 3{x^3} + 7x$
$p\left( {{{ - 7} \over 3}} \right)$ $= 3{\left( { - \frac{7}{3}} \right)^3} + 7\left( { - \frac{7}{3}} \right)\,\, = 3\left( { - \frac{{343}}{{27}}} \right) - \frac{{49}}{3}$
$ = - \frac{{343}}{9} - \frac{{49}}{3}\,\, = \frac{{ - 343 - 147}}{9}$
$= \frac{{ - 490}}{9}.$
We conclude that on dividing the polynomial $3{x^3} + 7x $ by 7 + 3x, we will get the remainder as $\frac{{ - 490}}{9}$ which is not 0.
Therefore, we conclude that 7 + 3x is not a factor of $ 3{x^3} + 7x$

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Question 261 Mark

Find the remainder when x³ + 3x² + 3x + 1 is divided by 5 + 2x.

Answer

Let p(x) = x³ + 3x² + 3x + 1
5 + 2x = 0
$\Rightarrow$ 2x = -5
$\Rightarrow$ x = -$\frac{5}{2}$
= (-$\frac{5}{2}$)³ + 3(-$\frac{5}{2}$)² + 3(-$\frac{5}{2}$) + 1
= -$\frac{125}{8}$ + $\frac{75}{4}$ - $\frac{15}{2}$ + 1 = -$\frac{27}{8}$

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Question 271 Mark

Find the remainder when x³ + 3x² + 3x + 1 is divided by x + $\pi$

Answer

Let p(x) = x³ + 3x² + 3x + 1
x + $\pi$ = 0
$\Rightarrow$ x = -$\pi$
$\therefore$ Remainder = (-$\pi$)³ + 3(-$\pi$)² + 3(-$\pi$) + 1 = $-\pi ^3$ + 3$\pi^2$ - 3$\pi$ + 1

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Question 281 Mark

Find the remainder when ${x^3} + 3{x^2} + 3x + 1$ is divided by x + 1

Answer

x + 1
We need to find the zero of the polynomial x + 1
$x + 1 = 0{\text{ }} \Rightarrow {\text{ }}x = - 1$
While applying the remainder theorem, we need to put the zero of the polynomial x + 1 in the polynomial
${x^3} + 3{x^2} + 3x + 1$,to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( -1 \right) = {\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + 3\left( { - 1} \right) + 1$

=-1+3-3+1
= 0
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1$ by x + 1, we will get the remainder as 0.

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Question 291 Mark

Find the zero of the polynomial in p(x) = 3x

Answer

p(x) = 3x
put, p(x) = 0
3x = 0
x = 0
Therefore, x = 0 is a zero of the polynomial p(x) = 3x.

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Question 301 Mark

Find the zero of the polynomial in p(x) = 3x - 2.

Answer

We have,
p(x) = 3x - 2
p(x) = 0
3x – 2 = 0
x = $\frac{2}{3}$ Therefore, x = $\frac{2}{3}$ is a zero of the polynomial p(x) = 3x - 2

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Question 311 Mark

Find the zero of the polynomial in p(x) = 2x + 5

Answer

The given polynomial is,
P(x) = 2x + 5
put P(x) = 0
2x + 5 = 0
2x = -5
$x =\frac{-5}{2}$
Therefore, $x =\frac{-5}{2}$ is a zero of the polynomial p(x) = 2x + 5

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Question 321 Mark

Find the zero of the polynomial in p(x) = x – 5

Answer

p(x) = x - 5
put p(x)=0
x - 5 = 0
x = 5
Therefore, x = 5 is a zero of the polynomial p(x) = x - 5

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Question 331 Mark

Find the zero of the polynomial in p(x) = x + 5.

Answer

Zero of a polynomial means the value of the variable at which the polynomial becomes zero.
p(x) = x + 5

p(x)=0
x + 5 = 0
x = -5
Thus, x = -5 is a zero of the polynomial p(x) = x + 5.

