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i) The value of p(x) at x = 1 is 3(1)² + 5(1) — 7 = 1.

ii) The value of q(z) at z = 2 is 

5(2)³ – 4(2) + √2 

= 40 – 8 + √2 

= 32 + √2

The value of r(t) at t = p is 4p⁴ + 3p³ – p² + 6

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as 

x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1) 

Let a = \(\frac{1}{2}\), b = 1 and c = –3 

Substituting the values in (1), we get

x³ – \((\frac{1}3+1-3)\) x² + \((\frac{1}3-3-\frac{3}2)\)x - \((\frac{-3}2)\)

⇒ x³ – \((\frac{-3}2)\)x² - 4x = \(\frac{3}2\)

⇒ 2x³ + 3x² – 8x + 3

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as 

x³ – (a + b + c)x² + (ab + bc + ca)x – abc ……(1) 

Let a = 2, b = –3 and c = 4 

Substituting the values in 1, we get 

x³ – (2 – 3 + 4)x² + (– 6 – 12 + 8)x – (–24) 

⇒ x³ – 3x² – 10x + 24

 (b) \(\frac{4}{3},\frac{-3}2\) 

Let f(x) = x² + \(\frac{1}6\) x – 2 = 0 

⇒ 6x² + x – 12 = 0 

⇒ 6x² + 9x – 8x – 12 = 0 

⇒ 3x (2x + 3) –4 (2x + 3) = 0 

⇒ (2x + 3) (3x – 4) = 0 

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

p(x) = (3x³ – 10x² – 27x + 10) 

p(5) = (3 × 5³ – 10 × 5² – 27 × 5 + 10) = (375 – 250 – 135 + 10) = 0 

p(–2) = [3 × (–2³) – 10 × (–2²) – 27 × (–2) + 10] = (–24 – 40 + 54 + 10) = 0

p \((\frac{1}3)\) = \(\{\) 3 x \((\frac{1}3)^3\) - 10 x \((\frac{1}3)^2\) - 27 x \(\frac{1}3\) + 10 \(\}\) = ( 3 x \(\frac{1}{27}\) - 10 x \(\frac{1}9\) - 9 + 10)

= \((\frac{1}9-\frac{10}9+1)\) = \((\frac{1-10-9}9)\) = \((\frac{0}9)\) = 0

∴ 5, –2 and \(\frac{1}3\) are the zeroes of p(x).

Let α = 5, β = –2 and γ =  \(\frac{1}3\). Then we have:

(α +β + γ) = (5 − 2 + \(\frac{1}3\)) = \(\frac{10}3\) = \(\frac{-(coefficient\,of\,x^2)}{(coefficient\,of\,x^3)}\)

(αβ +βγ + γα) = (– 10 – \(\frac{2}3\) + \(\frac{5}3\)) = \(\frac{-27}3\) = \(\frac{coefficient\,of\,x}{coefficient\,of\,x^3}\)

αβγ = \(\{\) 5 x (-2) x \(\frac{1}3\)\(\}\) = \(\frac{-10}3\) = \(\frac{-(constant\,term)}{(coefficient\,of\,x^3)}\)

 (a) \(\frac{2}{3}\), \(\frac{-1}{7}\)

Let f(x) = 7x² – \(\frac{11}3\) x – \(\frac{2}3\) = 0 

⇒21x² – 11x – 2 = 0 

⇒21x² – 14x + 3x – 2 = 0 

⇒7x (3x – 2) + 1(3x – 2) = 0 

⇒ (3x – 2) (7x + 1) = 0

⇒ x = \(\frac{2}3\) or x = \(\frac{-1}7\)

(c) x² – 3x – 10 

Given: 

Sum of zeroes, α + β = 3 

Also, product of zeroes, αβ = –10 

∴ Required polynomial = x² – (α + β) + αβ = x² – 3x – 10

(c) x² – 2x – 15 

Here, 

the zeroes are 5 and –3. 

