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(i) (x + 4) (x + 10) 

Suitable identity is (x + a) (x + b) 

= x² + (a + b)x + ab (x + 4) (x + 10) 

= x² + (4 + 10)x + (4 × 10) 

= x² + 14x + 40 

(ii) (x + 8) (x- 10) 

Suitable identity is (x + a) (x + b) 

= x² + (a + b)x + ab (x + 8) (x – 10)

= (x)² + (8 – 10)x + (8 × -10) 

= x² + (-2)x + (-80) 

= x² – 2x – 80 

(iii) (3x + 4) (3x – 5) 

Suitable identity is (x + a) (x + b) 

= x² + (a + b)x + ab (3x + 4) (3x – 5) 

= (3x)² + (+4 – 5)3x + (4 x -5) 

= 9x² + (-1)3x + (-20) 

= 9x² – 3x – 20 

(iv) Suitable identity is (a + b) (a – b) 

= a² – b² 

(v) (3 – 2x) (3 + 2x) 

Suitable identity is (a + b) (a – b) 

= a² – b² (3 – 2x) (3 + 2x) 

= (3)² – (2x)² = 9 – 4x² .

(i) x³ – 2x² – x + 2 

= x²(x – 2) – 1 (x – 2) 

= (x – 2) (x² – 1) 

= (x – 2) (x + 1) (x- 1) 

(ii) x³ – 3x² – 9x – 5 

= x³ – 5x² + 2x² – 10x + x – 5 

= x²(x – 5) + 2x(x – 5) + 1 (x – 5) 

= (x – 5) (x² + 2x + 1) 

= (x – 5) {x² + x + x + 1} 

= (x – 5) (x(x + 1) + 1(x + 1)} 

= (x- 5)(x + 1) (x + 1) 

(iii) x³ + 13x² + 32x + 20 

= x³ + 10x² + 3x² + 30x + 2x + 26 

= x²(x + 10) + 3x(x + 10) + 2(x + 10) 

= (x + 10) (x² + 3x + 2) 

= (x + 10) {x² + 2x + x + 2) 

= (x + 10) {x(x + 2) + 1 (x + 2)) 

= (x + 10) (x + 2) (x + 1) 

(iv) 2y³ + y² – 2y – 1 

= y²(2y + 1) – 1(2y + 1) 

= (2y + 1) (y²– 1)

= (2y + 1) {(y)² – (1)²} 

= (2y+ 1) (y + 1) (y- 1)

Let x be the shorter side, then the other side will be (x + 1).

i. Perimeter = 2[x + (x + 1)] = 2(2x + 1) = 4x + 2

That is, p(x) = 4x + 2

ii. Area = x(x + 1)

a(x) = x² + x

Area, a(x) = x² + x

iii. p(x) = 4x + 2

p(1) = 4 × 1 + 2 = 6 

p(2) = 4 × 2 + 2 = 10 

p(3) = 4 × 3 + 2= 14 

p(4) = 4 × 4 + 2= 18 

p(5) = 4 × 5 + 2 = 22

Perimeter is a sequence increasing by 4.

iv. a(x) = x² + x

a(1) = 1² + 1 = 2 = 1 × 2 = 2 

a(2) = 2² + 2 = 6 = 2 × 3 = 6 

a(3) = 3² + 3 = 12 = 3 × 4 = 12 

a(4) = 4² + 4 = 20 = 4 × 5 = 20 

a(5) = 5² + 5 = 30 = 5 × 6 = 30

Area is the product of x and the number one more than x.

i. 1 m = 100 cm

If one side is x cm, then the other side is 50 – x.

