The sum of first n odd number is Sn = .....

The nth mean is given by  mn = a + .....

The nth term of an AS is given by tn = a + (n - .....

i) Statement of factor theorem: "If a polynomial p(x) is divided by (x – a) and f(a) = R = 0 then (x – a) is a factor of p(x). ii) Here, p(x) = x3 – 8x + 3 and d(x) = x + 3 Comparing (x + 3) with (x – a) then, a = .....

Here,  p(x) = x3 + px2 + 4x + 5 and a factor = (x – 5) Comparing x - 5 with x - a, we get x = 5 . Now, Using factor theorem,        p(5) =  0 or,  53 + p × 52 + 4 × 5 + 5=0 or,  125 + .....

Here, p(x) = x3 + 6x2 + kx + 10,  d(x) = x + 2, R = 4  Using remainder theorem,      p(-2) =4 or, (-2)3 + 6 × (-2)2+ k (-2) + 10 = 4. or, 2k = 22  or, 2k = .....

Here,  y3 – 6y2 + 11y – 6 = 0      The factors of 6 are: ±1, ±2, ±3 and ±6 Now, testing the factors by using the synthetic division method: For, y = 1 using synthetic .....

Here,         2x3 + 6 - 3x2 – 11x = 0      Or,  2x3 - 3x2 – 11x + 6 = 0. The factors of 6 are: ±1, ±2, ±3 and ±6 At x = -2 using synthetic division,  R = .....

Here,  x3 - 3x2 – 10x + 24 = 0     The factors of 24 are: ±1, ±2, ±3, ±6, ±8, ±12, ±24 For, x = 2 Since R = 0, so (x - 2) is a factor & Quotient Q(x) = x2 – x .....

Here,  Let, P(x) = 2x3 + 3x2 – 11x - 6 = 0  The factors of 6 are: ±1, ±2, ±3 and ±6. When, x = 2, p(2) = 2 × 23 + 3 × 22 – 11 × 2 – 6         = 16 + .....

Here,  x3 = 7x2 - 36         or, x3 - 7x2 + 36 = 0. The factors of 36 are: ±1, ±2, ±3, ±4, ±9, ±12, ±18, ±36 Using synthetic division taking x = .....

Here,  x3 -21x – 20 = 0 = P(x). Factors of 20 are ±1, ±2, ±4, ±5, ±10, ±20 For, x = -1, Since R = 0 so (x + 1) is a factor, And Quotient Q(x) = x2 – x – 20. Now, Factor × .....

Here,  x3 - 4x2 + x + 6 = 0  The factors of 6 are: ±1, ±2, ±3 and ±6. For, x = -1. R = 0, so (x + 1) is a factor. So, Quotient = x2  - 5x + 6  We have, P(x) = Factor × Quotient or,0 = (x + 1) .....

Here, p(x) = 2x3 – x2 + 10x – k and a factor = (x – 3). Compare x - 3 with  x - a, we get  a = 3 . Now, Usingfactor  theorem.     p(3) = 0 or, 2  × 33 – 32 + 10  × 3 .....

Here, x3 – 19x – 30 = (x + 2). Q(x)  Comparing (x + 2) with (x – a), then a = -2.  Using synthetic division,  ∴ Q(x) = x2 – 2x – .....

Here, p(x) = 8x3 – 4x + 2x – 5 and d(x) = 2x – 1= 2( x - 1/2) Comparing x - 1/2 with x- a, we get, a = 1/2. Now,Using remainder theorem,       R = P ( 1/2)= 8 × (12 )3 – 4 × (12 )2 + 4 .....

Here, p(x) = x3 – 19x – p and x + 2 is factor of p(x), Comparing x + 2 with x - a, we get  a  = - 2. By factor theorem, So, p(-2) = 0  or, (-2)3 – 19 (-2) – p = 0 or, -8 + 38 – p = 0 ∴  .....

Here,  Let, p(x) = 6x3 + x2 – 19x + 6 = 0     The factors of 6 are: ±1, ±2, ±3 and ±6. Now, testing the factors by using the synthetic division method: At x = -2 using synthetic .....

Here,  6x3 = 4 - 13x2         6x3 + 13x2 – 4 = 0 The factors of 4 are: ±1, ±2, ±4. Using synthetic division taking x = -2.  R = 0, shows that (x + 2) is a factor and 6x2  - x - .....

Here, x3 – 19x – 30 = (x + 2). Q(x)  Comparing (x - a) with (x + 1), then a = -1.  Using synthetic division,  ∴ Q(x) = x2 – x – .....

Here,        p(x) = 2x3 – 4x2 + kx + 10,         d(x) = x + 2 and R = 4 . Comparing x + 2 with x - a , we get         a = -2  Now, Using remainder theorem,     .....

Here, p(x) = 2x3 – 7px + (p – 12) and factor = x – 5. Comparing x - 5 with x - a, we get, a = 5. Now,By using factor theorem     p(5) = 0 or,  2 × 53 – 7p × 5 + p – 12 = 0 or, 250 .....