Let α = 5 and β = –3 

So, 

sum of the zeroes, α + β = 5 + (–3) = 2 

Also, product of the zeroes, αβ = 5 × (–3) = –15 

The polynomial will be x² – (α + β) x + αβ

∴ The required polynomial is x² – 2x – 15.

4y² – 4y + 1 

= (2y)² – 2 (2y) (1) + (1)² 

[∵ (x -y)² = x² – 2xy + y²] 

= (2y -1)² = (2y – 1) (2y-1)

f(x) = 2x³ - 13x² + 17x + 12 

 f(-2) = 2(-2)³ - 13(-2)² + 17 (-2) + 12 

 = - 16 - 52 - 34 + 12 = - 90 

f(3) = 2(3)³ - 13(3)² + 17(3) + 12 

 = 54 - 117 + 51 + 12 = 0

-8 + 9(a - b)⁶ - (a - b)¹² 

 Let (a - b)⁶ = x 

Then -8 + 9x - x² = - (x² - 9x + 8) = - (x² - 8x - x + 8) 

= - (x - 8)(x - 1) 

 = - [(a - b)⁶ - 8][(a - b)⁶ - 1] 

 = [1 - (a - b)⁶ ][(a - b)⁶ - 8] 

 = [(1)³ - {(a - b)² } ³ ][{(a - b)² } ³ - (2)³ ] 

 = [1 - (a - b)² ][1 + (a - b)⁴ + (a - b)² ][(a - b)² - 2][(a - b)⁴ + 4 + 2(a - b)²]

6x² + x[1 - 5y] - [4y² - 17y + 15] 

= 6x² + x[1 - 5y] - [4y² - 17y + 15] 

= 6x² + x[1 - 5y] - [4y(y - 3) -5(y - 3)] 

= 6x² + x[1 - 5y] - (4y - 5)(y - 3) 

= 6x² + 3(y - 3)x - 2(4y - 5)x - (4y -)(y - 3) 

= 3x[2x + y 3] - (4y - 5)(2x + y - 3) 

= (2x + y - 3)(3x - 4y + 5)

Correct option is (A) –1

(x+1) & (x+2) are two factors of \(p(x)=x^3+3x^2-2\alpha x+\beta\)

\(\therefore\) x = -1 & x = -2 are two roots of equation p(x) = 0

\(\therefore\) \((-1)^3+3(-1)^2-2\alpha(-1)+\beta=0\)

and \((-2)^3+3(-2)^2-2\alpha(-2)+\beta=0\)

\(\Rightarrow\) \(-1+3+2\alpha+\beta=0\) and \(-8+12+4\alpha+\beta=0\)

\(\Rightarrow\) \(2\alpha+\beta=-2\) and \(4\alpha+\beta=-4\)

\(\Rightarrow\) \((4\alpha+\beta)-(2\alpha+\beta)\) = -4 - (-2) = -2

\(\Rightarrow\) \(2\alpha=-2\)

\(\Rightarrow\) \(\alpha=\frac{-2}2=-1\)

\(\therefore\) \(\beta=-2-2\alpha=-2-2\times-1\) = -2+2 = 0

\(\therefore\) \(\alpha+\beta\) = -1+0 = -1

Correct option is A) – 1

(i) A Bionomial of degree 35 

E.g. f(x) = – x³⁵ + 10 

(ii) A binomial of degree 100 

E.g. f(y) = – y¹⁰⁰ .

Degree of a polynomial is the highest power of the variable in the polynomial. Binomial has two terms in it. Therefore, binomial of degree 35 can be written as x¹⁵ + x¹⁴

Monomial has only one term in it. Therefore, monomial of degree 100 can be written as x¹⁰⁰.

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively.

(i) x² +x is a quadratic polynomial as its degree is 2.

(ii) x - x³ is a cubic polynomial as its degree is 3.

(iii) y + y² +4 is a quadratic polynomial as its degree is 2.

(iv) 1 + x is a linear polynomial as its degree is 1.