Area of rectangle = a(x) = x(50 – x)

= 50x – x² cm²

a(x) = 50x – x²

ii. a(10) = 50 × 10 – 10² = 500 – 100 = 400

a(40) = 50 × 40 – 40² = 2000 – 1600 = 400 

10 and 40 are the sides of rectangle, so a(10) and a(40) are same .

iii. Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

Given a + b + c = 0 

⇒ a + b = -c ……………..(1)

Cubing on both sides 

(a + b)³ = (-c)³ 

a³ + b³ + 3ab (a + b) = -c³

a³ + b³ + 3ab (-c) =-c³ From(l) 

a³ + b³ – 3abc = -c³ 

a³ + b³ + c³ – 3abc

(A) 4/3

Explanation:

According to the question,

-3 is one of the zeros of quadratic polynomial (k-1)x²+kx+1

Substituting -3 in the given polynomial,

(k-1)(-3)²+k(-3)+1=0

(k-1)9+k(-3)+1 = 0

9k-9-3k+1=0

6k-8=0

k=8/6

Therefore, k = 4/3

Hence, option (A) is the correct answer.

Let p(x) = x³ – ax² + 2x + a – 1 

Since x – a is a factor of p(x), so p(a) = 0. 

i.e., a³ – a(a)² + 2a + a – 1 = 0 

a³ – a³ + 2a + a – 1 = 0 

3a = 1 

Therefore, a = 1/3

Correct option is (B) \(\frac{x^2+6x-11}{x^2+x-6}\)

Let p(x) should be subtracted from

\(\left(\frac{2x^2+2x-7}{x^2+x-6}\right)\) to get \(\left(\frac{x-2}{x+3}\right)\)

i.e., \(\frac{2x^2+2x-7}{x^2+x-6}-p(x)\) \(=\frac{x-2}{x+3}\)

\(\Rightarrow\) \(p(x)=\frac{2x^2+2x-7}{x^2+x-6}\) \(-\frac{x-2}{x+3}\)

\(=\frac{2x^2+2x-7-(x-2)^2}{(x+3)(x-2)}\)

\(=\frac{2x^2+2x-7-x^2+4x-4}{(x-2)(x+3)}\)

\(=\frac{x^2+6x-11}{x^2+x-6}\)

Correct option is B) \(\frac{x^2+6x-11}{x^2+x-6}\)

Correct option is (A) x² + 2x

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

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

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

\(=1+x+x^2+x-1\)

\(=2x+x^2\)

Correct option is A) x² + 2x

(-12)³ + (7)³ + (5)³

Let x = −12, y = 7, and z = 5

It can be observed that,

x + y + z = − 12 + 7 + 5 = 0

It is known that if x + y + z = 0, then

x³ + y³ +z³

(-12)³ + (7)³ + (5)³

= 3(-12)(7)(5)

= -1260

Correct option is (D) (x - 2)/y

\(\frac{x^2-4}{xy+2y}\) \(=\frac{(x-2)(x+2)}{y\,(x+2)}\)

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

Correct option is D) (x - 2)/y

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. 

Here, (x – y) + (y – z) + (z – x) = 0 

Therefore, (x – y)³ + (y – z)³ + (z – x)³ 

= 3(x – y) (y – z) (z – x).

(b) x² + x + 1

\(x^3 - 1 = (x - 1)(x^2 + x +1)\)

\(x^4 + x^2 + 1 = x^4 + 2x^2 + 1 - x^2 = (x^2 + 1) - x^2\) 

= \((x^2+1-x)(x^2+1+x)\)

∴ Reqd. HCF = x² + x + 1

Given equation is (b – c)x² + (c – a)x + (a – b) = 0

Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b)

Discriminant (D) = b² – 4ac

= (c – a)² – 4(b – c)(a – b)

= (c² + a² – 2ac) – 4(ab – ac – b² + bc)

= c² + a² – 2ac – 4ab + 4ac + 4b² – 4bc

= c² + a² + 2ac – 4ab – 4bc + 4b²

= (a + c)² – 4 b(a + c) + 4b²

= (a + c)² – 2 × (a + c) × 2b + (2b)²

= [(a + c) – 2b]²

Roots of equation are real and equal.

