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(x + 5) (X + 5) = (x + 5)² 

= x² + 2(x) (5) + 5² 

= x² + 10x + 25

(p – 3) (p + 3) 

= p² – 3² 

= p² – 9

Given, area of rectangle = 4a² + 6a - 2a - 3

= 4a² + 4a – 3 [by splitting middle term]

= 2a(2a + 3) -1 (2a + 3)

= (2a – 1)(2a + 3) 

Hence, possible length = 2a -1 and breadth = 2a + 3.

102 x 98 = (100 + 2) (100 -2) 

= 100² – 2² 

= 10000 – 4 

= 9996

49a² + 70ab + 25b² 

= (7a)² + 2 (7a) (5b) + (5b)² 

= (7a + 5b)² 

= (7a + 5b)(7a + 5b)

Given that (0.2)³ – (0.3)³ + (0.1)³ 

= (0.2) + (- 0.3) + (0.1) 

Sum of the bases 

= 0.2 – 0.3 + 0.1 = 0

∴ (0.2)³ + (-0.3)³ + (0.1)³ 

= 3 x (0.2) (- 0.3) (0.1) 

= -0.018

= (t)² – 2(t) (1) + (1)² 

= (t – 1)² = (t – 1) (t – 1)

– 7x¹⁰ is a monomial of degree 10. 

3x² y⁸ + 7xy – 8 is a trinomial of degree 10.

Given (28)³ + (- 15)³ + (- 13)³

Sum of the bases = 28 + (- 15) + (- 13) = 0 

∴ (28)³ + (- 15)³ + (- 13)³ 

= 3 x 28 x (- 15) x (- 13) 

= 16380

Let f(x) = x² + 3x ˗ 10

= x² + 5x ˗ 2x ˗ 10

= x(x + 5) ˗ 2(x + 5)

= (x ˗ 2) (x + 5)

To find the zeroes, set f(x) = 0, then

either x ˗ 2 = 0 or x + 5 = 0

⇒ x = 2 or x = −5.

So, the zeroes of f(x) are 2 and −5.

Again,

Sum of zeroes = 2 + (-5) = -3 = (-3)/1

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = (2)(-5) = -10 = (-10)/1

= c/a

= Constant term / Coefficient of x²

Let f(x) = x² ˗ 2x ˗ 8

= x² ˗ 4x + 2x ˗ 8

= x(x ˗ 4) + 2(x ˗ 4)

= (x ˗ 4) (x + 2)

To find the zeroes, set f(x) = 0, then

either (x ˗ 4) = 0 or (x+2) = 0

x = 4 or x = -2

Again,

Sum of zeroes = (4 – 2) = 2 = 2/1

= -b/a

= (-Coefficient of x)/(Cofficient of x²)

Product of zeroes = (4) (-2) = -8/1

= c/a

= Constant term / Coefficient of x²

Constant polynomials: The polynomial of the degree zero.

Linear polynomials: The polynomial of degree one.

Quadratic polynomials: The polynomial of degree two.

Cubic polynomials: The polynomial of degree three.

(i) 2 – x² + x³

Powers of x = 2, and 3 respectively.

Highest power of the variable x in the given expression = 3

Hence, degree of the polynomial = 3

Since it is a polynomial of the degree 3, it is a cubic polynomial.

(ii) 3x³

Power of x = 3.

Highest power of the variable x in the given expression = 3

Hence, degree of the polynomial = 3

Since it is a polynomial of the degree 3, it is a cubic polynomial.

(iii) 5t – √7

Power of t = 1.

Highest power of the variable t in the given expression = 1

Hence, degree of the polynomial = 1

Since it is a polynomial of the degree 1, it is a linear polynomial.

(iv) 4 – 5y²

Power of y = 2.

Highest power of the variable y in the given expression = 2

Hence, degree of the polynomial = 2

Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(v) 3

There is no variable in the given expression.

Let us assume that x is the variable in the given expression.

3 can be written as 3x⁰.

i.e., 3 = x⁰

Power of x = 0.

Highest power of the variable x in the given expression = 0

Hence, degree of the polynomial = 0

Since it is a polynomial of the degree 0, it is a constant polynomial.

(vi) 2 + x

Power of x = 1.

Highest power of the variable x in the given expression = 1

Hence, degree of the polynomial = 1

Since it is a polynomial of the degree 1, it is a linear polynomial.