Here,      (x + 1)(x + 3)(x + 5)(x + 7) + 16  = [(x + 1)(x + 7][(x + 5)(x + 3)] + 16 = [ x2 + 8x + 7] [ x2 + 8x + 15] + 16 = [ x2 + 8x + 7] [ x2 + 8x + 7 + 8] + 16 Let   y = x2 + 8x + .....

Here, x + 1 is the factor of p(x) = 6x3 + ax2 - 10x - 3 So, f(-1) = 0 or, 6(-1)3 + a(-1)2 - 10(-1) - 3 = 0 or, -6 + a + 10 - 3 = 0 or, a = -1 Thus,  p(x) = 6x3 + ax2 - 10x - 3        = 6x3 -  x2 - 10x - .....

Here,        3x3 = 7x2 – 4 or, 3x3 – 7x2 + 0.x + 4 = 0  Factors of 4 are ±1, ±2, ±4. At x = 1, then   = R. R = 0 shows that (x – 1) is a factor.  ∴ Q(x) = .....

Here, x - a is the factor of p(x) = x3 - ax2 - 2x + a + 4, So, p(a) = 0 or,  a3 - a.a2 - 2a + a + 4 = 0, or, a3 - a3 - a + 4 = 0 or, a = 4. Thus, x - a = x - 4 is the factor of p(x) = x3 - ax2 - 2x + a + 4 = .....

Here, When, f(x) =  x3 + 4x2 - 2x + 1 is divided by x - a, remainder = f(a) i.e;  Remainder = f(a) =  a3 + 4a2 - 2a + 1. When, g(x) =  x3 + 3x2 - x + 7 is divided by x - a, remainder = g(a) i.e;  Remainder = g(a) =  .....

Here, Let f(x) =  x3 + ax2 + bx + 6. x - 2 is factor of f(x). So,      f(2) = 0 or,  23 + a.22 + b.2 + 6 = 0 or, 4a + 2b = - 14 or, b = - 7 - 2a................(i) Again, when f(x) is divided by x - 3, remainder is .....

Let f(x) = x3 + ax2 + bx - 6, If x - 1 is the factor of f(x),      f(1) = 0 or, 13 + a(1)2 + b.1 - 6 = 0, or, a + b = 5 or, b = 5 - a....................(i) Again,  If x - 2 is the factor of f(x),    .....

We have given:        x = 2 - y  ⟹ y = 2 - x .......................(i) And y = 2x3 + x2 - 3x + 1 or, 2 - x = 2x3 + x2 - 3x + 1        [ Since, from (i).] or, .....

Here, f(x) = 4x3 - 8x2 and g(x) = 3x3 - 2x2 - 11x + 6, Now, f(x) - g(x) = 0. or, 4x3 - 8x2 - 3x3 + 2x2 + 11x - 6 = 0, or, x3 - 6x2 + 11x - 6 = 0, Let, h(x) = x3 - 6x2 + .....

We have given:        y = x + 4........................(i) And y = x3 - 3x2 - 9x + 28 or, x + 4 = x3 - 3x2 - 9x + 28        [ Since, from (i).] or, x3 - 3x2 - 10x + .....

Polynomial = x2(x - 2) = x3 - .....

Remainder = .....

Another polynomial is (x + .....

Another polynomial is (x - .....

Degree of polynomial f(x)/g(x) = 6n - 2n = .....

Degree of f(x).g(x) = 6n + 2n = .....

Remainder = .....

Remainder = .....

Here,  Remainder = f(2) = 22 - 4 = 0 Thus, the remainder = .....

Polynomial = p(x) = d(x).q(x) + .....

Here, f(1) = 13 - 1 = 0 So, x - 1 is factor of .....

Here, f (x)=2x+5 and  g(x) =3x-1 So,    gof (x) = g(f(x))                = g(2x+5)                =3(2x+5) -1             .....

Here,  f (x)=x+1 and  g(x) =2x+1. So,       gof(x) = g (f (x))                  = g(x+1)                  =2(x+1)+1   .....

Here,       f (x)=2x-3  and  g(x) =x2 +1. We have, fog (x) (3)  = f(g (3))                                      = .....

Here,       f-1 (x) = 2x-3. Let       y = f-1 (x) or,  y= 2x-3. Interchanging the Value of x and y      x= 2y-3 or, x+3 = 2y  or, y = x+32 Thus,f(x)= x+32 .....

Here,      h (x)=2x+3 and  g(x) =3x-2. We know that,                         hog (5) = h(g(5))                     .....

Here,      f(x) = 2x + 7 and     fog(x) = 4x + 3. So,       fog(x) = 4x + 3 or, f(g(x)) = 4x + 3 or, 2.g(x) - 7 = 4x + 3  or, 2g(x) = 4x + 10  ∴   g(x) = 2x + 5 We .....

Here, f = {x, 5x – 13}, g = {x, 2x+73}   and g-1(x) = f of(x) Let y = g(x)                            or, y = 2x+73 Interchanging the position of x and y, .....