D = 0

⇒ [(a + c) – 2b]² = 0

⇒ (a + c) – 2b = 0

⇒ a + c = 2b

8\(x\)²y² = 2 x 2 x2 x \(x\) x \(x\) x y x y

12\(x\)³y² = 2 x 2 x 3 x \(x\) x \(x\) x \(x\) x y x y

24\(x\)⁴y³ z² = 2 x 2 x 2 x 3 x \(x\) x \(x\) x \(x\) x \(x\) x y x y x y x z x z.

HCF of 8x²y² , 12x³y² and 24x⁴y³ z² = 2 x 2 x \(x\) x \(x\) x y x y = 4x²y².

\(x\)² - 5\(x\) + 6 = \(x\)² - 2\(x\) - 3\(x\) + 6

= \(x\) (\(x\) –2) – 3 (\(x\) – 2) = (\(x\) –2) (\(x\) –3)

x² – 9 = (\(x\) –3) (\(x\) + 3) 

∴ HCF of (\(x\)² – 5\(x\) + 6) and (\(x\)² – 9) = \(x\) – 3

Correct option is (C) either a = b or a² + ab + b² = 0

Given that \(ax^2 + 2a^2 x + b^3\) is divisible by (x+a).

i.e., x = -a is a zero of \(ax^2 + 2a^2 x + b^3\)

\(\therefore\) \(a(-a)^2+2a^2\times-a+b^3=0\)

\(\Rightarrow\) \(a^3-2a^3+b^3=0\)

\(\Rightarrow\) \(b^3-a^3=0\)

\(\Rightarrow\) \((b-a)(b^2+ab+a^2)=0\)

\(\Rightarrow\) Either b - a = 0 or \(b^2+ab+a^2=0\)

\(\Rightarrow\) Either b = a or \(a^2 + ab + b^2 = 0\)

Correct option is C) either a = b or a² + ab + b² = 0

Correct option is (B) -b/2

\(\frac{b}{3}\) is a zero of expression (3x+2a) (2x+1)

\(\therefore(3\times\frac b3+2a)(\frac{2b}3+1)=0\)

\(\Rightarrow\) (b+2a) \(=\frac0{\frac{2b}3+1}=0\)

\(\Rightarrow\) 2a = -b

\(\Rightarrow\) a = \(\frac{-b}2\)

Correct option is B) -b/2

\(\frac{6p^2-150}{p^2-3x-40}\) = \(\frac{6(p^2-25)}{p^2-8x+5x-40}\)

= \(\frac{6(p-5)(p+5)}{p(p-8)+5(p-8)}\) = \(\frac{6(p-5)(p+5)}{(p-8)(p+8)}\) = \(\frac{6(p-5)}{p-8}.\)

x³ - 343 = x³ - (7)³ = (x - 7)(x² + 7x + 49)

x² - 9x + 14 = x² - 2x -7x + 14

= x(x - 2) - 7(x - 2) = (x - 7)(x - 2)

HCF of the polynomials = (x – 7)

∴ Reqd. LCM = \(\frac{(x^3 -343)\times(x^2-9x+14)}{HCF}\) 

= \(\frac{(x-7)(x^2+7x+49)\times(x-7)(x-2)}{(x-7)}\)

= (x - 7)(x - 2) x (x² + 7x + 49)

3y + 12 = 3 (y + 4)

y² – 16 = (y +4) (y – 4)

y⁴ - 64y = y(y³ - 64) - y(y - 4)(y² + 4y + 16)

∴ LCM of given polynomials = 3y ( y – 4) ( y + 4) (y² + 4y + 16)

Correct option is (A) x + 1

x = 1 & x = -3 are zeros of polynomial \(x^3 + 3x^2 – x – 3 .\)

Let the other zero be x.

