Math IOQM

Q1. Given that f ( x ) = x 4 + 3 x 3 + 8 x 2 − k x + 11 f(x) = x^4 + 3x^3 + 8x^2 - kx + 11 is divisible by x + 3 x + 3 , find the value of k k . Solution: Put f(-3) = 0. Solving will give you k = -83/3. Q2.  Given that f ( x ) = x 4 − a x 2 − b x + 2 f(x) = x^4 - ax^2 - bx + 2 is divisible by ( x + 1 ) ( x + 2 ) (x + 1)(x + 2) , find the values of a a and b b . Solution: f(-1) = 0 and f(-2) = 0 Plug those values and solve. You would get: a = 6, b = 3 Q3. Given that a polynomial f ( x ) f(x) has remainders 1, 2, 3 when divided by ( x − 1 ) , ( x − 2 ) , ( x − 3 ) (x - 1), (x - 2), (x - 3) , respectively. Find the remainder of f ( x ) f(x) when it is divided by ( x − 1 ) ( x − 2 ) ( x − 3 ) (x - 1)(x - 2)(x - 3) . Solution: By the remainder theorem: f(1) = 1, f(2) = 2, f(3) = 3 Now, when we divide f(x) by (x-1)(x-2)(x-3) remainder's degree will be 1 less than divisor. => Remainder = ax^2 + bx + c f(x) = Q(x).(x-1).(x-2).(x-3) + ax^2 + bx + c f(1) = 1 = a + b + c f(2) = ...

Q1. x 2 + ( 2 a − 1 ) x + a 2 = 0 x^2 + (2a - 1)x + a^2 = 0 α , β > 0 \alpha, \beta > 0 α \alpha & β \beta are real roots of above quadratic & a a is integer, Find value of ∣ α − β ∣ |\sqrt{\alpha} - \sqrt{\beta}| . Solution: S 1 = α + β = − ( 2 a − 1 ) 1 S_1 = \alpha + \beta = -\frac{(2a - 1)}{1} α + β = 1 − 2 a \alpha + \beta = 1 - 2a S 2 = α β = a 2 1 S_2 = \alpha \beta = \frac{a^2}{1} α β = a 2 \alpha \beta = a^2 ( α − β ) 2 = ( α ) 2 + ( β ) 2 − 2 α β (\sqrt{\alpha} - \sqrt{\beta})^2 = (\sqrt{\alpha})^2 + (\sqrt{\beta})^2 - 2\sqrt{\alpha\beta} = α + β − 2 α β = \alpha + \beta - 2\sqrt{\alpha\beta} = 1 − 2 a − 2 a 2 = 1 - 2a - 2\sqrt{a^2} = 1 − 2 a − 2 ∣ a ∣ = 1 - 2a - 2|a| D ≥ 0 D \geq 0 a x 2 + b x + c = 0 ax^2 + bx + c = 0 1 x 2 + ( 2 a − 1 ) x + a 2 = 0 1x^2 + (2a - 1)x + a^2 = 0 b 2 − 4 a c ≥ 0 b^2 - 4ac \geq 0 ( 2 a − 1 ) 2 − 4 ( 1 ) ( a 2 ) ≥ 0 (2a - 1)^2 - 4(1)(a^2) \geq 0 4 a 2 − 4 a + 1 − 4 a 2 ≥ 0 4a^2 - 4a + 1 - 4a^2 \geq 0 1 − 4 a ≥ ...

1. If f(x), g(x) are polynomials with degree m,n each: A. Degree of f(x) + g(x) = max(m,n) B. Degree of f(x).g(x) = m + n C. Degree of f(g(x)) = m.n 2. Leading coefficient = coefficient of degree defining term 3. Monic polynomial => leading coefficient = 1 4. Dividend = Divisor * Quotient + Remainder f(x) = h(x).Q(x) + R(x) where h(x) is divisor of f(x) leaving quotient as Q(x) and remainder as R(x). f(x) has degree m. h(x) has degree n. then: degree of Q(x) = m - n. degree of R(x) < n 5. Factor theorem: A polynomial f(x) has factor (x-a) if f(a) = 0. f(x) = (x-a).Q(x) + r If r = 0 then (x-a) is a factor of f(x). f(a) = 0.Q(a) + 0 f(a) = 0 6. Remainder theorem: Remainder when f(x) is divided by (x-a) is f(a). Proof: f(x) = (x-a).Q(x) + R(x) But R(x) needs to have degree 0 since divisor is of degree 1. So R(x) = r = constant. f(a) = (a-a).Q(a) + r = 0 + r = r => f(a) = r. Hence proved. Ex1:  f(x) = 3x^2 - 2x + 1 Divide f(x) by x-2 and find the remainder. Solution: By remain...

Common identities: Common Algebraic Identities : (i) ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 (ii) ( a − b ) 2 = a 2 − 2 a b + b 2 (a - b)^2 = a^2 - 2ab + b^2 (iii) a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a - b)(a + b) (iv) ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 a b + 2 b c + 2 c a (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca (v) ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 (vi) ( a − b ) 3 = a 3 − 3 a 2 b + 3 a b 2 − b 3 (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 (vii) a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) a^3 + b^3 = (a + b)(a^2 - ab + b^2) (viii) a 3 − b 3 = ( a − b ) ( a 2 + a b + b 2 ) a^3 - b^3 = (a - b)(a^2 + ab + b^2) (ix) a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) Let's spend some time on the last (ix) identity: a + b + c = 0   ⇒   a² + b² + c² = 3abc a³ + b³ + c³ = 3ab...