(vii) y³ – y

Powers of y = 3 and 1, respectively.

Highest power of the variable x in the given expression = 3

Hence, degree of the polynomial = 3

Since it is a polynomial of the degree 3, it is a cubic polynomial.

(viii) 1 + x + x²

Powers of x = 1 and 2, respectively.

Highest power of the variable x in the given expression = 2

Hence, degree of the polynomial = 2

Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(ix) t²

Power of t = 2.

Highest power of the variable t in the given expression = 2

Hence, degree of the polynomial = 2

Since it is a polynomial of the degree 2, it is a quadratic polynomial.

(x) √2x – 1

Power of x = 1.

Highest power of the variable x in the given expression = 1

Hence, degree of the polynomial = 1

Since it is a polynomial of the degree 1, it is a linear polynomial.

i) 5x² + x – 7 : degree 2 hence quadratic polynomial.

ii) x – x³ , : degree 3 hence cubic polynomial. 

iii) x² + x + 4 : degree 2 hence quadratic polynomial. 

iv) x – 1 : degree 1 hence linear polynomial. 

v) 3p : degree 1 hence linear polynomial. 

vi) πr² : degree 2 hence quadratic polynomial.

Correct option is (B) \(\frac{8x^2}{x^4-1}\)

\(\frac{x+1}{x-1}+\frac{x-1}{x+1}-\frac{2x^2-2}{x^2+1}\)

\(=\frac{(x+1)^2(x^2+1)+(x-1)^2(x^2+1)-2(x^2-1)(x^2-1)}{(x-1)(x+1)(x^2+1)}\)

\(=\frac{(x^2+1)(x^2+2x+1+x^2-2x+1)-2(x^2-1)^2)}{(x^2-1)(x^2+1)}\)

\(=\frac{2(x^2+1)^2\div2(x^2-1)^2}{z^4-1}\)

\(=\frac{2(x^4+2x^2+1-(x^4-2x^2+1))}{z^4-1}\)

\(=\frac{8x^2}{x^4-1}\)

Correct option is B) \(\cfrac{8x^2}{x^4-1}\)

Correct option is (C) 0

\(\frac{x^2-(y-2z)^2}{x-y+2z}+\frac{y^2-(2x-z)^2}{y+2x-z}+\frac{z^2-(x-2y)^2}{z-x+2y}\)

\(=\frac{(x-(y-2z))(x+y-2z)}{x-y+2z}+\frac{(y-(2x-z))(y+2x-z)}{y+2x-z}+\frac{(z-(x-2y))(z+x-2y)}{z-x+2y}\) \((\because(a+b)(a-b)=a^2-b^2)\)

\(=\frac{(x-y+2z)(x+y-2z)}{x-y+2z}+\frac{(y-2x+z)(y+2x-z)}{y+2x-z}+\frac{(z-x+2y)(z+x-2y)}{z-x+2y}\)

= (x+y-2z) + (y-2x+z) + (z+x-2y)

= (x+x-2x) + (y+y-2y) + (-2z+z+z)

= 0+0+0 = 0

Correct option is C) 0

Correct option is (B) 4ab

\(\because\) \((a^2 + 2ab + b^2)\) \(-(a^2 – 2ab +b^2)\) = 4ab

\(\therefore\) \(a^2 + 2ab + b^2\) is 4ab more than \(a^2 – 2ab +b^2.\)

Correct option is B) 4ab

Correct option is (C) 1

\(\frac{a^2-(b-c)^2}{(a+c)^2-b^2}+\frac{b^2-(a-c)^2}{(a+b)^2-c^2}+\frac{c^2-(a-b)^2}{(b+c)^2-a^2}\)

\(=\frac{(a-(b-c))(a+b-c)}{(a+c-b)(a+c+b)}+\frac{(b-(a-c))(b+a-c)}{(a+b-c)(a+b+c)}+\frac{(c-(a-b))(c+a-b)}{(b+c-a)(b+c+a)}\)

\(=\frac{(a+c-b)(a+b-c)}{(a+c-b)(a+b+c)}+\frac{(b-a+c)(a+b-c)}{(a+b-c)(a+b+c)}+\frac{(c-a+b)(a+c-b)}{(c-a+b)(a+b+c)}\)