(v) 3t is a linear polynomial as its degree is 1.

(vi) r² is a quadratic polynomial as its degree is 2.

(vii) 7x³ is a cubic polynomial as its degree is 3.

(i) p(x) = 5x - 4x² + 3

p(0) = 5(o) +4(0)² + 3

= 3

(ii) p(x) = 5x - 4x² + 3

p(-1) = 5(-1) -4(-1) + 3

= -5 - 4(1) +3 = -6

(iii) p(x) = 5x - 4x² +3

p(2) = 5(2) - 4(2)2 + 3

= 10 - 16 + 3 = -3

If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

p(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1

= 1 − 3 + 3 − 1 + 1 = 1 As p(−1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

(i) p(x) = x³ p(0) = (0)³ = 0

p(1) = (1)³ = 1

p(2) = (2)³ = 8

(iv) p(x) = (x − 1) (x + 1)

p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1

p(1) = (1 − 1) (1 + 1) = 0 (2) = 0

p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3

(i) If (x + 1) is a factor of p(x) = x³ + x² + x + 1, then p (−1) must be zero,

otherwise (x + 1) is not a factor of p(x).

p(x) = x³ + x² + x + 1

p(−1) = (−1)³ + (−1)² + (−1) + 1

= − 1 + 1 − 1 − 1 = 0

(ii) If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

p(x) = x⁴ + x³ + x² + x + 1

p(−1) = (−1)⁴ + (−1)³ + (−1)² + (−1) + 1

= 1 − 1 + 1 −1 + 1 = 1 As p(− 1) ≠ 0,

Therefore, x + 1 is not a factor of this polynomial.

If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must be 0.

p(x) = x³ +3x² + 3x + 1

p(−2) = (−2)³ + 3(−2)² + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As p(−2) ≠ 0,

Hence, g(x) = x + 2 is not a factor of the given polynomial.

3x² − x − 4

We can find two numbers such that pq = 3 × (− 4) = −12 and p + q = −1.

3x² − x − 4 = 3x² − 4x + 3x − 4

= x (3x − 4) + 1 (3x − 4)

= (3x − 4) (x + 1)

6x² + 5x − 6

We can find two numbers such that pq = −36 and p + q = 5. They are p = 9 and q = −4.

6x² + 5x − 6 = 6x² + 9x − 4x − 6

= 3x (2x + 3) − 2 (2x + 3)

= (2x + 3) (3x − 2)

Correct option is (A) a = 3, b = -3

Let \(p(x)=2x^3 + ax^2 + bx – 2\)

\(\because\) p(x) leaves a remainder 7 when divided by (2x – 3).

\(\therefore\) \(p(\frac32)=7\)

\(\Rightarrow\) \(2(\frac32)^3+a(\frac32)^2+b\times(\frac32)–2=7\)

\(\Rightarrow\) \(\frac{27}4+\frac{9a}4+\frac{3b}2-2=7\)

\(\Rightarrow\) 27+9a+6b-8 = 28

\(\Rightarrow\) 9a+6b = 9

\(\Rightarrow\) 3a+2b = 3   __________(1)

Also, p(x) leaves remainder 0 when divided by (x+2).

\(\therefore\) p(-2) = 0

\(\Rightarrow2(-2)^3+a(-2)^2+b\times-2-2=0\)

\(\Rightarrow\) -16 + 4a - 2b - 2 = 0

\(\Rightarrow\) 4a - 2b = 18  __________(2)

Adding (1) & (2), we get

7a = 21 \(\Rightarrow\) \(a=\frac{21}7=3\)

\(\therefore\) 2b = 3 - 3a = 3 - 9 = -6 \(\Rightarrow\) \(b=\frac{-6}2=-3\)

Thus, a = 3, b = -3

Correct option is A) a = 3, b = -3

The correct option is: (D) - 4

Explanation:

Given polynomial is 2x³ - 3x² + 4tx - 5

Sum of product of roots taken two at a time is 4t/2

. ^.. 4t/2 = -8 => = -4

The correct option is: (D) 7 or -7

Explanation:

Given equation is x² + px + 12 = 0

Now, if a and b are its roois, then

sum of roots, 6 + b = -p and

product of roots, a x b = 12

Also,a - b = 1 (Given)

We know that, (a - b)² = (a + b)² - 4ab

=> 1 = p² - 4 x 12 => 1 = p² - 48 => p² = 49 = p = ±7

The correct option is: (B) 1- c

Explanation:

Given equation is x² + k(x- 1) - c

= x² + kx - (k + c)

Since, p and q are the zeroes,

... p + q = -k and pq = -(k+ c)

Now, (p - 1) (q - 1) = Pq - q - p + 1

= pq - (p + q)+ 1 = -(k+ c)-(-k) + 1

= -k - c + k + 1 = 1 - c

Let x and x + 2 are two consecutive odd positive integers.

According to question,

(x)² + (x + 2)² = 290

⇒ x² + x² + 4x + 4 = 290

⇒ 2x² + 4x + 4 = 290

⇒ 2x² + 4x – 286 = 0

⇒ x² + 2x – 143 = 0

⇒ x² + 13x – 11x – 143 = 0

⇒ x(x + 13) – 11(x + 13) = 0

⇒ (x + 13) (x – 11) = 0

⇒ x – 11 = 0

⇒ x = 11

Consecutive odd numbers x = 11

x + 2 = 11 + 2 = 13

Hence, two consecutive odd positive integers are 11, 13

(c) c² – ab

If a + b + c = 0, then a³ + b³ + c³ – 3abc = 0 

⇒ (a + b) (a² – ab + b²) + c³ = 3abc 

⇒ (– c) (a² – ab + b²) + c³ = 3abc           [∵ (a + b) = – c] 

⇒ a² – ab + b² – c² = – 3ab 

⇒ a² – ab + b² + 2ab – c² = – 3ab + 2ab = a² + ab + b² = c² – ab

Correct option is (A) 5

sol: Here, degree of first polynomial = 2 and

degree of second polynomial = 3

∴ Degree of polynomial obtained by multiplication

= 2 + 3 = 5

(c) \(\frac13\)

a + b + c = 0 ⇒ (a + b + c)2 = 0 

⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0 

⇒ a² + b² + c² = – (2ab + 2bc + 2ca)

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

= \(\frac{a^2+b^2+c^2}{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca}\)

= \(\frac{a^2+b^2+c^2}{2(a^2+b^2+c^2)-(2ab+2bc+2ca)}\)

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

= \(\frac{a^2+b^2+c^2}{3(a^2+b^2+c^2)}\) = \(\frac13\)

(c)  – \(x\).

f(x) = x¹⁹ + x¹⁷ + x¹³ + x¹¹ + x⁷ + x⁵ + x³ 

Putting x² = – 1, we get 

f(x) = (x²)⁹.\(x\) + (x²)⁸.\(x\) + (x²)⁶.\(x\) + (x²)⁵.\(x\) + (x²)².\(x\) + x².\(x\) 

= (–1)⁹\(x\) + (–1)⁸.\(x\) + (–1)⁶.^x + (–1)⁵.\(x\) + (–1)².\(x\) + (–1).\(x\) 

= – \(x\) + \(x\) + \(x\) – \(x\) + \(x\) – \(x\) = – \(x\).

Let f(x) = x³ – 6x² – 19x + 84 

To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem. 

Let f(- 4), f(3) and f(7) 

f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84 

= -64 – 96 + 76 + 84 = 0 . 

∴ (x + 4) is a factor of f(x). 

f(3) = 3³ – 6(3)² – 19(3) + 84 

= 27 – 54 – 57 + 84

= 0 

∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84 

= 343 – 294 – 133 + 84 

= 427 – 427 = 0 

∴ (x – 7) is a factor of f(x).