\(\therefore\) Sum of zeros \(=\frac{\text{-coefficient of }x^2}{\text{coefficient of }x^3}\) \(=\frac{-3}1\) = -3

\(\therefore\) 1 + (-3) + \(\alpha\) = -3

\(\Rightarrow\) \(\alpha\) = -1

\(\therefore\) x = -1 is another zero of polynomial \(x^3 + 3x^2 – x – 3 .\)

\(\Rightarrow\) (x+1) is a other factor of polynomial \((x^3+3x^2-x-3).\)

Correct option is A) x + 1

Correct option is (D) \(\frac{x^2-3x+2}{x-3}\)

Quadratic polynomial whose zeros are 1 and 2 is

(x - 1) (x - 2) \(=x^2-3x+2\)

Monomial with zero 3 is (x-3).

\(\therefore\) Required rational expression \(=\frac{x^2-3x+2}{x-3}\)

Correct option is D) \(\frac{x^2-3x+2}{x-3}\)

Correct option is (C) (9x² + 5) (9x² – 5)

\(81x^4-25=(9x^2)^2-5^2\)

= \((9x^2+5)\) \((9x^2-5)\)

C) (9x² + 5) (9x² – 5) 

Correct option is (A) 2x⁵ – x⁴ – 15x³ + 25x² – 23x + 4

\((x^3 + 2x^2 – 3x + 4)\) \((2x^2 – 5x + 1)\)

\(=2x^5+4x^4-6x^3+8x^2-5x^4-10x^3\) \(+15x^2-20x+x^3+2x^2-3x+4\)

\(=2x^5-x^4-15x^3+25x^2-23x+4\)

Correct option is A) 2x⁵ – x⁴ – 15x³ + 25x² – 23x + 4

Correct option is (A) x² + x – 6

Additive inverse of \(\frac{x^2-9}{2+x}\) is \(\frac{9-x^2}{2+x}.\)

\(\therefore\) Product \(=(\frac{x^2-4}{3-x})(\frac{9-x^2}{2+x})\)

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

= (x-2) (x+3)

\(=x^2-2x+3x-6\)

\(=x^2+x-6\)

Correct option is A) x² + x – 6

HCF = x,

LCM = x³ - 9x = x(x² - 9) = x(x + 3)(x - 3)

Given expression = x² + 3x = x(x + 3)

∴ Other expression = \(\frac{\text{HCF X LCM}}{\text{Given expression}}\)

= \(\frac{x\times{x}\times(x+3)\times(x-3)}{x\times(x+3)}\)

= x(x - 3) = x² - 3x

Correct option is (C) \(-3x + 4 - \frac{x}{2x-1}\)

The additive inverse of 3x - 4 + \(\frac{x}{2x-1}\) is \(-3x + 4 - \frac{x}{2x-1}.\)

\((\because\) Additive inverse of p(x) is -p(x))

Correct option is C) -3x + 4 - x/2x - 1

  Correct option is D) \(\cfrac{4x^2+5x+28}{x^3+4x^2-16x-64}\)

Correct option is (D) 9x² + 4x – 4

\(\frac{81x^4 -16x^2 +32x-16}{9x^2-4x+4}\) \(=\frac{81x^4-(16x^2-32x+16)}{9x^2-4x+4}\)

\(=\frac{(9x^2)^2-(4x-4)^2}{9x^2-4x+4}\)

\(=\frac{(9x^2-(4x-4))(9x^2+4x-4)}{9x^2-4x+4}\)

\(=\frac{(9x^2+4x-4)(9x^2-4x+4)}{9x^2-4x+4}\)

\(=9x^2+4x-4\)

Correct option is D) 9x² + 4x – 4

The least common denominator (LCD) of the given expression is 30 xyz.

\(\frac{a}{3xy}+\frac{2b}{6yz}+\frac{3c}{15xz}\) = \(\frac{10az + 10bx+6cy}{30xyz}\)

= \(\frac{5az+5bx+3cy}{15xyz}.\)

Correct option is (C) \(\frac{x^2-5x+6}{x-3}\)

(A) \(3x^2-2\sqrt{5x}+7\) is not a polynomial because it consists term \(\sqrt x.\)

(B) \(\frac{x^3-2x+1}{2x+5}\) \(=\frac{(x-1)(x^2+x-1)}{2x+5}\) which is a rational expression not a polynomial.