\(=\frac{a+b-c}{a+b+c}+\frac{b-a+c}{a+b+c}+\frac{a+c-b}{a+b+c}\)

\(=\frac{a+b-c+b-a+c+a+c-b}{a+b+c}\)

\(=\frac{2a-a+2b-b+2c-c}{a+b+c}\)

\(=\frac{a+b+c}{a+b+c}\) = 1

Correct option is C) 1

Correct option is (A) 2

\(C=A\times-B\)

\(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right)(\frac{x^4-1}{4x})\)

\(=\left(\frac{(x+1)^2(x^2+1)-(x-1)^2(x^2+1)-4x(x^2-1)}{(x-1)(x+1)(x^2+1)}\right)(\frac{x^4-1}{4x})\)

\(=\left(\frac{(x^2+1)(x^2+2x+1-(x^2-2x+1))-4x(x^2-1)}{(x^2-1)(x^2+1)}\right)(\frac{x^4-1}{4x})\)

\(=\frac{4x(x^2+1-x^2+1)}{x^4-1}\times\frac{x^4-1}{4x}\)

= 2

Correct option is A) 2

A) The sum of two rational expressions is always a rational expression

Correct option is (A) x² - 1/x

\(x- \frac{1}{x}\) = \(\frac{x^2-1}{x}\)

Correct option is A) x² - 1/x

We know that x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx). 

If x + y + z = 0, then x³ + y³ + z³ – 3xyz = 0 or x³ + y³ + z³ = 3xyz. 

We have to find the value of 

48³ – 30³ – 18³

= 48³ + (–30)³ + (–18)³ . 

Here, 48 + (–30) + (–18) = 0 

So, 48³ + (–30)³ + (–18)³ 

= 3 × 48 × (–30) × (–18)

= 77760

x³ + y³ 

= (x + y) (x² – xy + y² ) 

= (x + y) [(x + y)² – 3xy]

= 12 [122 – 3 × 27] 

= 12 × 63 

= 756

α + ß = \(\frac{-b}a=\frac{-5}k\)

αß = \(\frac ca=\frac 2k\)

Now, \(\frac1{\alpha^2} + \frac1{\beta^2}\) = \(\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\frac{(α+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}\)

= \(\cfrac{(\frac{-5}k)^2-4/k}{(\frac 2k)^2}\) = \(\cfrac{\frac{25}{k^2}-\frac 4k}{\frac4{k^2}}\)

 = \(\frac{25-4k}4=\frac{17}4\)

⇒ 25 - 4k = 17

⇒ 4k = 25 - 17 = 8

⇒ k = 2

Given;

f(x) = (k² + 4) x² + 13x + 4k,

One zero of the polynomial is reciprocal of the other,

Let a be the one zero,

∴ The other zero will be 1/a

As we know that,

Product of the zeros = c/a = 4k/k² + 4

∴ 4k/k² + 4 = 1

⇒ 4k = k² + 4

⇒ k ² + 4 – 4k = 0

⇒ (k – 2)² = 0

⇒ k = 2

So the value of k is 2

(c) \(\frac{-9}2\)

Given: 

α and β be the zeroes of 2x² + 5x – 9. 

If α + β are the zeroes, then x² – (α + β) x + αβ is the required polynomial. 

The polynomial will be x² – \(\frac{5}2x\) - \(\frac{9}2\).

∴ αβ = \(\frac{-9}2\)

Correct option is (D) -4/3

Product of zeros \(=\frac{\text{constant term}}{\text{coefficient of }x^2}=\frac{-4}3\)

\(\therefore\alpha\beta\) \(=\frac{-4}3\)

Correct option is D) -4/3

Given : α + β = -6 = sum of zeros and αβ = 5 = Product of zeroes 

We know that, if α and β are the zeros of the polynomial then the quadratic polynomial can be x² – (α + β) x + αβ 

Now substituting the values, we get 

x² – (-6)x + 5

= x² + 6x + 5, which is required polynomial.

Correct option is (B) 5

Let \(\alpha\;and\;\frac1\alpha\) are zeros of given polynomial f(x).

\(\therefore\) Product of zeros \(=\frac{\text{constant term}}{\text{coefficient of }x^2}=\frac k5\)

\(\Rightarrow\) \(\alpha.\frac1\alpha\) \(=\frac k5\)

\(\Rightarrow\) \(\frac k5=1\)

\(\Rightarrow\) k = 5

Correct option is B) 5

Let breadth of a plot is x m.