(a) \(\frac{2abc}{ac+bc-ab}\)

Given, c = \(\frac{yz}{y+z}\) ⇒ cy + cz = yz ⇒ yz - cz = cy ⇒ z (y - c) = cy 

⇒ z = \(\frac{cy}{y-c}\)

Also b = \(\frac{xz}{x+z}\) ⇒ z = \(\frac{bx}{x-b}\)

∴ \(\frac{cy}{y-c}\) = \(\frac{bx}{x-b}\) ⇒ cyx - cyb = bxy - bxc

⇒ cyx – cyb – bxy = – bxc 

⇒ – y(bx + bc – cx) = – bxc

⇒ y = \(\frac{bxc}{bx+bc-cx}\)

Now, a = \(\frac{yz}{y+z}\) ⇒ y = \(\frac{ax}{x-a}\)

∴ \(\frac{bxc}{bx+bc-cx}\) = \(\frac{ax}{x-a}\)

⇒ abx² + abcx – acx² = bx²c – abcx 

⇒ 2abcx = x² (bc + ac – ab) ⇒ x = \(\frac{2abc}{(ac+bc-ab)}\)

Correct option is (A)  x + 5

(c) 0

Let a^1/3 = x, b^1/3 = y, c^1/3 = z. Then, 

a + b + c + 3a^1/3 b^2/3 + 3a^2/3 b^1/3 = x³ + y³ + z³ + 3xy² + 3x²y 

and a^1/3 + b^1/3 + c^1/3 = x + y + z.

Now x³ + y³ + z³ + 3xy² + 3x²y = x³ + 3xy² + 3x²y y³ + z³

= (x + y)³ + z³ = [x + y + z] [(x + y)² - (x + y)z + z²].

∴ Given expression is completely divisible by (x + y + z), 

i.e., by a^1/3 + b^1/3 + c^1/3.

i. 7x⁴ – x³ + 4x² – x + 9

 ii. 5p⁴ + 2p³ + 10p² + p – 8

(a) (x⁴ + 1 – x²) (x² + 1 + x) (x² + 1 – x)

x⁸ + x⁴ + 1 = x⁸ + 2x⁴ + 1 – x⁴ (Adding and subtracting x⁴)

= (x⁴ + 1)² – (x²)² = (x⁴ + 1 + x²) (x⁴ + 1 – x²) 

= [(x⁴ + 2x² + 1) – x²] (x⁴ + 1 – x²) 

= [(x² + 1)² – (x)²] (x⁴ + 1 – x²) 

= (x² + 1 + x) (x² + 1 – x) (x⁴ + 1 – x²)

(c) 10

f(x) = x⁶ + px⁵ + qx⁴ – x² – x – 3 

= x⁴ . x² + p.x⁴ x + q.x⁴ – x² – x – 3 

As (x⁴ – 1) is a factor of f(x), so putting x⁴ = 1, we get 

x² + px + q – x² – x – 3 = 0 

⇒ (p – 1)x + (q – 3) = 0 ⇒ p – 1 = 0 and q – 3 = 0 

⇒ p = 1 and q = 3.

∴ p² + q² = 1 + 9 = 10.

(b) (x³ + y³ + z³) only

Hint. x⁴ + xy³ + x³y + xz³ + y⁴ + yz³ 

= (x⁴ + xy³ + xz³) + (x³y + y⁴ + yz³) 

= x(x³ + y³ + z³) + y(x³ + y³ + z³) 

= (x + y) (x³ + y³ + z³)

(b) 5a² – 9a – 1 = 0

Let f (x) = x⁴ + 9x³ + 7x² + 9ax + 5a². 

If (x + 1) is a factor of f (x), then 

f(–1) = 0 ⇒ (–1)⁴ + 9 (–1)³ + 7 (–1)² + 9a (–1) + 5a² = 0 

⇒ 1 – 9 + 7 – 9a + 5a² = 0 ⇒ 5a² – 9a – 1 = 0.