(C) \(\frac{x^2-5x+6}{x-3}\) \(=\frac{(x-3)(x-2)}{x-3}\) = x - 2 which is a polynomial of degree 1.

(D) \(\frac{x^2-2x+1}{2x+7}\) \(=\frac{(x-1)^2}{2x+7}\) which is a rational expression not a polynomial.

Correct option is C) \(\frac{x^2-5x+6}{x-3}\)

i. Number of coefficients = 3 

∴ Degree = 3 – 1 = 2 

∴ Taking x as variable, the index form is x² + 2x + 3 

ii. Number of coefficients = 5 

∴ Degree = 5 – 1=4 

∴ Taking x as variable, the index form is 5x⁴ + 0x³+ 0x² + 0x – 1

iii. Number of coefficients = 4 

∴Degree = 4 – 1 = 3 

∴Taking x as variable, the index form is -2x³ + 2x²– 2x + 2

Monomial: x⁵ 

Binomial: x⁵ + x 

Trinomial: 2x⁵ – x² + 5

Correct option is (D) \(\frac{4x}{1+4x^2}\)

We have P = \(\frac{1+2x}{1-2x}\) and Q = \(\frac{1-2x}{1+2x}\)

\(\therefore\) \(\frac{P-Q}{P+Q}\) \(=\cfrac{\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}}{\frac{1+2x}{1-2x}+\frac{1-2x}{1+2x}}\) \(=\cfrac{\frac{(1+2x)^2-(1-2x)^2}{(1-2x)(1+2x)}}{\frac{(1+2x)^2+(1-2x)^2}{(1-2x)(1+2x)}}\)

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

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

Correct option is A) \(-\frac{4x}{1+4x^2}\)

i. Quadratic polynomial: x² ; 2x² + 5x + 10; 3x²+ 5x 

ii. Cubic polynomial: x³ + x² + x + 5; x³ + 9 

iii. Linear polynomial: x + 7 

iv. Binomial: x + 7; x³ + 9; 3x² + 5x

v. Trinomial: 2x² + 5x + 10 

vi. Monomial: x²

Correct option is (C) -b

\(\because\) \(\frac{1}{2}\) is a zero of expression (ax+b) (3x+2).

\(\therefore\) \((a\times\frac{1}{2}+b)\,(3\times\frac{1}{2}+2)=0\)

\(\Rightarrow\) \(\frac{a+2b}2\times\frac72=0\)

\(\Rightarrow\) a+2b = 0

\(\Rightarrow\) a+b+b = 0

\(\Rightarrow\) a+b = -b

Correct option is C) -b

(i) f(x) = 5x – 4x² + 3 x = 0 then, 

f(0) = 5(0) – 4(0)² + 3 

=0 – 0 + 3 

f(0) = 3 

(ii) f(x) = 5x – 4x² + 3 

x = -1 then, f(-1) = 5(-1) – 4(-1)² + 3 

= -5 – 4(+1) + 3 = -5 – 4 + 3 

= -9 + 3 f(-1) = -6 

(iii) f(x) = 5x – 4x² + 3 

x = 2 then, 

f(2) = 5(2) – 4(2)² + 3 

= 5(2) – 4(4) + 3 

= 10 – 16 + 3 

= 13 – 16 

f(2) = -3

Correct option is (C) f(4) = -152.52, f(-5) = 659.28, f(3.2) = -63.304

\(\because f(x)=6.2x^2-4x^3+4.28\)

\(f(4)=6.2\times4^2-4\times4^3+4.28\)

= 99.2 - 256 + 4.28

= 103.48 - 256.00

= -152.52

\(f(-5)=6.2\times(-5)^2-4(-5)^3+4.28\)

= 155 + 500 + 4.28

= 659.28

\(f(3.2)=6.2\times(3.2)^2-4(3.2)^3+4.28\)

= 63.488 - 131.072 + 4.28

= - 63.304

C) f(4) = -152.52, f(-5) = 659.28, f(3.2) = -63.304