According to question,

Length of plot = (2 × breadth) + l 

= (2 × x + 1) = (2x + 1) m.

Area of rectangular plot = l × b 

= (2x + 1) × x = (2x² + x) sq. m.

Given : Area of plot = 528 sq. m.

2x² + x = 528

⇒ 2x² + x – 528 = 0

Required quadratic equation is :

2x² + x – 528 = 0

⇒ 2x² + 33x – 32x – 528 = 0

⇒ x(2x + 33) – 16(2x + 33) = 0

⇒ (2x + 33) (x – 16) = 0

⇒ x – 16 = 0 or 2x + 33 = 0

⇒ x = 16 or x = -33/2 (impossible)

Hence, Length of plot is 2x + 1 

= 2 × 16 + 1 = 33 m

and breadth is 16 m.

Given polynomial p(x) = x² – 4x + 3 

We have, x² – 4x + 3 = x² – 3x – x + 3 

= x(x – 3) – 1 (x – 3) 

= (x – 3) (x – 1) 

So, the value of x² – 4x + 3 is zero when x – 3 = 0

or x – 1 =0, i.e., 

when x = 3 or x = 1 So, the zeroes of x² – 4x + 3 are 3 and 1

∴ Sum of the zeroes = 3 + 1 = 4

\(=-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2} = \frac{-(-4)}{1}=4\)

And product of the zeroes = 3 × 1 = 3

\(=\frac{Constant\,term}{Coefficient\,of\,x^2}= \frac{3}{1}=3\)

Correct option is (D) a natural number

\(x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y\) \(+x^{n-3}y^2+....+xy^{n-2}+y^{n-1}),\) where n is a natural number.

Thus, \(x^n-y^n\) is divisible by x - y when n is a natural number.

Correct option is D) a natural number

Correct option is (B) \(\frac{x-1}{x+1}\)

Let p(x) should be added to \(\frac{4x}{x^2-1}\) to get \(\frac{x+1}{x-1}.\)

i.e., \(\frac{4x}{x^2-1}+p(x)\) \(=\frac{x+1}{x-1}\)

\(\Rightarrow\) p(x) \(=\frac{x+1}{x-1}-\frac{4x}{x^2-1}\)

\(=\frac{(x+1)^2-4x}{x^2-1}\) \(=\frac{x^2+2x+1-4x}{(x-1)(x+1)}\)

\(=\frac{x^2-2x+1}{(x-1)(x+1)}\) \(=\frac{(x-1)^2}{(x-1)(x+1)}\)

\(=\frac{x-1}{x+1}\)

Correct option is B) \(\cfrac{x-1}{x+1}\)

Correct option is (A) \(\frac{1}{x^2-5x+6}\)

Let p(x) should be added to \(\frac{1}{x^2 -7x+12}\) to get \(\frac{2}{x^2 -6x+8}\)

\(\therefore\) \(p(x)+\frac{1}{x^2 -7x+12}\) \(=\frac{2}{x^2 -6x+8}\)

\(\Rightarrow\) \(p(x)=\frac{2}{x^2 -6x+8}\) \(-\frac{1}{x^2 -7x+12}\)

\(=\frac{2}{(x-4)(x-2)}-\frac{1}{(x-4)(x-3)}\)

\(=\frac{2(x-3)-(x-2)}{(x-2)(x-3)(x-4)}\) \(=\frac{2x-6-x+2}{(x-2)(x-3)(x-4)}\)

\(=\frac{x-4}{(x-2)(x-3)(x-4)}\)

\(=\frac{1}{(x-2)(x-3)}\) \(=\frac{1}{x^2-5x+6}\)

Correct option is A) \(\cfrac{1}{x^2-5x+6}\)

Correct option is (A) \(\frac{2x-6}{x}\)

\(\frac{2x^3 -54}{x^3 +3x^2 +9x}\) \(=\frac{2(x^3-27)}{x(x^2+3x+9)}\)

\(=\frac{2(x-3)(x^2+3x+9)}{x(x^2+3x+9)}\)

\(=\frac{2(x-3)}{x}\) \(=\frac{2x-6}{x}\)

Correct option is A) \(\cfrac{2x-6}{x}\)