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Question 341 Mark

Find p(0), p(1) and p(2) for the polynomial: p(x) = x³

Answer

p(x) = x³
$\therefore$ p(0) = (0)³= 0,
p(1) = (1)³ = 1
p(2) = (2)³ = 8

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Question 351 Mark

Find p(0), p(1) and p(2) for the polynomial: p(y) = y² – y + 1

Answer

p(y) = y² – y + 1
$\therefore$ p(0) = (0)² – (0) + 1 = 1,
p(1) = (1)² – (1) + 1 = 1,
p(2) = (2)² – (2) + 1 = 4 – 2 + 1 = 3

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Question 361 Mark

Find the value of the polynomial 5x – 4x² + 3 at x = 2

Answer

Let f(x) = 5x – 4x² + 3
The value of f(x) at x = 2
f(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3 = –3

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Question 371 Mark

Find the value of the polynomial 5x – 4x² + 3 at x = -1

Answer

Let f(x) = 5x – 4x² + 3
The value of f(x) at x = –1
f(– 1) = 5(– 1) – 4 (– 1)² + 3
= – 5 – 4 + 3 = – 6

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Question 381 Mark

Find the value of the polynomial 5x - 4x² + 3 at x = 0

Answer

Let f(x) = 5x - 4x² + 3
We need to substitute x = 0 in the polynomial f(x) = 5x - 4x² + 3 to get the value of given polynomial.
f(0) = 5(0) - 4(0)² + 3
= 0 - 0 + 3
= 3
Therefore, we conclude that at x = 0, the value of the polynomial 5x - 4x² + 3 is 3.

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Question 391 Mark

Classify as linear, quadratic and cubic polynomials:
$7{x^3}$

Answer

$7{x^3}$
We can observe that the degree of the polynomial $7{x^3}$ is 3.
Therefore, we can conclude that the polynomial $7{x^3}$ is a cubic polynomial.

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Question 401 Mark

Classify as linear, quadratic and cubic polynomials:
${r^2}$

Answer

${r^2}$
We can observe that the degree of the polynomial ${r^2}$ is 2.
Therefore, we can conclude that the polynomial ${r^2}$ is a quadratic polynomial.

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Question 411 Mark

Classify as linear, quadratic and cubic polynomials:
3t

Answer

3t
We can observe that the degree of the polynomial $\left( {3t} \right)$is 1. Therefore, we can conclude that the polynomial 3t is a linear polynomial.

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Question 421 Mark

Classify as linear, quadratic and cubic polynomials:
$1 + x{\text{ }}$

Answer

$1 + x{\text{ }}$
We can observe that the degree of the polynomial $\left( {1 + x{\text{ }}} \right)$ is 1.
Therefore, we can conclude that the polynomial $1 + x{\text{ }}$ is a linear polynomial.

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Question 431 Mark

Classify as linear, quadratic and cubic polynomials:
$y + {y^2} + 4$

Answer

$y + {y^2} + 4$
We can observe that the degree of the polynomial $y + {y^2} + 4$ is 2.
Therefore, the polynomial $y + {y^2} + 4$ is a quadratic polynomial.

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Question 441 Mark

Classify the linear, quadratic and cubic polynomials: $x - {x^3}$

Answer

$x - {x^3}$
We can observe that the degree of the polynomial $x - {x^3}$ is 3.
Therefore, we can conclude that the polynomial $x - {x^3}$ is a cubic polynomial.

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Question 451 Mark

Classify as linear, quadratic and cubic polynomials:
${x^2} + x{\text{ }}$

Answer

${x^2} + x{\text{ }}$
We can observe that the degree of the polynomial ${x^2} + x{\text{ }}$ is 2.
Therefore, we can conclude that the polynomial ${x^2} + x{\text{ }}$ is a quadratic polynomial.

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Question 461 Mark

Write the degree of the polynomial : 3

Answer

It is a non-zero constant. So the degree of this polynomial is zero.

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Question 471 Mark

Write the degree of the polynomial: 5t - $\sqrt{7}$

Answer

Term with the highest power of t = 5t
Exponent of t in this term = 1
$\therefore$ Degree of this polynomial = 1

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Question 481 Mark